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$$ \begin{align} \int_0^\infty e^{-x^2}\,\mathrm{d}x &=\int_0^\infty e^{-\left(x-ia\right)^2}\,\mathrm{d}x\\ \end{align} $$

For any real number a.

I understand that the key is Cauchy's Integral Theorem and that taking the integral $e^{-z^2}$ around the contour of a rectangle -b, b, b+ai, -b+ai results in zero as there are no poles in that zone. However, I'm unable to quite figure out how that key works here. I've found some pages that try to explain this, but I don't seem to be getting it. Can I just discard the Imaginary portion of the exponent? Apparently it'd be different if there was a pole inside this rectangle.

Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$

How to do this integral $\int_{-\infty}^{\infty}{\rm e}^{-x^{2}}\cos\left(\,kx\,\right)\,{\rm d}x$

How to prove $\int_0^\infty e^{-x^2}cos(2bx) dx = \frac{\sqrt{\pi}}{2} e^{-b^2}$

Alan Ox
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    Your rectangle is faulty. One of the edges should lie on the imaginary axis exactly, so any points referencing "$-b$" are wrong. – Ninad Munshi May 12 '24 at 12:49
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    It's simpler to use uniqueness of holomorphic extension. You already have the identity for $a \in i\mathbb{R}$. Now you just have to show that the RHS is holomorphic on some neighborhood of $i\mathbb{R}$. – Mason May 12 '24 at 13:45
  • "Your rectangle is faulty." That's a good point, as I've been going off a problem that took the integral from negative infinity to positive infinity but the problem I presented was from 0 to positive infinity. I should have also clarified that a goes to infinity. One question though is that I'd been considering the value of b to be arbitrary - the height of the rectangle. Does it matter or does it have to match the exponent? There are no poles anywhere anyway. – Alan Ox May 12 '24 at 18:35
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    Perhaps you meant $$\int\limits_{-\infty}^{\infty} e^{-x^2} , dx=\int\limits_{-\infty}^{\infty} e^{-(x-i a)^2} , dx=\sqrt{\pi }?$$ – Steven Clark May 13 '24 at 00:26
  • I'll have to look up the uniqueness of holomorphic extension, that's new to me. The book chapter I'm in specifically deals with using complex integrals to solve real equations. – Alan Ox May 13 '24 at 22:47

1 Answers1

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Let $b>0,a>0$

$$\gamma_1(t) = t, \quad t\in(-b,b) $$

$$\gamma_2(t) = b+it, \quad t\in(0,a) $$

$$\gamma_3(t) = -t+ai, \quad t\in(-b,b) $$

$$\gamma_4(t) = -b-it, \quad t\in(-a,0) $$

$$ \oint_{\gamma_1} e^{-z^2}dz = \int_{-b}^{b}e^{-t^2}dt $$

$$ \oint_{\gamma_2} e^{-z^2}dz = i\int_{0}^{a}e^{-(b+it)^2}dt $$

$$ \oint_{\gamma_3} e^{-z^2}dz = -\int_{-b}^{b}e^{-(-t+ai)^2}dt $$

$$ \oint_{\gamma_4} e^{-z^2}dz = -i\int_{-a}^{0}e^{-(-b-it)^2}dt $$

Clearly,

\begin{eqnarray*} \left| \oint_{\gamma_2} e^{-z^2}dz \right|= \left|\int_{0}^{a}e^{-(b+it)^2}dt \right| &\leq& \int_{0}^{a}\left|e^{-b^2-t^2} e^{-2itb}\right| dt\\ & =& \int_{0}^{a}e^{-b^2-t^2} dt \end{eqnarray*}

Then if $b\to \infty$

$$\left| \oint_{\gamma_2} e^{-z^2}dz \right| \to 0$$

Similarly

$$\left| \oint_{\gamma_4} e^{-z^2}dz \right| \to 0$$

by the Cauchy's integral theorem

$$\lim_{b\to \infty} \oint_{\gamma_1+...+\gamma_4}e^{-z^2}dz = \int_{-\infty}^{\infty }e^{-t^2}dt -\int_{-\infty}^{\infty}e^{-(-t+ai)^2}dt = 0 $$

Hence

$$\boxed{ \int_{0}^{\infty }e^{-t^2}dt = \frac{1}{2} \int_{-\infty}^{\infty}e^{-(t+ai)^2}dt \quad a>0}$$

The case $a=-c<0$ is analogous.

The statement

$$ \int_{0}^{\infty }e^{-t^2}dt = \int_{0}^{\infty}e^{-(t+ai)^2}dt $$

is false. We know

$$\int_{0}^{\infty }e^{-t^2}dt = \frac{\sqrt{\pi}}{2}$$

However

$$\int_{0}^{\infty }e^{-(t+i)^2}dt = \frac{\sqrt{\pi}}{2}\left(1-i\operatorname{erfi}(1) \right)$$

while

$$\frac{1}{2}\int_{-\infty }^{\infty }e^{-(t+i)^2}dt = \frac{\sqrt{\pi}}{2} $$

Bertrand87
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    thanks a lot for taking the time to walk me through this. And even though I asked the wrong question, you were able to answer the right question. One thing - after "clearly," should the t^2 on the right side of the inequality be added rather than subtracted? – Alan Ox May 17 '24 at 20:25
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    Hi, these integrals are equivalent, by the change of variable $t \mapsto -w$

    $$ \int_{-\infty}^{\infty}e^{-(t+ai)^2}dt = \int_{-\infty}^{\infty}e^{-(-w+ai)^2}dw$$

    Hope this answer your question.

     – Bertrand87 May 17 '24 at 20:58