# Why the substitute part of integration of

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

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The title refers to integration but the body doesn't appear to mention integration. Please clarify. –  joriki Nov 10 '12 at 11:51
For not writing clearly with LaTeX mathematics, your question is pretty misterious. If by "$\,xe^{cx}\,$ from o to $\,\infty\,$" you meant $$\int_0^\infty xe^{cx}\,dx$$ then the solution is not $\,\left.xe^{cx}\right|_0^\infty\,$ but in fact $$\left.\frac{e^{cx}}{c}\left(x-\frac{1}{c}\right)\right|_0^\infty$$so the question is: what did you mean?? –  DonAntonio Nov 10 '12 at 12:08
My question a part of the expectation of exponential function which is the integration of $\lambda x e^{-\lambda x}$, where $\lambda$ is positive. So the $c$ above is a negative number. And the above expression is the substitute part of integration by part. –  RHS Nov 10 '12 at 12:59
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 $e^0=1$.