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Does the following contour integral have a closed form solution:

$$\int^{+\infty}_{-\infty} \frac{\exp (-a\tau^2)}{\tau- b i} \, d\tau$$

$i$ is the imaginary number. $a$ and $b$ are real coefficients.

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  • $\begingroup$ what "contour" are you talking about? $\endgroup$ – tired Aug 27 '16 at 18:54
  • $\begingroup$ It's quite possibile there is. Try using a rectangular contour with height $2b$. (of course $a$ must be positive for the integral to converge) $\endgroup$ – b00n heT Aug 27 '16 at 18:56
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Usually integrals with a square in the exponential are difficult to tackle by contour methods. Instead i will reduce the problem to a differential equation which is way easier to solve then the original integral.

By symmetry (the real part is an odd function integrated over an even intercal) the integral in question reduces to

$$ I(a,b)=\int_{-\infty}^{\infty}dx\frac{e^{-a x^2}}{x-ib}=ib\int_{-\infty}^{\infty}dx\frac{e^{-a x^2}}{x^2+b^2} $$

clearly we need $a>0$ for the integral to exist. Now, a straightforward differentation yields

$$ \partial_aI(a,b)=ib\int_{-\infty}^{\infty}dx\frac{e^{-a x^2}(-x^2)}{x^2+b^2} $$

which means that

$$ \partial_aI(a,b)=-ib\int_{-\infty}^{\infty}dxe^{-ax^2}+I(a,b) $$ or

$$ \partial_aI(a,b)-b^2I(a,b) =-ib\sqrt{\frac{\pi}{a}} $$

using the method of integrating factors we find for the particular solution ($\text{erf}(z)$ denotes the error function)

$$ I_p(a,b)=-i\sqrt{\pi}be^{b^2a}\int da \frac{e^{-b^2a}}{\sqrt{a}}=-\pi ie^{b^2a}\text{erf}(\sqrt{a}b) $$

the homegenous solution is trivial $I_h(a,b)=Ce^{ab^2}$ where $C$ is a constant. It can be fixed by observing that $I(\infty,b)=0$ which yields $C=i\pi $ and we finally obtain

$$ I(a,b)=i\pi e^{b^2a}\text{erfc}(\sqrt{a}b) $$

where $\text{erfc}(z)$ denotes the complimentary error function.

This agrees with WA


Edit:

I missed the obvious Feynman

writing $I(a,b)=ib e^{ab^2}J(a,b)$ with $J(a,b)=\int_{\mathbb{R}} dx\frac{e^{-a (b^2+x^2)}}{b^2+x^2}$.

Now $\partial_aJ(a,b)=\sqrt{\pi}e^{-ab^2}/\sqrt{a}$.

Integrating back w.r.t a follows the same line as the calculation for $I_p(a,b)$.

Done

Edit 2:

A third way would be to exploit Parsevals theorem

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  • $\begingroup$ +1. Neat. I was trying to square the integral to follow the 'old $\,\mathrm{e}^{-x^{2}}$ trick' but it becomes quite messy. So, I left it. $\endgroup$ – Felix Marin Aug 27 '16 at 21:42
  • $\begingroup$ thanks @FelixMarin the squareing trick seems to be quite limited to go beyond standard gaussians $\endgroup$ – tired Aug 28 '16 at 2:49
  • $\begingroup$ According to OP $a$ and $b$ are real coefficients. But if $a<0$ then the integral doesn't converge. If $b<0$ then the imaginary part of $I(a,b)$ is negative. $\endgroup$ – JanG Aug 28 '16 at 8:43
  • $\begingroup$ @JanG yeah sure..i assume $a>0$. furthermore $\text{erfc}$ is an odd function so indeed if you change $b\rightarrow-b$ we obtain a quantity which has opposit sign $\endgroup$ – tired Aug 28 '16 at 10:05

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