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

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The discussion focuses on evaluating the integral of the function f(z) = e^(-z^2)/z around a specific contour to demonstrate that for a > 0, the integral from 0 to infinity of e^(-x^2) multiplied by (a cos(2ax) + x sin(2ax))/(x^2 + a^2) equals (π/2)e^(-a^2). Participants emphasize the importance of providing complete solutions rather than hints, adhering to community guidelines. The mathematical approach involves contour integration techniques to derive the result. The conversation underscores the significance of clarity and thoroughness in problem-solving within the forum. Overall, the integral evaluation is framed within the context of complex analysis and its applications.
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By integrating $ \displaystyle f(z) = \frac{e^{-z^{2}}}{z}$ around the appropriate contour, or otherwise, show that for $a>0$,

$$ \int_{0}^{\infty} e^{-x^{2}} \ \frac{a \cos (2ax) + x \sin(2ax)}{x^{2}+a^{2}} \ dx = \frac{\pi}{2}e^{-a^{2}}.$$
 
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Random Variable said:
By integrating $ \displaystyle f(z) = \frac{e^{-z^{2}}}{z}$ around the appropriate contour, or otherwise, show that for $a>0$,

$$ \int_{0}^{\infty} e^{-x^{2}} \ \frac{a \cos (2ax) + x \sin(2ax)}{x^{2}+a^{2}} \ dx = \frac{\pi}{2}e^{-a^{2}}.$$
Let u = 2 a x. The rest is trivial and is left for the interested student.

-Dan
 
topsquark said:
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
 
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.
 
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