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Is there any way to solve the following integral?

$$\int_0^\infty \exp\Bigl(-\frac{(y^2 - x^2)^2}{x^2}\Bigr) \mathrm{d}x$$

Assume $y>0$.

Nikd0
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  • Can you please post your attempt as well? – QFi Apr 09 '19 at 19:14
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    This integral is equal to $\frac{\sqrt{\pi}}2$ for all $y\in\mathbb{R}$ – Peter Foreman Apr 09 '19 at 19:33
  • This is the prettiest form I got from original problem using substitutions. Then I tried several methods I know. The integrand does not fit any common function with simple integral transform (Laplace, Fourier) that would allow to solve it using properties of the transform. – Nikd0 Apr 09 '19 at 19:35
  • I attempted to use complex analysis - the integrand has a removable singularity at zero only, $\lim_0 \exp(−\frac{(y^2−x^2)^2}{x^2})\mathrm{d}x=0$. Also $\lim_\infty \exp(−\frac{(y^2−x^2)^2}{x^2})\mathrm{d}x=0$. I suggested a few contours to integrate along including the interval $\mathbb{R}^+$. Resulted to integrals that were useless or that I was not able to solve. – Nikd0 Apr 09 '19 at 19:35
  • @PeterForeman That's what WolframAlpha told me. I'm wondering how to get it. – Nikd0 Apr 09 '19 at 19:36
  • See here: https://math.stackexchange.com/q/496088/515527 and note that: $$\frac{(y^2-x^2)^2}{x^2}=\left(\frac{y^2-x^2}{x}\right)^2=\left(\frac{y^2}{x}-x\right)^2$$ – Zacky Apr 11 '19 at 20:19
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    @Zacky Thanks, got it. – Nikd0 Apr 11 '19 at 21:23

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Following the aproach of How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$ I got

$$\begin{align}I=&\int_0^\infty \exp\bigl(-\frac{(y^2-x^2)^2}{x^2}\bigr)\mathrm{d}x\\ =& \begin{Vmatrix}t =\frac{x}{y}\end{Vmatrix}=y\int_0^\infty \exp\bigl(-(t-t^{-1})^2y^2\bigr)\mathrm{d}t = y\biggl(\int_0^1+\int_1^\infty\biggr) \exp\bigl(-(t-t^{-1})^2y^2\bigr)\mathrm{d}t\\ =&\begin{Vmatrix}s=\frac{1}{t}\text{ in first integral}\end{Vmatrix} =\int_1^\infty \frac{1}{s^2}\exp\bigl(-(s-s^{-1})^2y^2\bigr)\mathrm{d}s+\int_1^\infty \exp\bigl(-(t-t^{-1})^2y^2\bigr)\mathrm{d}x\\ =&\int_1^\infty (1+t^{-2})\exp\bigl(-(t-t^{-1})^2y^2\bigr)\mathrm{d}x =\begin{Vmatrix}z=t-\frac{1}{t}\end{Vmatrix}=y\int_0^\infty \exp(-z^2y^2)\mathrm{d}z\\ =&y\frac{1}{2}\sqrt{\frac{\pi}{y^2}}=\frac{\sqrt{\pi}}{2} \end{align}$$

Thanks again, @Zacky.

Nikd0
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