When to use the property $int_{0}^{2a}f(x)dx=int_{0}^{a}f(x)dx+int_{0}^{a}f(2a-x)dx$?
Consider the property: $displaystyleint_{0}^{2a}f(x)dx=int_{0}^{a}f(x)dx+int_{0}^{a}f(2a-x)dx$
In some definite integral problems this property is used while in others it is not used even though it is usable. In some problems, if we do not use this property we get infinity after substituting the limits whereas in some cases if we use this property we get infinity or the wrong answer.
For example let $I=displaystyleint_{0}^{pi}frac{x}{a^2cos^2x+b^2sin^2x}dx$
$=displaystyleint_{0}^{pi}frac{pi-x}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= int_{0}^{pi}frac{pi}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{1}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{sec^2x}{a^2+b^2tan^2x}dx$
Case 1: The property is not used
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi, z=0$
$displaystyle therefore 2I= piint_{0}^{0}frac{dz}{a^2+b^2z^2}$
$therefore I=0$
Case 2:The property is used
$displaystyle 2I= 2piint_{0}^{pi/2}frac{sec^2x}{a^2+b^2tan^2x}dx$
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi/2, z=infty$
$displaystyle therefore 2I= 2piint_{0}^{infty}frac{dz}{a^2+b^2z^2}$
This on evaluation yields $pi^2/2ab$.
This is very confusing. How do i know when to use this property and when to not?
calculus integration definite-integrals
asked Dec 27 '18 at 22:41
MrAP
1,14721331
add a comment |
Consider the property: $displaystyleint_{0}^{2a}f(x)dx=int_{0}^{a}f(x)dx+int_{0}^{a}f(2a-x)dx$
In some definite integral problems this property is used while in others it is not used even though it is usable. In some problems, if we do not use this property we get infinity after substituting the limits whereas in some cases if we use this property we get infinity or the wrong answer.
For example let $I=displaystyleint_{0}^{pi}frac{x}{a^2cos^2x+b^2sin^2x}dx$
$=displaystyleint_{0}^{pi}frac{pi-x}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= int_{0}^{pi}frac{pi}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{1}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{sec^2x}{a^2+b^2tan^2x}dx$
Case 1: The property is not used
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi, z=0$
$displaystyle therefore 2I= piint_{0}^{0}frac{dz}{a^2+b^2z^2}$
$therefore I=0$
Case 2:The property is used
$displaystyle 2I= 2piint_{0}^{pi/2}frac{sec^2x}{a^2+b^2tan^2x}dx$
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi/2, z=infty$
$displaystyle therefore 2I= 2piint_{0}^{infty}frac{dz}{a^2+b^2z^2}$
This on evaluation yields $pi^2/2ab$.
This is very confusing. How do i know when to use this property and when to not?
calculus integration definite-integrals
asked Dec 27 '18 at 22:41
MrAP
1,14721331
2
It's naturally splitting the integration bounds in half when useful, nothing more
– qwr
Dec 27 '18 at 22:51
@qwr, It may be the case that one is getting a different answer when he is using the property than when he is not using it. For example in the definite integral given in the question, if one uses the property then the value comes out to be $pi^2/2ab$ otherwise the value comes out to be 0 on correct evaluation.
– MrAP
Dec 27 '18 at 23:07
Well, getting a different result using a certain method of integration actually means that some mistake is done during the calculations, so, could you share with use the specific calculations you are reffering to? Since, if your question remains like that, I cannot think of any better/more appropriate answer than that provided by @qwr.
– Βασίλης Μάρκος
Dec 27 '18 at 23:30
@ΒασίληςΜάρκος, Check my question now.
– MrAP
Dec 27 '18 at 23:52
add a comment |
Consider the property: $displaystyleint_{0}^{2a}f(x)dx=int_{0}^{a}f(x)dx+int_{0}^{a}f(2a-x)dx$
In some definite integral problems this property is used while in others it is not used even though it is usable. In some problems, if we do not use this property we get infinity after substituting the limits whereas in some cases if we use this property we get infinity or the wrong answer.
For example let $I=displaystyleint_{0}^{pi}frac{x}{a^2cos^2x+b^2sin^2x}dx$
$=displaystyleint_{0}^{pi}frac{pi-x}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= int_{0}^{pi}frac{pi}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{1}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{sec^2x}{a^2+b^2tan^2x}dx$
Case 1: The property is not used
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi, z=0$
$displaystyle therefore 2I= piint_{0}^{0}frac{dz}{a^2+b^2z^2}$
$therefore I=0$
Case 2:The property is used
$displaystyle 2I= 2piint_{0}^{pi/2}frac{sec^2x}{a^2+b^2tan^2x}dx$
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi/2, z=infty$
$displaystyle therefore 2I= 2piint_{0}^{infty}frac{dz}{a^2+b^2z^2}$
This on evaluation yields $pi^2/2ab$.
This is very confusing. How do i know when to use this property and when to not?
calculus integration definite-integrals
asked Dec 27 '18 at 22:41
MrAP
1,14721331
Consider the property: $displaystyleint_{0}^{2a}f(x)dx=int_{0}^{a}f(x)dx+int_{0}^{a}f(2a-x)dx$
In some definite integral problems this property is used while in others it is not used even though it is usable. In some problems, if we do not use this property we get infinity after substituting the limits whereas in some cases if we use this property we get infinity or the wrong answer.
For example let $I=displaystyleint_{0}^{pi}frac{x}{a^2cos^2x+b^2sin^2x}dx$
$=displaystyleint_{0}^{pi}frac{pi-x}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= int_{0}^{pi}frac{pi}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{1}{a^2cos^2x+b^2sin^2x}dx$
$displaystyle 2I= piint_{0}^{pi}frac{sec^2x}{a^2+b^2tan^2x}dx$
Case 1: The property is not used
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi, z=0$
$displaystyle therefore 2I= piint_{0}^{0}frac{dz}{a^2+b^2z^2}$
$therefore I=0$
Case 2:The property is used
$displaystyle 2I= 2piint_{0}^{pi/2}frac{sec^2x}{a^2+b^2tan^2x}dx$
Let $z=tan x$
$therefore (sec^2x)dx=dz $
When $x=0, z=0$ and when $x=pi/2, z=infty$
$displaystyle therefore 2I= 2piint_{0}^{infty}frac{dz}{a^2+b^2z^2}$
This on evaluation yields $pi^2/2ab$.
This is very confusing. How do i know when to use this property and when to not?
calculus integration definite-integrals
calculus integration definite-integrals
asked Dec 27 '18 at 22:41
MrAP
1,14721331
asked Dec 27 '18 at 22:41
MrAP
1,14721331
edited Dec 28 '18 at 9:37
asked Dec 27 '18 at 22:41
MrAP
1,14721331
asked Dec 27 '18 at 22:41
MrAP
1,14721331
asked Dec 27 '18 at 22:41
MrAP
1,14721331
1,14721331
2
It's naturally splitting the integration bounds in half when useful, nothing more
– qwr
Dec 27 '18 at 22:51
@qwr, It may be the case that one is getting a different answer when he is using the property than when he is not using it. For example in the definite integral given in the question, if one uses the property then the value comes out to be $pi^2/2ab$ otherwise the value comes out to be 0 on correct evaluation.
– MrAP
Dec 27 '18 at 23:07
Well, getting a different result using a certain method of integration actually means that some mistake is done during the calculations, so, could you share with use the specific calculations you are reffering to? Since, if your question remains like that, I cannot think of any better/more appropriate answer than that provided by @qwr.
– Βασίλης Μάρκος
Dec 27 '18 at 23:30
@ΒασίληςΜάρκος, Check my question now.
– MrAP
Dec 27 '18 at 23:52
add a comment |
2
It's naturally splitting the integration bounds in half when useful, nothing more
– qwr
Dec 27 '18 at 22:51
@qwr, It may be the case that one is getting a different answer when he is using the property than when he is not using it. For example in the definite integral given in the question, if one uses the property then the value comes out to be $pi^2/2ab$ otherwise the value comes out to be 0 on correct evaluation.
– MrAP
Dec 27 '18 at 23:07
Well, getting a different result using a certain method of integration actually means that some mistake is done during the calculations, so, could you share with use the specific calculations you are reffering to? Since, if your question remains like that, I cannot think of any better/more appropriate answer than that provided by @qwr.
– Βασίλης Μάρκος
Dec 27 '18 at 23:30
@ΒασίληςΜάρκος, Check my question now.
– MrAP
Dec 27 '18 at 23:52
2
2
It's naturally splitting the integration bounds in half when useful, nothing more
– qwr
Dec 27 '18 at 22:51
It's naturally splitting the integration bounds in half when useful, nothing more
– qwr
Dec 27 '18 at 22:51
@qwr, It may be the case that one is getting a different answer when he is using the property than when he is not using it. For example in the definite integral given in the question, if one uses the property then the value comes out to be $pi^2/2ab$ otherwise the value comes out to be 0 on correct evaluation.
– MrAP
Dec 27 '18 at 23:07
@qwr, It may be the case that one is getting a different answer when he is using the property than when he is not using it. For example in the definite integral given in the question, if one uses the property then the value comes out to be $pi^2/2ab$ otherwise the value comes out to be 0 on correct evaluation.
– MrAP
Dec 27 '18 at 23:07
Well, getting a different result using a certain method of integration actually means that some mistake is done during the calculations, so, could you share with use the specific calculations you are reffering to? Since, if your question remains like that, I cannot think of any better/more appropriate answer than that provided by @qwr.
– Βασίλης Μάρκος
Dec 27 '18 at 23:30
Well, getting a different result using a certain method of integration actually means that some mistake is done during the calculations, so, could you share with use the specific calculations you are reffering to? Since, if your question remains like that, I cannot think of any better/more appropriate answer than that provided by @qwr.
– Βασίλης Μάρκος
Dec 27 '18 at 23:30
@ΒασίληςΜάρκος, Check my question now.
– MrAP
Dec 27 '18 at 23:52
@ΒασίληςΜάρκος, Check my question now.
– MrAP
Dec 27 '18 at 23:52
add a comment |
2 Answers
2
active
oldest
votes
The formula is useful when you have either $f(2a-x)=f(x)$ or $f(2a-x)=-f(x)$. Mostly trigonometric functions have this property if $a=pi/2$ and the formula is really useful in such contexts.
The issue in case 1 is not related to this formula but rather is due to the fact that an invalid substitution has been used. Always check the theorem for substitution in definite integrals:
Theorem: Let the function $g$ have a continuous derivative on $[c, d] $ and function $f$ be continuous on $g([c, d]) $. Then we have $$int_{g(c)} ^{g(d)} f(x) , dx=int_{c} ^{d} f(g(t)) g'(t) , dt$$
In your case $[c, d] =[0, pi] $ and $g(x) =tan x$ and you can see that $g$ alone is discontinuous on $[0,pi]$ so that the requirements of the above theorem are not met.
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
add a comment |
Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$. When is it worthwhile to split the integral like that? When there's some sort of even/odd symmetry to exploit. Back in the original form, that's symmetry around $a$.
Now, if you're getting results that are different? That means you're making mistakes in at least one of the calculations. Now that you've edited, I can see what it is - you're making a substitution that has a singularity in the middle, and blindly crossing it. That $int_0^0$? It's really $int_0^{infty} +int_{-infty}^0$, as $tan$ goes to $infty$ in the middle of the interval, then comes back from $-infty$. See also this AoPS post of mine for a detailed explanation of a similar integral.
answered Dec 27 '18 at 23:56
jmerry
2,151210
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
add a comment |
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2 Answers
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2 Answers
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The formula is useful when you have either $f(2a-x)=f(x)$ or $f(2a-x)=-f(x)$. Mostly trigonometric functions have this property if $a=pi/2$ and the formula is really useful in such contexts.
The issue in case 1 is not related to this formula but rather is due to the fact that an invalid substitution has been used. Always check the theorem for substitution in definite integrals:
Theorem: Let the function $g$ have a continuous derivative on $[c, d] $ and function $f$ be continuous on $g([c, d]) $. Then we have $$int_{g(c)} ^{g(d)} f(x) , dx=int_{c} ^{d} f(g(t)) g'(t) , dt$$
In your case $[c, d] =[0, pi] $ and $g(x) =tan x$ and you can see that $g$ alone is discontinuous on $[0,pi]$ so that the requirements of the above theorem are not met.
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
add a comment |
The formula is useful when you have either $f(2a-x)=f(x)$ or $f(2a-x)=-f(x)$. Mostly trigonometric functions have this property if $a=pi/2$ and the formula is really useful in such contexts.
The issue in case 1 is not related to this formula but rather is due to the fact that an invalid substitution has been used. Always check the theorem for substitution in definite integrals:
Theorem: Let the function $g$ have a continuous derivative on $[c, d] $ and function $f$ be continuous on $g([c, d]) $. Then we have $$int_{g(c)} ^{g(d)} f(x) , dx=int_{c} ^{d} f(g(t)) g'(t) , dt$$
In your case $[c, d] =[0, pi] $ and $g(x) =tan x$ and you can see that $g$ alone is discontinuous on $[0,pi]$ so that the requirements of the above theorem are not met.
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
add a comment |
The formula is useful when you have either $f(2a-x)=f(x)$ or $f(2a-x)=-f(x)$. Mostly trigonometric functions have this property if $a=pi/2$ and the formula is really useful in such contexts.
The issue in case 1 is not related to this formula but rather is due to the fact that an invalid substitution has been used. Always check the theorem for substitution in definite integrals:
Theorem: Let the function $g$ have a continuous derivative on $[c, d] $ and function $f$ be continuous on $g([c, d]) $. Then we have $$int_{g(c)} ^{g(d)} f(x) , dx=int_{c} ^{d} f(g(t)) g'(t) , dt$$
In your case $[c, d] =[0, pi] $ and $g(x) =tan x$ and you can see that $g$ alone is discontinuous on $[0,pi]$ so that the requirements of the above theorem are not met.
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
The formula is useful when you have either $f(2a-x)=f(x)$ or $f(2a-x)=-f(x)$. Mostly trigonometric functions have this property if $a=pi/2$ and the formula is really useful in such contexts.
The issue in case 1 is not related to this formula but rather is due to the fact that an invalid substitution has been used. Always check the theorem for substitution in definite integrals:
Theorem: Let the function $g$ have a continuous derivative on $[c, d] $ and function $f$ be continuous on $g([c, d]) $. Then we have $$int_{g(c)} ^{g(d)} f(x) , dx=int_{c} ^{d} f(g(t)) g'(t) , dt$$
In your case $[c, d] =[0, pi] $ and $g(x) =tan x$ and you can see that $g$ alone is discontinuous on $[0,pi]$ so that the requirements of the above theorem are not met.
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
answered Dec 28 '18 at 6:33
Paramanand Singh
49k555160
49k555160
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
add a comment |
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
Is there any reason why you have written the equality (in the theorem) in that order?
– MrAP
Jan 1 at 7:25
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
@MrAP: the theorem is about the substitution $x=g(t) $ and $g$ need not necessarily be invertible. So don't try to think of it like $t=g^{-1}(x)$.
– Paramanand Singh
Jan 1 at 7:59
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
Can you help me with this question: math.stackexchange.com/questions/2756453/…?
– MrAP
Jan 1 at 9:06
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
@MrAP: in case of indefinite integrals you need the substitution $x=g(t) $ to be invertible because the final answer has to be in terms of the original variable $x$. Apart from that I don't see much difference.
– Paramanand Singh
Jan 1 at 10:20
add a comment |
Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$. When is it worthwhile to split the integral like that? When there's some sort of even/odd symmetry to exploit. Back in the original form, that's symmetry around $a$.
Now, if you're getting results that are different? That means you're making mistakes in at least one of the calculations. Now that you've edited, I can see what it is - you're making a substitution that has a singularity in the middle, and blindly crossing it. That $int_0^0$? It's really $int_0^{infty} +int_{-infty}^0$, as $tan$ goes to $infty$ in the middle of the interval, then comes back from $-infty$. See also this AoPS post of mine for a detailed explanation of a similar integral.
answered Dec 27 '18 at 23:56
jmerry
2,151210
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
add a comment |
Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$. When is it worthwhile to split the integral like that? When there's some sort of even/odd symmetry to exploit. Back in the original form, that's symmetry around $a$.
Now, if you're getting results that are different? That means you're making mistakes in at least one of the calculations. Now that you've edited, I can see what it is - you're making a substitution that has a singularity in the middle, and blindly crossing it. That $int_0^0$? It's really $int_0^{infty} +int_{-infty}^0$, as $tan$ goes to $infty$ in the middle of the interval, then comes back from $-infty$. See also this AoPS post of mine for a detailed explanation of a similar integral.
answered Dec 27 '18 at 23:56
jmerry
2,151210
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
add a comment |
Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$. When is it worthwhile to split the integral like that? When there's some sort of even/odd symmetry to exploit. Back in the original form, that's symmetry around $a$.
Now, if you're getting results that are different? That means you're making mistakes in at least one of the calculations. Now that you've edited, I can see what it is - you're making a substitution that has a singularity in the middle, and blindly crossing it. That $int_0^0$? It's really $int_0^{infty} +int_{-infty}^0$, as $tan$ goes to $infty$ in the middle of the interval, then comes back from $-infty$. See also this AoPS post of mine for a detailed explanation of a similar integral.
answered Dec 27 '18 at 23:56
jmerry
2,151210
Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$. When is it worthwhile to split the integral like that? When there's some sort of even/odd symmetry to exploit. Back in the original form, that's symmetry around $a$.
Now, if you're getting results that are different? That means you're making mistakes in at least one of the calculations. Now that you've edited, I can see what it is - you're making a substitution that has a singularity in the middle, and blindly crossing it. That $int_0^0$? It's really $int_0^{infty} +int_{-infty}^0$, as $tan$ goes to $infty$ in the middle of the interval, then comes back from $-infty$. See also this AoPS post of mine for a detailed explanation of a similar integral.
answered Dec 27 '18 at 23:56
jmerry
2,151210
answered Dec 27 '18 at 23:56
jmerry
2,151210
answered Dec 27 '18 at 23:56
jmerry
2,151210
answered Dec 27 '18 at 23:56
jmerry
2,151210
2,151210
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
add a comment |
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
So if there is a singularity in the middle then we have to use the property?. Could you please elaborate on the even/odd symmetry part and what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?Also could you please point out to a page where i can learn more about this.
– MrAP
Dec 28 '18 at 0:06
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
No, you don't have to use the property if there's a singularity in the middle. You have to pay attention and split the interval anywhere your substitutions break down. If that happens to be in the exact center of the interval (as in your example, but not my link), then the property here (splitting the interval at its midpoint) does that.
– jmerry
Dec 28 '18 at 0:12
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
And what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$"?
– MrAP
Dec 28 '18 at 0:19
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@MrAP, When substituting you need,in the first place, a substitution function which is defined over the entire interval you are integrating. As jmerry pointed out, you are using the substitution $z=tan x$ on the interval $[0,pi]$ while $tan x$ has a singularity at $pi/2$, which is a point in the interior of the interval over which you are integrating, which is the actual mistake.
– Βασίλης Μάρκος
Dec 28 '18 at 13:28
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
@jmerry, I am waiting for your reply on what do you mean by "Shift it over by $a$, so that it's $int_{-a}^a$ versus $int_0^a$ and $int_{-a}^0$".
– MrAP
Dec 28 '18 at 16:18
add a comment |
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2
It's naturally splitting the integration bounds in half when useful, nothing more
– qwr
Dec 27 '18 at 22:51
@qwr, It may be the case that one is getting a different answer when he is using the property than when he is not using it. For example in the definite integral given in the question, if one uses the property then the value comes out to be $pi^2/2ab$ otherwise the value comes out to be 0 on correct evaluation.
– MrAP
Dec 27 '18 at 23:07
Well, getting a different result using a certain method of integration actually means that some mistake is done during the calculations, so, could you share with use the specific calculations you are reffering to? Since, if your question remains like that, I cannot think of any better/more appropriate answer than that provided by @qwr.
– Βασίλης Μάρκος
Dec 27 '18 at 23:30
@ΒασίληςΜάρκος, Check my question now.
– MrAP
Dec 27 '18 at 23:52