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Evaluate : $$\int_{0}^{\infty }{\frac{{{\text{e}}^{-{{x}^{2}}}}}{{{\pi }^{2}}+{{\left( \gamma +x \right)}^{2}}}\text{d}x}$$

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The sign of $\gamma$ is? – Ron Gordon Feb 1 '13 at 15:19
Is $\gamma$ Euler's constant? Why should we evaluate this? – GEdgar Feb 1 '13 at 15:22
Why should we, indeed? – Did Feb 1 '13 at 15:37
@rlgordonma : $\gamma $ is Euler's constant – Ryan Feb 1 '13 at 22:55
up vote 1 down vote accepted

Before I begin and draw the reader in, I will say that this exposition outlines a general approach to integrals like these. Unfortunately, I was unable to evaluate the integral analytically, but I was able to break it into pieces, some of which did have a closed form.

A good approach with integrals like these is to use Parseval's theorem (or Plancherel's Theorem, depending on your preference). That is, I see two functions in the integrand whose Fourier transform I recognize. For the record, I define the FT of a function $f$ as

$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x}$$

Parseval's theorem then states that, for functions $f$ and $g$, both of whom have FT's $\hat{f}$ and $\hat{g}$, respectively:

$$\int_{-\infty}^{\infty} dx \: f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{f}(k) \hat{g}^*(k) $$

In this case, $f(x) = e^{-x^2} \theta(x)$ and $g(x) = [\pi^2 + (\gamma + x)^2]^{-1}$, where $\theta(x)$ is the Heaviside step function. The FT's of these functions are

$$\hat{f}(k) = \frac{\sqrt{\pi}}{2} e^{-\frac{k^2}{4}} - i F \left ( \frac{k}{2} \right ) $$

where $F$ is Dawson's integral

$$F(z) = e^{-z^2} \int_0^z dt \: e^{t^2}$$


$$\hat{g}(k) = e^{-i \gamma k} e^{-\pi |k|}$$

In multiplying the FT's together, we get a real and imaginary part. The imaginary part is an odd fucntion of $k$ and its integral over the real line is zero. The integral we want is then

$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \left [ \frac{\sqrt{\pi}}{2} e^{-\frac{k^2}{4}} \cos{\gamma k} + F \left ( \frac{k}{2} \right ) \sin{\gamma k} \right ] e^{-\pi |k|} $$

The first part of the integral may be evaluated in terms of Dawson's integral:

$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \frac{\sqrt{\pi}}{2} e^{-\frac{k^2}{4}} e^{-\pi |k|} \cos{\gamma k} = \frac{1}{2} e^{\pi^2 - \gamma^2} \cos{2 \pi \gamma} - \frac{1}{\sqrt{\pi}} \Im{F(\gamma + i \pi)}$$

The second part may be manipulated into the following form:

$$\begin{align}\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: F \left ( \frac{k}{2} \right ) e^{-\pi |k|} \sin{\gamma k} &= \frac{1}{4 \pi} \int_0^1 du \: \int_{-\infty}^{\infty} dk \: k \, e^{-\frac{(1-u^2) k^2}{4}} e^{-\pi |k|} \sin{\gamma k}\\ &= -\frac{1}{4 \pi} \frac{\partial}{\partial \gamma} \int_0^1 du \: \int_{-\infty}^{\infty} dk \: e^{-\frac{(1-u^2) k^2}{4}} e^{-\pi |k|} \cos{\gamma k}\\ \end{align}$$

Although the inner integral looks very much the the integral from the first piece above, I have not been able to evaluate it yet. I will leave the evaluation here for now.

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