# Is there a way to calculate the improper integral of $e^{-x^2}$ without the use of double integrals?

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.

• Please use MathJax Commented Apr 20, 2020 at 20:25
• 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 $N\to\infty$... Commented Apr 20, 2020 at 20:26
• 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. Commented Apr 20, 2020 at 20:27

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