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$f_A(x,y)=\int_0^\infty du \frac{u \left(e^{-\frac{(u-x)^2}{2 A}}-e^{-\frac{(u+x)^2}{2 A}} \right)}{\sqrt{2 \pi } \sqrt{A} x \left(y^2+u^2\right)} $ is there a closed form?

I was able to find it for $f_A^n(x,y)=\int_0^\infty du \frac{u \left(e^{-\frac{(u-x)^2}{2 A}}-e^{-\frac{(u+x)^2}{2 A}} \right)}{\sqrt{2 \pi } \sqrt{A} x}\left(y^2+u^2\right)^n $ for positive integer $n$, but for $n=-1$ mathematica doesn't simplify the integral.

For the special case $n=-1, y=0$ mathematica tells me

$-\frac{i \sqrt{\frac{\pi }{2}} e^{-\frac{x^2}{2 A}}}{\sqrt{A} x}$ that looks wrong (why is complex?), as it doesn't converge for $x\rightarrow0$.

EDIT: Calculating $\int_0^\infty du (e^{-u(u-x)}-e^{-u(u+x)}) u^{1+2n}$ and taking the limit for $n\rightarrow -1$ I get $\pi Erfi[x]$, so a real result that is also matching the behaviour that I am getting numerically.

If there is no closed form for the first integral, it is possible to find any approximated series expression for $y > x, \sqrt{A}$? For $y \gg x, \sqrt{A}$ then replacing $\frac{1}{y^2+u^2}\rightarrow \frac{1}{y^2}$ is a good approximation, but in the region $y \gtrsim x, \sqrt{A}$ is not. Expanding the numerator $\frac{1}{y^2+u^2}$ in the integral of course doesn't work as $u$ is integrated up lo infinite values.

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Hint:

With a linear transform of $u$, you can decompose as two indefinite integrals of the form

$$\int\frac{u}{u^2+c^2}e^{-u^2}du=\frac12\int e^{-u^2}d\ln(u^2+c^2)=\frac{e^{c^2}}{2}\int e^{-e^{t^2}}dt.$$

Because of the double exponential, you shouldn't expect a closed form, but thanks to the very fast decreasing behavior, numerical integration over a short range should do.

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Concerning the special case where $y=0$ the problem is $$I=\int_0^\infty \frac{e^{-\frac{(u-x)^2}{2 A}}-e^{-\frac{(u+x)^2}{2 A}}}{\sqrt{2 \pi } \sqrt{A}x \,u }\,du$$ Integration leads to $$I=\frac{e^{-\frac{x^2}{2 A}} \left(\log \left(\frac{x}{A}\right)-\log \left(-\frac{x}{A}\right)\right)}{\sqrt{2 \pi } \sqrt{A} x}$$ provided $\Re(A)\geq 0$.

Now $$\log \left(\frac{x}{A}\right)-\log \left(-\frac{x}{A}\right)=\log(-1)=i\pi$$ Replacing, then $$I=\frac{i \sqrt{\frac{\pi }{2}} e^{-\frac{x^2}{2 A}}}{\sqrt{A} x}$$

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  • $\begingroup$ yes, what I don't understand is, thinking the integral as an approximated sum, how can the result be complex if the integrand is always real. The integral $\int_0^\infty du \frac{e^{-u(u-x)}-e^{-u(u+x)}}{u}=i \pi$ if I try to compute it numerically i get an x-dependent result, and is real. $\endgroup$ – Giorgio Busoni Jul 22 '16 at 6:56
  • $\begingroup$ Calculating $\int_0^\infty du (e^{-u(u-x)}-e^{-u(u+x)}) u^{1+2n}$ and taking the limit for $n\rightarrow -1$ I get $\pi Erfi[x]$, so a real result that is also matching the behaviour that I am getting numerically. $\endgroup$ – Giorgio Busoni Jul 22 '16 at 7:22

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