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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:

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

Guy Fsone
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Jiew Meng
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3 Answers3

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

Andrew Miller
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  • Just noticed that and updated OP, but looks like my answer is still wrong? My answer factors to $(-\lambda e^{-\lambda x})(x-1)$ which I don't think I will get $\frac{1}{\lambda}$ from there? – Jiew Meng Sep 29 '13 at 06:48
  • Assume ${\lambda x} $ to be the function to be differentiated and $e^{ - \lambda x}$ to be integrated. – Andrew Miller Sep 29 '13 at 07:06
  • Just tried that, I seem to have made a mistake somewhere? My $v$ seems to have an extra $-1/\lambda$ term? See update 2 in OP. – Jiew Meng Sep 29 '13 at 07:16
  • if your $v= e^{- \lambda x} $ then your $dv = -{\lambda}e^{- \lambda x} dx$. – Andrew Miller Sep 29 '13 at 07:22
  • Did u mean $dv$ instead of $v$ and vice versa? We pick $dv$ not $v$ right? – Jiew Meng Sep 29 '13 at 07:34
  • OK I got it ... $du = \lambda , dx$ the $\lambda$s cancel – Jiew Meng Sep 29 '13 at 07:38
  • By the way $[-x e^{-\lambda x}]^\infty_0$ how do I reason it to be 0? Its because $\infty \cdot 0 = 0$ right? $e^{-\lambda \infty} = 0$? – Jiew Meng Sep 29 '13 at 07:42
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    One way to look at the answer is $ \lim_{x\to \infty} {xe^{- \lambda x}} $. Write ${xe^{- \lambda x}}$ as $\frac{x}{e^{\lambda x}}$ and apply L'Hospital's rule. – Andrew Miller Sep 29 '13 at 09:16
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How to integrate:

$$\int_0^\infty x \, \lambda e^{-\lambda x} \, dx \Longrightarrow -\frac{1}{\lambda}\int_0^\infty u\, e^{u} \, \Longrightarrow (-\frac{1}{\lambda})e^u(u + 1) + C \Longrightarrow -\frac{1}{\lambda}(e^{-\lambda x}(\lambda x - 1) + C)$$

1) Choose $u = -\lambda x$. $-du = \lambda dx$. No need for integration by parts so early.

Don Larynx
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Of course, $\lambda > 0$ must be assumed.

Another way: $$ \lambda x e^{-\lambda x} = - \lambda \dfrac{\partial}{\partial\lambda} e^{-\lambda x}$$

$$ \eqalign{\int_0^R \lambda x e^{-\lambda x} \; dx &= - \lambda \dfrac{d}{d\lambda} \int_0^R e^{-\lambda x}\; dx = -\lambda \dfrac{d}{d\lambda} \left(\dfrac{1}{\lambda} - \dfrac{e^{-\lambda R}}{\lambda}\right) = \dfrac{1}{\lambda} + (\ldots) e^{-\lambda R}\cr &\to \dfrac{1}{\lambda} \text{ as } R \to \infty\cr} $$

And yet another, and more general:

The Maclaurin series for $g(s) = \int_0^\infty e^{(s-1) \lambda x}\; dx$ is $$ g(s) = \int_0^\infty \sum_{n=0}^\infty \dfrac{s^n \lambda^n x^n}{n!} e^{-\lambda x}\; dx = \sum_{n=0}^\infty \dfrac{s^n}{n!} \int_0^\infty \lambda^n x^n e^{-\lambda x}\; dx$$

But $$g(s) = \dfrac{1}{(1-s)\lambda} = \frac{1}{\lambda} \sum_{n=0}^\infty s^n$$ so for nonnegative integers $n$ $$\int_0^\infty \lambda^n x^n e^{-\lambda x}\; dx = \dfrac{n!}{\lambda}$$

Robert Israel
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