I am interested in computing the following integral $$\int_0^\infty \frac{1}{\sqrt{t}}\exp\left(-\frac{(x-t)^2}{t} - t\right)\, dt. $$ A hint is good enough for me!
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Possibly useful: $\frac{(x-t)^2+t^2}{t}=\frac{2t^2-2tx+x^2}{t}=\frac{2(t-\frac{1}{2}x)^2+\frac{1}{2}x^2}{t}$ – πr8 Jan 01 '17 at 00:19
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Hint. Make the change of variable $$ u=\sqrt{t}, \quad du=\frac{dt}{2\sqrt{t}}, $$ giving $$ \int_0^\infty \frac{1}{\sqrt{t}}\exp\left(-\frac{(x-t)^2}{t} - t\right)\, dt=\int_{-\infty}^\infty \exp\left(-2u^2-\frac{x^2}{u^2}-2 x\right)\, du $$ then use G. Boole's result:
$$ \int_{-\infty}^{+\infty}f\left(u-\frac{a}{u}\right)\mathrm{d}u=\int_{-\infty}^{+\infty} f(u)\: \mathrm{d}u,\quad a>0. $$
Olivier Oloa
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What are you proposing for $f$? I tried $f(x)=-x^2$ but I still have an extra $-u^2$ term. – Chee Han Jan 01 '17 at 00:36
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1One may observe that the argument of $\exp\left(-2u^2-\frac{x^2}{u^2}-2 x\right) $ rewrites $-2\left(u-\frac{x}{\sqrt{2}u}\right)^2-2\left(\sqrt{2}-1\right)x$. The term $-2\left(\sqrt{2}-1\right)x$ is a constant going 'outside' the integral. Thus $f(u)=\exp (-2u^2)$ is fine and easy to integrate over $\mathbb{R}$. – Olivier Oloa Jan 01 '17 at 00:41
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Just did the calculation and it coincides with the given solution. Many thanks (: – Chee Han Jan 01 '17 at 01:01
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