Does the following contour integral have a closed form solution?

Question

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.

Answer 1

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

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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