Let's do some calculus! (15)

Consider the Gaussian integral $\displaystyle I=\int_{-\infty}^{\infty} e^{-x^2} dx$ , For this being an even function and symmetric over the line $x=0$ we consider $\displaystyle I=2\int_{0}^{\infty}e^{-x^2}dx$ and hence set $x^2=u$ that makes $\displaystyle I=\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}}du = \Gamma(\frac{1}{2})=\sqrt{\pi}$ which completes the derivation.

We know by Euler's formula that , $\displaystyle e^{i\pi}+1=0$ and so the given expression equals $\boxed{2}$

Relevant wiki: Integration Tricks

Consider the Gaussian integral $I = \displaystyle \int_0^\infty e^{-x^2} dx$ . It can be solved using the following identity involving polar integral (right).

$I^2 = \int_0^\infty \int_0^\infty e^{-x^2}e^{-y^2} \ dy \ dx = \int_0^\frac \pi 2 \int_0^\infty re^{-r^2} \ dr \ d\theta$

$\begin{aligned} I^2 & = \int_0^\frac \pi 2 \int_0^\infty re^{-r^2} \ dr \ d\theta = \int_0^\frac \pi 2 \frac {e^{-r^2}}2\bigg|_\infty^0 \ d\theta = \frac 12 \int_0^\frac \pi 2 \ d \theta = \frac \pi 4 \\ \implies I & = \frac {\sqrt \pi}2 \end{aligned}$

Now, we have:

$\begin{aligned} X & = \frac 1{\sqrt \pi} \color{#3D99F6}{\int_{-\infty}^\infty e^{-x^2} dx} - e^{i\pi} & \small \color{#3D99F6}{\text{As the integrand is even}} \\ & = \frac 1{\sqrt \pi} \cdot \color{#3D99F6}{2I} - (\cos \pi + i \sin \pi) \\ & = \frac 1{\sqrt \pi} \cdot \color{#3D99F6}{\sqrt \pi} - (-1 + 0) \\ & = \boxed{2} \end{aligned}$