# Integrating by parts - question on a limit

I want to integrate this function by parts:

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

And I arrive to the following expression:

$$\int_0^\infty x\lambda e^{-\lambda x} \, \mathrm{d}x = \left. -xe^{-\lambda x}\right|^\infty_0 + \int_0^\infty e^{-\lambda x} \, \mathrm{d}x$$

How do I evaluate this: $\left.-xe^{-\lambda x}\right|^\infty_0$ as it seems to yield $\frac{-\infty}{0}$

Thanks

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Since $-x\to-\infty$ and $e^{-\lambda x}\to 0$, it yields $-\infty\cdot0$, not $-\infty/0$. L'Hopital's rule handles that quickly but gives no insight. – Michael Hardy Sep 2 '14 at 15:25
I bet you're evaluating the mean of exponential distribution. The simple explanation for $\dfrac{x}{e^{\lambda x}}\to0$ as $x\to\infty$ is the exponential function goes to infinity faster than the linear/ polynomial function. – Tunk-Fey Sep 2 '14 at 19:23

(Assuming $\lambda > 0$) you can use L'Hôpital's rule:

$$\lim_{x\to\infty} -xe^{-\lambda x} = -\lim_{x\to\infty} \dfrac{x}{e^{\lambda x}} = -\lim_{x\to\infty} \dfrac{1}{\lambda e^{\lambda x}} = 0$$

The relevant insight is that, while $x$ and $e^{\lambda x}$ both grow as $x\to \infty$, $e^{\lambda x}$ grows much faster, and overall the fraction tends to zero.

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L'Hopital's rule gives the answer quickly but gives little if any insight. – Michael Hardy Sep 2 '14 at 15:22
@MichaelHardy The intuitive statement of L'Hopital's is "if you can't compare the functions directly, then compare their growth rates". If students are being taught the rule properly (i.e. the "why?" in addition to the "how?"), then its application should yield the answer and the intuition simultaneously. – BaronVT Sep 2 '14 at 15:41
@MichaelHardy What sort of insight were you looking for? This is a perfectly reasonable result and the last sentence gives some intuition about why $$\lim_{x\to\infty} -xe^{-\lambda x} = -\lim_{x\to\infty} \dfrac{x}{e^{\lambda x}} .$$ – alexqwx Sep 2 '14 at 18:56

$\lim_{x \to \infty}-xe^{-\lambda x}=\lim_{x \to \infty}\frac{-x}{e^{\lambda x}}$ has the indeterminate form $\frac{-\infty}{\infty}$ which sets up Hopital's Rule.

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