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I would like to compute an integral of the form ($a,b \neq 0$)

$$\int_{-\infty}^{\infty} e^{-(ax+ib)^2} dx = \frac{1}{a} \int_{-\infty+ib}^{\infty+ib} e^{-z^2} dz$$

where we made the substitution $z = ax+ib$. I know that the last integral is a gaussian integral and that it should equal $\sqrt{\pi}$, but I'm not sure how to compute it by hand. I tried to solve it in the complex plane via contour integration but I seem to go in a circle and do not arrive at $\sqrt{\pi}$.

Could somebody help out?

root
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  • Please take a look at this beautiful and comprehensive answer! – Eman Yalpsid Apr 19 '16 at 17:41
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    @Andrew : it is not beautiful and comprehensive. he starts with $\frac{\pi}{\sin \pi s} = \Gamma(s) \Gamma(1-s)$ one of the hardest complex analysis theorem to prove. the solution 2 is OK, the solution 3 is as hard to prove (the duplication formula for $\Gamma(s)$, the solution 4 is even worst (lol the functional equation for $\zeta(s)$...), the solution 5 and 6 are the same as solution 2, the solution 7 is ok. and he didn't write the original proof of Laplace : with the help of the Laplace/Fourier transform. – reuns Apr 20 '16 at 02:48
  • Thank you for the correction. – Eman Yalpsid Apr 20 '16 at 10:22

3 Answers3

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Consider the integral of $e^{-z^2}$ around a rectangular contour with corners at $-R, R, R+ib, -R+ib$. As $R \to +\infty$ the integrals over the vertical pieces go to $0$, while the integral over $(-R, R)$ goes to $\sqrt{\pi}$.

Robert Israel
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when $c,a \ne 0$ are real : $$F(c) = \int_{-\infty}^{+\infty} e^{-(ax+c)^2} dx = \frac{1}{a} \int_{-\infty}^{+\infty} e^{-y^2} dy = \frac{\sqrt{\pi}}{a}$$

the LHS is clearly analytic in $c$, as the RHS, hence by analytic continuation :

$$F(c) = \int_{-\infty}^{+\infty} e^{-(ax+c)^2} dx = \frac{\sqrt{\pi}}{a}$$

for every $c \in \mathbb{C}$

reuns
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We have $$ \int_{-\infty}^{\infty} e^{-(ax+ib)^2} dx = \int_{-\infty}^{\infty} e^{-a^2 x^2 - 2iabx +b^2} dx $$ Using the fact that $$ \int_{-\infty}^{\infty} e^{-\alpha y^2 +\beta y } dy = \sqrt{\frac{\pi}{\alpha}} e^{\frac{\beta^2}{4\alpha}} ; \quad \alpha>0, \beta\in \mathbb C$$ So, $$ \int_{-\infty}^{\infty} e^{-(ax+ib)^2} dx = e^{b^2}\int_{-\infty}^{\infty} e^{-a^2 x^2 - 2iabx} dx = \sqrt{\frac{\pi}{a^2}} e^{b^2} e^{\frac{-4 a^2 b^2}{4a^2}} = \sqrt{\frac{\pi}{a^2}} = \frac{\sqrt{\pi}}{|a|}; a\neq0.$$

Z. Alfata
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  • are you kidding ? using the fact that $\int_{-\infty}^{\infty} e^{-\alpha y^2 +\beta y } dy = \sqrt{\frac{\pi}{\alpha}} e^{\frac{\beta^2}{4\alpha}} ; \quad \alpha>0, \beta\in \mathbb C$ ? and how would you know that ? – reuns Apr 20 '16 at 02:46