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I tried to proof this as a (signal engineering) homework using binomal square, but the example answer was given using differential equations. I'd like to know if my approach was possible.

I tried the following:

$$ t, v \in \mathbb{R}$$ $$\int_{-\infty}^{\infty} e^{-\pi t^{2}} e^{- i 2 \pi t v} dt $$ Expand with $e^{-\pi v^{2}} e^{\pi v^{2}}$

$$ e^{-\pi v^{2}} \int_{-\infty}^{\infty} e^{-\pi t^{2}} e^{- i 2 \pi t v} e^{\pi v^{2}} dt $$ With $-v^{2} = (i v)^2$

$$ e^{-\pi v^{2}} \int_{-\infty}^{\infty} e^{-\pi (t^{2} + 2 t (i v) + (iv)^{2})} dt $$

And then the binomal square

$$ e^{-\pi v^{2}} \int_{-\infty}^{\infty} e^{-\pi (t + iv)^2} dt $$

Here I got stuck. I know $\int_{-\infty}^{\infty} e^{-\pi t^2} dt = 1$, but how to proof this is also true with a complex constant?

vahvero
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  • It is enough to invoke the residue theorem. $e^{-z^2}$ is an entire function and $\left|e^{-z^2}\right|\to 0$ pretty fast when $\left|\text{Re}(z)\right|\to +\infty$ and $\text{Im}(z)\in[0,|v|]$. – Jack D'Aurizio Aug 14 '18 at 08:50
  • There are several ways to derive the formula but your method doesn't work unless you invoke some results in complex analysis. – Kavi Rama Murthy Aug 14 '18 at 08:51
  • Why so much pain? $t + iv = y$ and you're done, since the differential involves only $t$. –  Aug 14 '18 at 09:01
  • @VonNeumann I was told by an assistant teacher that I cannot make real number to a complex one with simple substitution. – vahvero Aug 14 '18 at 09:33

1 Answers1

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By parity, on has $$\int_{-\infty}^{\infty} e^{-\pi t^{2}} e^{- i 2 \pi t v} dt =\int_{-\infty}^{\infty} e^{-\pi t^{2}} \cos{(2 \pi v t)} \, dt.$$

Now, you can use the well-known formula $$\int_{-\infty}^\infty e^{-at^2}\cos(bt)\,dt =\frac{\sqrt\pi}{\sqrt{a}}e^{-\frac{b^2}{4a}}.$$ If you don't know this formula, see here.

C. Dubussy
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  • Could you give example how this solves $\int_{- \infty}^{\infty} e^{-\pi (t + iv)^{2}} dt$? – vahvero Aug 14 '18 at 09:33
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    But... I did. Develop the square, then consider the real part and the imaginary part. By parity, the imaginary part is $0$ and the real part is what I wrote in my post. – C. Dubussy Aug 14 '18 at 09:35
  • I might have misunderstood this, but I think this means that I should/can not use binominal square to solve this. (By opening the square and "removing" the multiplication by 1.) – vahvero Aug 14 '18 at 09:47
  • Indeed. Without complex analysis, completing the square is not the good strategy. – C. Dubussy Aug 14 '18 at 09:59