Answer
Convergent $\;,\;$ 0
Work Step by Step
Let \[I=\int_{-\infty}^{\infty}xe^{-x^2}dx\]
\[I=\int_{-\infty}^{0}xe^{-x^2}dx+\int_{0}^{\infty}xe^{-x^2}dx\;\;\ldots (1)\]
Let \[I_1=\int_{-\infty}^{0}xe^{-x^2}dx\;\;\;\ldots(2)\]
\[I_1=\lim_{t_1\rightarrow -\infty}\int_{t_1}^{0}xe^{-x^2}dx\;\;\;\ldots(3)\]
Let \[I_2=\int xe^{-x^2}dx\]
Substitute $-x^2=t$ ____(4)
$\;\;\;\;\;\;\;\;\; -2xdx=dt$
Using (4)
\[I_2=\frac{-1}{2}\int e^tdt=\frac{-1}{2}e^{-x^2}\;\;\;\ldots (5)\]
Using (5) in (3)
\[I_1=\lim_{t_1\rightarrow -\infty}\left[\frac{-1}{2}e^{-x^2}\right]_{t_1}^{0}\]
\[I_1=\lim_{t_1\rightarrow -\infty}\left[\frac{-1}{2}+\frac{1}{2}e^{-t_{1}^2}\right]=\frac{-1}{2}\]
Since limit on R.H.S. of (3) exists so $I_1$ is convergent
$\Rightarrow I_1=\large\frac{-1}{2}$
Let \[I_3=\int_{0}^{\infty}xe^{-x^2}dx\]
\[I_3=\lim_{t_2\rightarrow\infty}\int_{0}^{t_2}xe^{-x^2}dx \;\;\;\ldots (6)\]
Using (5) in (6)
\[I_3=\lim_{t_2\rightarrow\infty}\left[\frac{-1}
{2}e^{-x^2}\right]_{0}^{t_2}\]
\[I_3=\lim_{t_2\rightarrow\infty}\left[\frac{-1}
{2}e^{-t_{2}^2}+\frac{1}{2}\right]=\frac{1}{2}\]
Since limit on R.H.S. of (6) exists so $I_3$ is convergent
\[I_3=\frac{1}{2}\]
Since $I_1$ and $I_3$ are convergent
So from (1)
$I$ is convergent and $I=I_1+I_3=\large\frac{-1}{2}+\frac{1}{2}=0$
