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To evaluate the integral $\int_0^{+\infty} x^{-\frac{3}{2}}e^{-x-\frac{k^2}{x}} dx$ (where $k$ is a positive real number) my idea is to use $\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi}$. In order to do that, I've tried to define $u=\sqrt{x}-\frac{k}{\sqrt{x}}$. So $du=\frac{x+k}{2\sqrt{x^3}}dx$ and $u^2=x+\frac{k^2}{x}-2k$. So after the substitution I got

$$\int_0^{+\infty} x^{-\frac{3}{2}}e^{-x-\frac{k^2}{x}} dx = \int^0_{+\infty} \frac{1}{\sqrt{x^3}} \frac{2\sqrt{x^3}}{x+k} e^{-u^2-2k} du = -2e^{-2k}\int_0^{+\infty} \frac{1}{x+k} e^{-u^2} du$$

but from there I don't know how to get rid of the extra $x$ term. The solution should be $\frac{e^{-2k}\sqrt{\pi}}{k}$ so it's really looks like I've got something wrong with that substitution but i can't figure out where. Can you please help me?

injo
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    Also do check this out - https://math.stackexchange.com/q/416122/1157207 – Amrut Ayan Aug 10 '24 at 06:41
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    It look really complicated for a exercise presented as low difficulty with "Hint: use a substitution" but everything is clear. Thanks for the answer. – injo Aug 10 '24 at 07:50
  • https://www.wolframalpha.com/input?i=integrate+x%5Eb%2Aexp%28-x-c%5E2%2Fx%29+from+x%3D0+to+infinity – JJacquelin Aug 12 '24 at 17:06

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