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