Finding the value of integral.
$begingroup$
If $ int_{-infty}^{infty} f(x) dx= 1,$
Find value of $ int_{-infty}^{infty} f(x-frac{1}{x}) dx$.
I tried substituting $x$ as $frac{1}{t}$, but nothing is happening.
In the denominator, $1+x^2$ is left.
definite-integrals
edited Jan 5 at 17:38
Gnumbertester
558112
asked Jan 5 at 17:22
Math_centricMath_centric
161
$endgroup$
add a comment |
$begingroup$
If $ int_{-infty}^{infty} f(x) dx= 1,$
Find value of $ int_{-infty}^{infty} f(x-frac{1}{x}) dx$.
I tried substituting $x$ as $frac{1}{t}$, but nothing is happening.
In the denominator, $1+x^2$ is left.
definite-integrals
edited Jan 5 at 17:38
Gnumbertester
558112
asked Jan 5 at 17:22
Math_centricMath_centric
161
$endgroup$
$begingroup$
Hint: Try substituting $x=e^u$.
$endgroup$
– John Doe
Jan 5 at 17:49
$begingroup$
@JohnDoe Your substitution only works for $xin(0,infty)$.
$endgroup$
– Matemáticos Chibchas
Jan 13 at 6:01
$begingroup$
@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part.
$endgroup$
– John Doe
Jan 14 at 2:20
add a comment |
$begingroup$
If $ int_{-infty}^{infty} f(x) dx= 1,$
Find value of $ int_{-infty}^{infty} f(x-frac{1}{x}) dx$.
I tried substituting $x$ as $frac{1}{t}$, but nothing is happening.
In the denominator, $1+x^2$ is left.
definite-integrals
edited Jan 5 at 17:38
Gnumbertester
558112
asked Jan 5 at 17:22
Math_centricMath_centric
161
$endgroup$
If $ int_{-infty}^{infty} f(x) dx= 1,$
Find value of $ int_{-infty}^{infty} f(x-frac{1}{x}) dx$.
I tried substituting $x$ as $frac{1}{t}$, but nothing is happening.
In the denominator, $1+x^2$ is left.
definite-integrals
definite-integrals
edited Jan 5 at 17:38
Gnumbertester
558112
asked Jan 5 at 17:22
Math_centricMath_centric
161
edited Jan 5 at 17:38
Gnumbertester
558112
asked Jan 5 at 17:22
Math_centricMath_centric
161
edited Jan 5 at 17:38
Gnumbertester
558112
edited Jan 5 at 17:38
Gnumbertester
558112
edited Jan 5 at 17:38
Gnumbertester
558112
558112
asked Jan 5 at 17:22
Math_centricMath_centric
161
asked Jan 5 at 17:22
Math_centricMath_centric
161
asked Jan 5 at 17:22
Math_centricMath_centric
161
161
$begingroup$
Hint: Try substituting $x=e^u$.
$endgroup$
– John Doe
Jan 5 at 17:49
$begingroup$
@JohnDoe Your substitution only works for $xin(0,infty)$.
$endgroup$
– Matemáticos Chibchas
Jan 13 at 6:01
$begingroup$
@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part.
$endgroup$
– John Doe
Jan 14 at 2:20
add a comment |
$begingroup$
Hint: Try substituting $x=e^u$.
$endgroup$
– John Doe
Jan 5 at 17:49
$begingroup$
@JohnDoe Your substitution only works for $xin(0,infty)$.
$endgroup$
– Matemáticos Chibchas
Jan 13 at 6:01
$begingroup$
@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part.
$endgroup$
– John Doe
Jan 14 at 2:20
$begingroup$
Hint: Try substituting $x=e^u$.
$endgroup$
– John Doe
Jan 5 at 17:49
$begingroup$
Hint: Try substituting $x=e^u$.
$endgroup$
– John Doe
Jan 5 at 17:49
$begingroup$
@JohnDoe Your substitution only works for $xin(0,infty)$.
$endgroup$
– Matemáticos Chibchas
Jan 13 at 6:01
$begingroup$
@JohnDoe Your substitution only works for $xin(0,infty)$.
$endgroup$
– Matemáticos Chibchas
Jan 13 at 6:01
$begingroup$
@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part.
$endgroup$
– John Doe
Jan 14 at 2:20
$begingroup$
@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part.
$endgroup$
– John Doe
Jan 14 at 2:20
add a comment |
1 Answer
1
active
oldest
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$begingroup$
One may recall the following property
$$
int_{-infty}^{+infty} f(x) , dx = int_{-infty}^{+infty} fleft( x - frac{1}{x} right) , dx
$$ proved here.
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One may recall the following property
$$
int_{-infty}^{+infty} f(x) , dx = int_{-infty}^{+infty} fleft( x - frac{1}{x} right) , dx
$$ proved here.
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
One may recall the following property
$$
int_{-infty}^{+infty} f(x) , dx = int_{-infty}^{+infty} fleft( x - frac{1}{x} right) , dx
$$ proved here.
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
add a comment |
$begingroup$
One may recall the following property
$$
int_{-infty}^{+infty} f(x) , dx = int_{-infty}^{+infty} fleft( x - frac{1}{x} right) , dx
$$ proved here.
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
$endgroup$
One may recall the following property
$$
int_{-infty}^{+infty} f(x) , dx = int_{-infty}^{+infty} fleft( x - frac{1}{x} right) , dx
$$ proved here.
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
answered Jan 5 at 17:30
Olivier OloaOlivier Oloa
108k17177294
108k17177294
add a comment |
add a comment |
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$begingroup$
Hint: Try substituting $x=e^u$.
$endgroup$
– John Doe
Jan 5 at 17:49
$begingroup$
@JohnDoe Your substitution only works for $xin(0,infty)$.
$endgroup$
– Matemáticos Chibchas
Jan 13 at 6:01
$begingroup$
@matematicoschibchas hmm, you're right. You can split it the integral range into positive and negative $x$, then use $x=-e^{-u}$ for the negative part.
$endgroup$
– John Doe
Jan 14 at 2:20