[Math] Finding $\int_0^{\infty}xe^{-\lambda x} \, dx$

calculusintegrationreal-analysis

How to integrate:

$$\int_0^\infty x \, \lambda e^{-\lambda x} \, dx$$

I tried using integration by parts:

Let $u = x$ and $dv = \lambda e^{-\lambda x} \, dx$

Then $du = dx$

And $v = – \lambda e ^{-\lambda x}$

Correct so far?

Then

$$\begin{aligned}
uv – \int v \, du &= -\lambda x e^{-\lambda x} – \int (- \lambda e^{- \lambda x}) dx \\
&= -\lambda x e^{-\lambda x} – \lambda e^{-\lambda x}
\end{aligned}$$

The correct answer from lecture notes

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UPDATE 2

Let $u=\lambda x$ and $dv = e^{-\lambda x} dx$

Then $du = \lambda \, dx$

$$dv = e^{-\lambda x} \, dx$$

Let $y = -\lambda x$

Then $dy = -\lambda \, dx$

So $dx = -\frac{1}{\lambda} dy$

$$\begin{aligned}
v &= \int e^u \cdot – \frac{1}{\lambda} du \\
&= – \frac{1}{\lambda} e^{-\lambda x}
\end{aligned}$$

But here I seem to have an extra $- \frac{1}{\lambda}$ in $v$?

If I continue using integration by parts, I get:

$$-x e^{-\lambda x} + \color{red}{\frac{1}{\lambda}} \int e^{-\lambda x} \, dx$$

Best Answer

Answer in not just $ -xe^{-\lambda x} $. Correct answer with limits is $ [-xe^{-\lambda x}]^{\infty} _0$ + ${\int^{\infty}_0e^{-\lambda x}dx}$ which turns out to be $1/\lambda.$

Explanation:Applying integration by parts (the correct way)

$${\int{x\lambda}e^{-\lambda x}dx} = {}\lambda x {\int e^{-\lambda x}dx} - {\lambda}{\int}[{d(x)/dx}. {\int e^{-\lambda x}}]dx$$

Now apply limit and calculate $$ = [-xe^{-\lambda x}]^{\infty} _0 + {\int^{\infty}_0e^{-\lambda x}dx} $$ $$=1/\lambda$$

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