Evaluating $\displaystyle \int f(z)dz$ with $a>0$

[polygamma]

By integrating $ \displaystyle f(z) = \frac{e^{-z^{2}}}{z}$ around the appropriate contour, or otherwise, show that for $a>0$,

$$ \int_{0}^{\infty} e^{-x^{2}} \ \frac{a \cos (2ax) + x \sin(2ax)}{x^{2}+a^{2}} \ dx = \frac{\pi}{2}e^{-a^{2}}.$$

 

[topsquark]

By integrating $ \displaystyle f(z) = \frac{e^{-z^{2}}}{z}$ around the appropriate contour, or otherwise, show that for $a>0$,

$$ \int_{0}^{\infty} e^{-x^{2}} \ \frac{a \cos (2ax) + x \sin(2ax)}{x^{2}+a^{2}} \ dx = \frac{\pi}{2}e^{-a^{2}}.$$

Spoiler

Let u = 2 a x. The rest is trivial and is left for the interested student.

-Dan

 

[MarkFL]

Spoiler

Let u = 2 a x. The rest is trivial and is left for the interested student.

-Dan

Dan, Dan, Dan...we ask that hints be left up to the OP to give, and that all others post complete solutions. (Mmm)

http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html

 

[polygamma]

Spoiler

By integrating $\frac{e^{-z^{2}}}{z}$ around a rectangle with vertices at $\pm R \pm ia$ and then letting $R \to \infty$, we get $$\int_{-\infty}^{\infty} \frac{e^{-(x-ia)^{2}}}{x-ia} \, dx - \int_{-\infty}^{\infty}\frac{e^{-(x+ia)^{2}}}{x+ia} \, dx = 2 \pi i \, \text{Res} \left[ \frac{e^{-z^{2}}}{z}, 0\right] = 2 \pi i $$

Combining the first two integrals,

$$ e^{a^{2}}\int_{-\infty}^{\infty} e^{-x^{2}} \, \frac{ e^{2iax}(x+ia) - e^{-2iax}(x-ia)}{x^{2}+a^{2}} \, dx = e^{a^{2}}\int_{-\infty}^{\infty} e^{-x^{2}} \frac{2ix \sin(2ax) +2ia \cos(2ax)}{x^{2}+a^{2}} \, dx = 2 \pi i $$

So

$$\int_{-\infty}^{\infty} e^{-x^{2}} \frac{x \sin(2ax) +a \cos(2ax)}{x^{2}+a^{2}} \, dx = \pi e^{-a^{2}} $$

The result then follows since the integrand is even.