Integrate $frac{1}{ sqrt{T^2 - tau^2}}expleft(-frac{a^2}{4 (T + tau )} - frac{b^2}{4 (T-tau )}right)$
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I want to compute the integral
$int_t^T int_{-infty}^infty frac{1}{ sqrt{tau - t} (T-tau)} expleft(-frac{(z-x)^2}{2(tau - t)} -frac{(z-v)^2 + (z-w)^2}{2(T-tau)} right) d z d tau$
First integrating with respect to $z$ I get
$ int_t^T frac{sqrt{2pi}}{ sqrt{T - tau}sqrt{-2 t + tau + T}}expleft(-frac{(v + w - 2 x)^2}{4 (-2 t + tau + T)} - frac{(v-w)^2}{4 (T-tau )}right) d tau$
So to go further I think I should be able to integrate
$int_0^T frac{1}{ sqrt{T^2 - tau^2}}expleft(-frac{a^2}{4 (T + tau )} - frac{b^2}{4 (T-tau )}right) d tau$
integration definite-integrals
asked Nov 22 at 21:35
mschauer
1014
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mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
down vote
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I want to compute the integral
$int_t^T int_{-infty}^infty frac{1}{ sqrt{tau - t} (T-tau)} expleft(-frac{(z-x)^2}{2(tau - t)} -frac{(z-v)^2 + (z-w)^2}{2(T-tau)} right) d z d tau$
First integrating with respect to $z$ I get
$ int_t^T frac{sqrt{2pi}}{ sqrt{T - tau}sqrt{-2 t + tau + T}}expleft(-frac{(v + w - 2 x)^2}{4 (-2 t + tau + T)} - frac{(v-w)^2}{4 (T-tau )}right) d tau$
So to go further I think I should be able to integrate
$int_0^T frac{1}{ sqrt{T^2 - tau^2}}expleft(-frac{a^2}{4 (T + tau )} - frac{b^2}{4 (T-tau )}right) d tau$
integration definite-integrals
asked Nov 22 at 21:35
mschauer
1014
New contributor
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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A integral over the whole line was meant. I edited the question and added a missing factor.
– mschauer
Nov 22 at 22:15
I see, makes more sense now
– Yuriy S
Nov 22 at 22:22
The obvious next substitution would be $$tau=Tu$$ then I'm not sure. If $a=b$, maybe it could have a closed form, but in this case? Can't say for sure
– Yuriy S
Nov 22 at 22:24
Should I settle for an expansion? It looks like the coefficients of the Taylor expansion of the integral at $tau = 0$ of the integral in can be expressed by polynomials of the form $prod_i (a-b + z_i)(a+b + z_i) e^{-a^2 - b^2}$ some equidistant numbers $z_i$.
– mschauer
Nov 23 at 13:09
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to compute the integral
$int_t^T int_{-infty}^infty frac{1}{ sqrt{tau - t} (T-tau)} expleft(-frac{(z-x)^2}{2(tau - t)} -frac{(z-v)^2 + (z-w)^2}{2(T-tau)} right) d z d tau$
First integrating with respect to $z$ I get
$ int_t^T frac{sqrt{2pi}}{ sqrt{T - tau}sqrt{-2 t + tau + T}}expleft(-frac{(v + w - 2 x)^2}{4 (-2 t + tau + T)} - frac{(v-w)^2}{4 (T-tau )}right) d tau$
So to go further I think I should be able to integrate
$int_0^T frac{1}{ sqrt{T^2 - tau^2}}expleft(-frac{a^2}{4 (T + tau )} - frac{b^2}{4 (T-tau )}right) d tau$
integration definite-integrals
asked Nov 22 at 21:35
mschauer
1014
New contributor
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to compute the integral
$int_t^T int_{-infty}^infty frac{1}{ sqrt{tau - t} (T-tau)} expleft(-frac{(z-x)^2}{2(tau - t)} -frac{(z-v)^2 + (z-w)^2}{2(T-tau)} right) d z d tau$
First integrating with respect to $z$ I get
$ int_t^T frac{sqrt{2pi}}{ sqrt{T - tau}sqrt{-2 t + tau + T}}expleft(-frac{(v + w - 2 x)^2}{4 (-2 t + tau + T)} - frac{(v-w)^2}{4 (T-tau )}right) d tau$
So to go further I think I should be able to integrate
$int_0^T frac{1}{ sqrt{T^2 - tau^2}}expleft(-frac{a^2}{4 (T + tau )} - frac{b^2}{4 (T-tau )}right) d tau$
integration definite-integrals
integration definite-integrals
asked Nov 22 at 21:35
mschauer
1014
New contributor
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 22 at 21:35
mschauer
1014
New contributor
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 22 at 22:13
asked Nov 22 at 21:35
mschauer
1014
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mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 22 at 21:35
mschauer
1014
asked Nov 22 at 21:35
mschauer
1014
1014
New contributor
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mschauer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A integral over the whole line was meant. I edited the question and added a missing factor.
– mschauer
Nov 22 at 22:15
I see, makes more sense now
– Yuriy S
Nov 22 at 22:22
The obvious next substitution would be $$tau=Tu$$ then I'm not sure. If $a=b$, maybe it could have a closed form, but in this case? Can't say for sure
– Yuriy S
Nov 22 at 22:24
Should I settle for an expansion? It looks like the coefficients of the Taylor expansion of the integral at $tau = 0$ of the integral in can be expressed by polynomials of the form $prod_i (a-b + z_i)(a+b + z_i) e^{-a^2 - b^2}$ some equidistant numbers $z_i$.
– mschauer
Nov 23 at 13:09
add a comment |
A integral over the whole line was meant. I edited the question and added a missing factor.
– mschauer
Nov 22 at 22:15
I see, makes more sense now
– Yuriy S
Nov 22 at 22:22
The obvious next substitution would be $$tau=Tu$$ then I'm not sure. If $a=b$, maybe it could have a closed form, but in this case? Can't say for sure
– Yuriy S
Nov 22 at 22:24
Should I settle for an expansion? It looks like the coefficients of the Taylor expansion of the integral at $tau = 0$ of the integral in can be expressed by polynomials of the form $prod_i (a-b + z_i)(a+b + z_i) e^{-a^2 - b^2}$ some equidistant numbers $z_i$.
– mschauer
Nov 23 at 13:09
A integral over the whole line was meant. I edited the question and added a missing factor.
– mschauer
Nov 22 at 22:15
A integral over the whole line was meant. I edited the question and added a missing factor.
– mschauer
Nov 22 at 22:15
I see, makes more sense now
– Yuriy S
Nov 22 at 22:22
I see, makes more sense now
– Yuriy S
Nov 22 at 22:22
The obvious next substitution would be $$tau=Tu$$ then I'm not sure. If $a=b$, maybe it could have a closed form, but in this case? Can't say for sure
– Yuriy S
Nov 22 at 22:24
The obvious next substitution would be $$tau=Tu$$ then I'm not sure. If $a=b$, maybe it could have a closed form, but in this case? Can't say for sure
– Yuriy S
Nov 22 at 22:24
Should I settle for an expansion? It looks like the coefficients of the Taylor expansion of the integral at $tau = 0$ of the integral in can be expressed by polynomials of the form $prod_i (a-b + z_i)(a+b + z_i) e^{-a^2 - b^2}$ some equidistant numbers $z_i$.
– mschauer
Nov 23 at 13:09
Should I settle for an expansion? It looks like the coefficients of the Taylor expansion of the integral at $tau = 0$ of the integral in can be expressed by polynomials of the form $prod_i (a-b + z_i)(a+b + z_i) e^{-a^2 - b^2}$ some equidistant numbers $z_i$.
– mschauer
Nov 23 at 13:09
add a comment |
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mschauer is a new contributor. Be nice, and check out our Code of Conduct.
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A integral over the whole line was meant. I edited the question and added a missing factor.
– mschauer
Nov 22 at 22:15
I see, makes more sense now
– Yuriy S
Nov 22 at 22:22
The obvious next substitution would be $$tau=Tu$$ then I'm not sure. If $a=b$, maybe it could have a closed form, but in this case? Can't say for sure
– Yuriy S
Nov 22 at 22:24
Should I settle for an expansion? It looks like the coefficients of the Taylor expansion of the integral at $tau = 0$ of the integral in can be expressed by polynomials of the form $prod_i (a-b + z_i)(a+b + z_i) e^{-a^2 - b^2}$ some equidistant numbers $z_i$.
– mschauer
Nov 23 at 13:09