Why the substitute part of integration of

Question

$xe^{cx}$ from 0 to $\infty$ which is $\left.xe^{cx}\right|_0^\infty$ is 0? Where $c$ is a const.

Answer 1

If $c<0$, then $e^{cx}\to 0$ whence $x\to\infty$, and much faster than $x$ grows, so to speak, also $xe^{cx}\to 0$. More precisely, let $k:=-c>0$, then $$xe^{cx}=\frac x{e^{kx}} $$ and its limit as $x\to\infty$ is of the form $\displaystyle\frac\infty\infty$, so we can apply the L'Hospital rule to see it really tends to $\displaystyle\frac1\infty=0$.

And, in $0$, it is $0\cdot 1=0$, because $e^0=1$.