# Problem of the Week #206 - May 10, 2016

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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By method of contour integration, find the values of the integrals

$$\int_{-\infty + i\alpha}^{\infty + i\alpha} e^{-x^2}\, dx$$

for all $\alpha \ge 0$.
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#### Euge

##### MHB Global Moderator
Staff member
No one answered this week's problem. You can view my solution below.

For $\alpha = 0$, the integral is $\int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}$. Let $\alpha > 0$ and $R > 0$. Consider the contour integral $\int_{\Gamma(R,\alpha)} e^{-z^2}\, dz$, where $\Gamma(R,\alpha)$ is the rectangle $\{z = x + iy: -R \le x \le R, 0 \le y \le \alpha\}$. Since $z\mapsto e^{-z^2}$ is an entire function, Cauchy's theorem gives $\int_{\Gamma(R,\alpha)} e^{-z^2}\, dz = 0$. Furthermore, the integrals of $e^{-z^2}$ along the vertical edges of $\Gamma(R,\alpha)$ are dominated by $Ce^{-R^2}$, where $C = \int_0^\alpha e^{t^2}\, dt$. Hence

$$\int_{-R +i\alpha}^{R + i\alpha} e^{-z^2}\, dz = \int_{-R}^R e^{-x^2}\, dx + O(e^{-R^2})\quad\text{as}\quad R\to \infty$$

Letting $R\to \infty$ yields

$$\int_{-\infty + i\alpha}^{\infty +i\alpha} e^{-z^2}\, dz = \int_{-\infty}^\infty e^{-x^2} = \sqrt{\pi}$$

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