Answer
The improper integral converges to $e$.
Work Step by Step
We will solve the integral $I=\int 2x e^{-x^2} \ dx$ by using u-substitution method.
Let us consider that $u=-x^2 \implies dx=\dfrac{-1}{2x}da$
So, we have, $I=\int 2x e^u (\dfrac{-1}{2x}) du=-\int e^u du=-e^u+C=-e^{-x^2}+C$
Now, $\lim\limits_{a \to \infty} \int_1^a 2x e^{-x^2} \ dx=\lim\limits_{a \to + \infty} [-e^{-x^2}]_1^a$
or, $=-\lim\limits_{a \to + \infty} [e^{-(a^2)}-e]$
or, $=-\lim\limits_{a \to + \infty} [e^{-(+\infty)^2}-e]$
or, $=-(0-e)$
or, $=e$
Therefore, the given improper integral converges to $e$.
