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I'm trying to do this integral essentially:

$$\int^\infty_0 \frac{e^{-ax^2}\sin(kx)}{x}dx$$ which I realized to be $$\frac{1}{2}\operatorname{Re}\left[F\left(\frac{e^{-ax^2}}{x}\right)\right]$$ where $F$ denotes the ordinary Fourier Transform (where I am using the convention $F[f] = \int^\infty_{-\infty}e^{ikx}f(x) \, dx)$.

This answer is supposed to be something related to the error function $\operatorname{erf}(k)$ but I don't have any ideas about how to get it to look like one. Could I somehow apply the convolution theorem?

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Do you know how to use contour integration in the complex plane to evaluate real integrals? –  icurays1 Nov 13 '12 at 3:22

1 Answer 1

I assume $\text{Re}(a) > 0$, so your integral exists. If your integral is $F(k)$, then $F'(k) = \int_0^\infty e^{-ax^2} \cos(kx)\ dx$. Now this is the real part of $J = \int_0^\infty e^{-ax^2} e^{ikx}\ dx$. An antiderivative of $g(x) = e^{-ax^2} e^{ikx} = e^{-k^2/(4a)} e^{-a(x-ik/(2a))^2}$ is $$G(x) = -\sqrt{\dfrac{\pi}{4a}}e^{-k^2/(4a)} \text{erf}\left(-\sqrt{a} x + \frac{i k}{2 \sqrt{a}}\right)$$ So $$J = \lim_{x \to \infty} G(x) - G(0) = \sqrt{\frac{\pi}{4a}} e^{-k^2/(4a)} \left(1 + \text{erf}\left(\frac{ik}{2\sqrt{a}}\right)\right) $$ Now $\text{erf}(it)$ is imaginary for real $t$, so $$F'(k) =\sqrt{\frac{\pi}{4a}} e^{-k^2/(4a)}$$ and using $F(0) = 0$, we integrate this to get $$F(k) = \frac{\pi}{2} \text{erf}\left(\frac{k}{2 \sqrt{a}}\right)$$

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