Answer
$$
\int_{-\infty}^{\infty} x e^{-x^{2}} d x
$$
The given improper integral
$$
\begin{aligned}
\int_{-\infty}^{\infty} x e^{-x^{2}} d x=0
\end{aligned}
$$
Work Step by Step
$$
\int_{-\infty}^{\infty} x e^{-x^{2}} d x
$$
The given improper integral
$$
\begin{aligned}
\int_{-\infty}^{\infty} x e^{-x^{2}} d x&=\lim _{a \rightarrow-\infty}\left(-\frac{1}{2} \int_{a}^{0}-2 x e^{-x^{2}} d x\right)+ \\
&+\lim _{b \rightarrow \infty}\left(-\frac{1}{2} \int_{0}^{b}-2 x e^{-x^{2}} d x\right)\\
&=\left.\lim _{a \rightarrow-\infty}\left(-\frac{1}{2} e^{-x^{2}}\right)\right|_{a} ^{0}\\
&\quad+\left.\lim _{b \rightarrow \infty}\left(-\frac{1}{2} e^{-x^{2}}\right)\right|_{0} ^{b}\\
&=\lim _{a \rightarrow -\infty}\left(-\frac{1}{2}+\frac{1}{2 e^{a^{2}}}\right)\\
&\quad+\lim _{b \rightarrow \infty}\left(-\frac{1}{2 e^{b^{2}}}+\frac{1}{2}\right) \\
&=-\frac{1}{2}+\frac{1}{2}\\
&=0
\end{aligned}
$$
