4
$\begingroup$

So I was just wondering whether I can calculate the integral $\int_{-\infty}^{\infty} e^{-x^2}$ without the famous trick of the use of double integrals. '

If it helps, this problem is an exercise in a book I bought about “definite integrals, integration techniques and integration of series”.

Thank you in advance for any help.

$\endgroup$
3
  • $\begingroup$ Please use MathJax $\endgroup$ Commented Apr 20, 2020 at 20:25
  • 2
    $\begingroup$ You can use the Taylor series for $e^u$, then do the integral over $[0,N]$ term by term and let $N\to\infty$, and over $[-N,0]$ and let $N\to\infty$... $\endgroup$ Commented Apr 20, 2020 at 20:26
  • $\begingroup$ I don't know of a simple way, but you can relate it to $\Gamma(\frac12)$ by substituting $x=\sqrt{u}$, and then the reflection formula $\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin{\pi x}}$ can be used to find $\Gamma(1/2)$. Of course, the reflection formula is non-trivial to prove in its own right, but I think you could do it without ever invoking a double integral. $\endgroup$ Commented Apr 20, 2020 at 20:27

1 Answer 1

9
$\begingroup$

I believe you can use Feynman's differentiation under the integral sign trick. Keith Conrad's notes do a good job, as per usual, explaining this method (https://kconrad.math.uconn.edu/blurbs/analysis/diffunderint.pdf). First, notice that the integrand, $e^{-x^2}$, is even, so you can instead study $I = \int_{0}^{\infty}e^{-x^2}dx$, and then double the result. Second, let: $$ F(t) = \int_{0}^{\infty}\frac{e^{-t^2(1+x^2)}}{1+x^2}dx $$ Then, $$F(0) = \int_{0}^{\infty}\frac{1}{1+x^2}dx = \left.\tan^{-1}(x)\vphantom{$\dfrac12$}\right\vert_{0}^{\infty}=\frac{\pi}{2} $$ At $t=\infty$, at every point the integrand becomes $0$. Calculating the derivative of $F(t)$, we arrive at: $$ F'(t) = \int_{0}^{\infty}\frac{e^{-t^2(1+x^2)}(-2t(1+x^2))}{1+x^2} = -2te^{-t^2}\int_{0}^{\infty}e^{(-tx)^2}dx$$ Letting $y = tx \implies dy = tdx$, where $y(0)=0$, and $y(\infty)=\infty$. Then we arrive at: $$\begin{aligned} &= -2te^{-t^2}\int_{0}^{\infty}e^{-y^2}\frac{1}{t}dy\\ &= -2e^{-t^2}\int_{0}^{\infty}e^{-y^2}dy\\ F'(t) &= -2e^{-t^2}I \end{aligned}$$ Then, using the FTC: $$ \begin{aligned} \int_{0}^{\infty}F'(t)dt &=\int_{0}^{\infty} -2e^{-t^2}Idt\\ F(\infty) - F(0) &= -2I \int_{0}^{\infty}e^{-t^2}dt \end{aligned}$$ Seeing that $I$ appears again on the right side, we have $$0 - \frac{\pi}{2} = -2I^2 \implies I^2 = \frac{\pi}{4} \implies I = \frac{\sqrt{\pi}}{2}$$ Thus, after doubling the result,

$$\int_{\mathbb{R}}^{}e^{-x^2}dx = \sqrt{\pi}$$

$\endgroup$
4
  • $\begingroup$ Hey. Thank you very much for spending time in answering this. I will have to give a lot of effort to completely understand everything but you have helped me a lot. $\endgroup$
    – Grigoris
    Commented Apr 21, 2020 at 14:43
  • $\begingroup$ Of course, the link I provided is a good starting place, and there are plenty of other references that talk about this technique in general @Grigoris $\endgroup$
    – teddy
    Commented Apr 21, 2020 at 14:44
  • $\begingroup$ Well I just sat down studied the notes you attached and with some sneak peeks of your solution I managed to do it! It’s the most interesting think I’ve ever done as a high school student thanks!!! $\endgroup$
    – Grigoris
    Commented Apr 21, 2020 at 15:49
  • $\begingroup$ You're welcome! $\endgroup$
    – teddy
    Commented Apr 21, 2020 at 15:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .