Integration by parts on $\int_{0}^\infty t \lambda e^{-\lambda t} dt $

Question

I'd like to integrate

\begin{align}
\int_{0}^\infty t \lambda e^{-\lambda t} dt
\end{align}

Which looks like something that needs to solved by integration by parts.

Integration by parts claims

\begin{align}
\int_{0}^\infty udv = uv – \int_{0}^\infty vdu
\end{align}

If I let $u= e^{-\lambda t}$ and $v=t\lambda$

Then

$$
\begin{align}
t \lambda e^{-\lambda t}- \int_0^\infty t \lambda du &= t \lambda e^{-\lambda t}- \int_0^\infty t \lambda du \\
&= t \lambda e^{-\lambda t} – (t\lambda e^{-\lambda t})\big|_0^\infty
\end{align}
$$

But I don't know what it means to evaluate at $e^{\lambda t} = \infty$ and it equal to $0$?

Some help understanding what the proper approach is would be awesome.

---

Thanks everybody for the help. It makes much more sense now 🙂

Answer 1

Hint

The improper integral $I = \int_0^{\infty}f(t) \,dt$ can be defined as $$ I = \lim_{a\to \infty} \int_0^a f(t) \, dt$$.

So you need to calculate $\int_0^a t\lambda e^{-\lambda t} \, dt$ and take the limit $a \to \infty $.