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I'm reading Keith Conrad's article "Differentiating Under the Integral Sign", exercise 8 is:

If you are familiar with integration of complex-valued functions, show for $y \in R$ that $$\int_{-\infty}^{\infty}e^{-(x+iy)^2}dx = \sqrt{\pi}$$.* In other words, show the integral on the left side is independent of $y$. (Hint: Use diffrentiation under the integral sign to compute the $y$-derivative of the left side.)

Note: the original article is asking to prove $\int_{-\infty}^{\infty}e^{-(x+iy)^2}dx = \sqrt{2\pi}$, but I guess here the "2" is a typo.

I can prove the conclusion: $$I = \int_{-\infty}^{\infty}e^{-(x+iy)^2}dx = \int_{-\infty}^{\infty}e^{-x^2+y^2-2xyi}dx= \int_{-\infty}^{\infty}e^{-x^2+y^2}[\cos{(2xy)}-i\sin{(2xy)}]dx$$

, as $e^{-x^2+y^2}$ and $\cos{(2xy)}$ are even and $\sin{(2xy)}$ odd function of $x$,

$$I=2\int_{0}^{\infty}e^{-x^2+y^2}\cos{(2xy)}dx$$

, substitue with $x=u/\sqrt{2}$,

$$I=\sqrt{2}e^{y^2}\int_{0}^{\infty}e^{-u^2/2}\cos{(\sqrt{2}yu)}du$$

, since $\int_0^\infty\cos{(tx)}e^{-x^2/2}dx = \sqrt{\frac{\pi}{2}}e^{-t^2/2}$,

$$I=\sqrt{2}e^{y^2} \sqrt{\frac{\pi}{2}}e^{-(\sqrt{2}y)^2/2}=\sqrt{\pi}$$

But, how to prove the conclusion following the hint, using the "differentiating under the integral sign" trick?

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  • $\begingroup$ You simply verify that $\partial_y \int_{-\infty}^{+\infty} e^{-(x+iy)^2} dx = 0$ by applying Leibniz's integral rule. $\endgroup$ – Hyperplane Jan 15 '17 at 11:02
  • $\begingroup$ Note that $$\int_0^\infty\cos{(tx)}e^{-x^2/2}dx = \sqrt{\frac{\pi}{2}}e^{-t^2/2}$$ is precisely proved by differentiation under the integral sign. $\endgroup$ – C. Dubussy Jan 15 '17 at 11:04
  • $\begingroup$ @Hyperplane is it that similar to my prove, $e^{y^2}\cos{2xy}$'s derivative of $y$ becomes $2ye^{y^2}\cos{2xy} - 2x e^{y^2}\sin{2xy}$, then $e^{y^2}$ is even function of $y$, while $y$ and $\sin{2xy}$ is odd, so integration becomes $0$? $\endgroup$ – athos Jan 15 '17 at 11:08
  • $\begingroup$ Use Cauchy's theorem by integrating $\exp{\left (-z^2 \right )}$ over a rectangle with vertices $\pm R$ and $\pm R+i y$, then take the limit as $R \to \infty$. $\endgroup$ – Ron Gordon Jan 15 '17 at 19:37
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\begin{align} \partial_y \int_{-\infty}^{+\infty} e^{-(x+iy)^2} dx &= \int_{-\infty}^{+\infty} \partial_y e^{-(x+iy)^2} dx = \int_{-\infty}^{+\infty} -2iy(x+iy)e^{-(x+iy)^2} dx \end{align}

Here we can substitute $z = x+iy$ and the integral becomes (with $\gamma = \{t+iy\mid t\in\mathbb R \}$)

\begin{align} iy \int_\gamma -2ze^{-z^2} dz = iy \int_\gamma \frac{d}{dz}e^{-z^2} dz = 0 \end{align}

Hence $\partial_y \int_{-\infty}^{+\infty} e^{-(x+iy)^2} dx = 0$ which means that the integral value is invariant with respect to $y$.

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  • $\begingroup$ Yes. Even after 5 years of studying math I still manage to f up substitution sometimes. Embarrassing. $\endgroup$ – Hyperplane Jan 15 '17 at 12:21

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