Calculating the integral $int_{-infty}^inftyprodlimits_{i=1}^ne^{-(y-v_i)}e^{-e^{-(y-v_i)}}dy$
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I'm a little rusty with my calculus so I was hoping if anyone would help me with the following integral:
$$int_{-infty}^{infty} prod _{i=1}^ne^{-(y-v_i)}e^{-e^{-(y-v_i)}}dy$$
calculus integration
edited Jan 11 at 22:07
Did
248k23225463
asked Mar 23 '12 at 15:28
user25720user25720
1112
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add a comment |
$begingroup$
I'm a little rusty with my calculus so I was hoping if anyone would help me with the following integral:
$$int_{-infty}^{infty} prod _{i=1}^ne^{-(y-v_i)}e^{-e^{-(y-v_i)}}dy$$
calculus integration
edited Jan 11 at 22:07
Did
248k23225463
asked Mar 23 '12 at 15:28
user25720user25720
1112
$endgroup$
add a comment |
$begingroup$
I'm a little rusty with my calculus so I was hoping if anyone would help me with the following integral:
$$int_{-infty}^{infty} prod _{i=1}^ne^{-(y-v_i)}e^{-e^{-(y-v_i)}}dy$$
calculus integration
edited Jan 11 at 22:07
Did
248k23225463
asked Mar 23 '12 at 15:28
user25720user25720
1112
$endgroup$
I'm a little rusty with my calculus so I was hoping if anyone would help me with the following integral:
$$int_{-infty}^{infty} prod _{i=1}^ne^{-(y-v_i)}e^{-e^{-(y-v_i)}}dy$$
calculus integration
calculus integration
edited Jan 11 at 22:07
Did
248k23225463
asked Mar 23 '12 at 15:28
user25720user25720
1112
edited Jan 11 at 22:07
Did
248k23225463
asked Mar 23 '12 at 15:28
user25720user25720
1112
edited Jan 11 at 22:07
Did
248k23225463
edited Jan 11 at 22:07
Did
248k23225463
edited Jan 11 at 22:07
Did
248k23225463
248k23225463
asked Mar 23 '12 at 15:28
user25720user25720
1112
asked Mar 23 '12 at 15:28
user25720user25720
1112
asked Mar 23 '12 at 15:28
user25720user25720
1112
1112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using $x=mathrm e^{-y}$, the integral you are interested in is
$$
I=int_0^{+infty}prod_{i=1}^nleft(xmathrm e^{v_i}cdotmathrm e^{-mathrm e^{v_i}x}right)cdotfrac{mathrm dx}x=int_0^{+infty}ax^{n-1}mathrm e^{-bx}mathrm dx,
$$
for some suitable positive $a$ and $b$, depending on $(v_i)_{1leqslant ileqslant n}$, that I will let you discover. Thus,
$$
I=int_0^{+infty}ab^{-n}x^{n-1}mathrm e^{-x}mathrm dx=(n-1)!cdot ab^{-n}.
$$
answered Mar 23 '12 at 15:38
DidDid
248k23225463
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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
$begingroup$
Using $x=mathrm e^{-y}$, the integral you are interested in is
$$
I=int_0^{+infty}prod_{i=1}^nleft(xmathrm e^{v_i}cdotmathrm e^{-mathrm e^{v_i}x}right)cdotfrac{mathrm dx}x=int_0^{+infty}ax^{n-1}mathrm e^{-bx}mathrm dx,
$$
for some suitable positive $a$ and $b$, depending on $(v_i)_{1leqslant ileqslant n}$, that I will let you discover. Thus,
$$
I=int_0^{+infty}ab^{-n}x^{n-1}mathrm e^{-x}mathrm dx=(n-1)!cdot ab^{-n}.
$$
answered Mar 23 '12 at 15:38
DidDid
248k23225463
$endgroup$
add a comment |
$begingroup$
Using $x=mathrm e^{-y}$, the integral you are interested in is
$$
I=int_0^{+infty}prod_{i=1}^nleft(xmathrm e^{v_i}cdotmathrm e^{-mathrm e^{v_i}x}right)cdotfrac{mathrm dx}x=int_0^{+infty}ax^{n-1}mathrm e^{-bx}mathrm dx,
$$
for some suitable positive $a$ and $b$, depending on $(v_i)_{1leqslant ileqslant n}$, that I will let you discover. Thus,
$$
I=int_0^{+infty}ab^{-n}x^{n-1}mathrm e^{-x}mathrm dx=(n-1)!cdot ab^{-n}.
$$
answered Mar 23 '12 at 15:38
DidDid
248k23225463
$endgroup$
add a comment |
$begingroup$
Using $x=mathrm e^{-y}$, the integral you are interested in is
$$
I=int_0^{+infty}prod_{i=1}^nleft(xmathrm e^{v_i}cdotmathrm e^{-mathrm e^{v_i}x}right)cdotfrac{mathrm dx}x=int_0^{+infty}ax^{n-1}mathrm e^{-bx}mathrm dx,
$$
for some suitable positive $a$ and $b$, depending on $(v_i)_{1leqslant ileqslant n}$, that I will let you discover. Thus,
$$
I=int_0^{+infty}ab^{-n}x^{n-1}mathrm e^{-x}mathrm dx=(n-1)!cdot ab^{-n}.
$$
answered Mar 23 '12 at 15:38
DidDid
248k23225463
$endgroup$
Using $x=mathrm e^{-y}$, the integral you are interested in is
$$
I=int_0^{+infty}prod_{i=1}^nleft(xmathrm e^{v_i}cdotmathrm e^{-mathrm e^{v_i}x}right)cdotfrac{mathrm dx}x=int_0^{+infty}ax^{n-1}mathrm e^{-bx}mathrm dx,
$$
for some suitable positive $a$ and $b$, depending on $(v_i)_{1leqslant ileqslant n}$, that I will let you discover. Thus,
$$
I=int_0^{+infty}ab^{-n}x^{n-1}mathrm e^{-x}mathrm dx=(n-1)!cdot ab^{-n}.
$$
answered Mar 23 '12 at 15:38
DidDid
248k23225463
answered Mar 23 '12 at 15:38
DidDid
248k23225463
answered Mar 23 '12 at 15:38
DidDid
248k23225463
answered Mar 23 '12 at 15:38
DidDid
248k23225463
248k23225463
add a comment |
add a comment |
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