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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: $\frac{(x-t)^2+t^2}{t}=\frac{2t^2-2tx+x^2}{t}=\frac{2(t-\frac{1}{2}x)^2+\frac{1}{2}x^2}{t}$ – 2017-01-01

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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:

$$ \int_{-\infty}^{+\infty}f\left(u-\frac{a}{u}\right)\mathrm{d}u=\int_{-\infty}^{+\infty} f(u)\: \mathrm{d}u,\quad a>0. $$

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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. – 2017-01-01
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    One 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 $\mathbb{R}$. – 2017-01-01
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    Just did the calculation and it coincides with the given solution. Many thanks (: – 2017-01-01
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    @Chee Han You are very welcome. – 2017-01-01
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    (Boole-)Glasser saves the day (again)...$(+1)$ – 2017-01-01