# Integral yielding part of a harmonic series

Why is this true? $$\int_0^\infty x \frac{M}{c} e^{(\frac{-x}{c})} (1-e^{\frac{-x}{c}})^{M-1} \,dx = c \sum_{k=1}^M \frac{1}{k}.$$

I already tried substituting $u = \frac{-x}{c}$. Thus, $du = \frac{-dx}{c}$ and $-c(du) = dx$. Then, the integral becomes (after cancellation) $\int_0^\infty c u M e^u (1-e^u)^{M-1}\,du$.

I looked at integral-table.com, and this wasn't there, and I tried wolfram integrator and it told me this was a "hypergeometric integral".

Thanks,

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if you substitute $t=1-e^{-x/c}$ your integral becomes $I_M=-c\int_0^1\log(1-t)Mt^{M-1}dt=-c\int_0^1\log(1-t)(t^M)'dt$. Hence $I_M-I_{M-1}=-c\int_0^1\log(1-t)(t^{M-1}(t-1))'dt=c\int_0^1 t^{M-1} dt$ (by per partes) $=c/M$. So you only need to compute $I_1$, which is $c$.

(there should be a nicer solution)

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That was pretty nice, I think. :) –  Hans Lundmark Mar 31 '11 at 10:34
Hi user8268, I read your explanation and I think I understand. This is my interpretation: The original integral was to solve $I_M$. We have found the quantity $I_M - I_{M-1}$. We also know the quantities $I_{M-1}-I_{M-2}$, and so on, all the way down to infinity. By using telescoping sum, if we add all the increments, we get $I_M$. The sum stops partway at $I_0 = 0$. That is, all the increments "above" $I_0$, we need for the sum, but all the increments "below" $I_0$, they all sum up to zero. (Won't there be a term $I_{-\infty}$, though?) –  jrand Apr 1 '11 at 14:58
$I_M = (I_M - I_{M-1}) + (I_{M-1} + I_{M-2}) + \cdots + I_{-\infty} =$. $=(I_M - I_{M-1}) + (I_{M-1} + I_{M-2}) + \cdots + (I_1 - I_0) + (I_0 - I_{-1}) + (I_{-1} - I_{-2}) + \cdots + I_{-\infty}$. But $I_0 = 0$ and $I_{-1} = (I_{-1} - I_{-2}) + (I_{-2} - I_{-3}) + (I_{-3} - I_{-4}) + \cdots + I_{-\infty}$. So, $I_M = (I_M - I_{M-1}) + (I_{M-1} + I_{M-2}) + \cdots + (I_1 - 0)$. –  jrand Apr 1 '11 at 15:21
@jrand: right: $I_M=(I_M-I_{M-1})+(I_{M-1}-I_{M-2})+\dots+(I_1-I_0)+I_0$. And $I_0=0$. –  user8268 Apr 1 '11 at 20:04

Careful, the substitution changes the limits of integration. It needs to be the integral going over negative real numbers.

You could just expand $(1-e^u)^{M-1}$ in $$\int_0^{-\infty} c u M e^u (1-e^u)^{M-1}\,du$$ to get

$$cM \int_0^{-\infty} u \sum_{n=0}^{M-1} \binom{M-1}{n} (-1)^n e^{(n+1)u}du.$$ Rearrange the integral and sum to find

$$cM \sum_{n=0}^{M-1} \binom{M-1}{n} (-1)^n \int_0^{-\infty} u e^{(n+1)u}du.$$ The anti-derivative of $xe^{rx}$ is $\frac{x}{r}e^{rx}-\frac{1}{r^2}e^{rx}$.

Can you solve it from here?

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Hm. After calculating the integral, and doing substituting the factorials in, I get this: $c \sum_{n=0}^{M-1} \frac{M!}{n!(M-1-n!)} \frac{(-1)^n}{(n+1)^2}$ –  jrand Mar 31 '11 at 5:26

Suppose that $X_1,\ldots,X_M$ are independent exponential random variables with mean $c$, so that their pdf and cdf are given by $f(x)=c^{-1}e^{-x/c}$ and $F(x)=1-e^{-x/c}$, $x \geq 0$, respectively. Let $Y_M=\max \{X_1,\ldots,X_M\}$. Then, $Y_M$ has cdf $F_M (x) = F(x)^M$ and hence pdf $f_M (x) = M F(x)^{M-1} f(x)$. Thus, the expectation of $Y_M$ is given by $${\rm E}[Y_M ] = \int_0^\infty {xMF(x)^{M - 1} f(x)dx} = \int_0^\infty {x\frac{M}{c}e^{ - x/c} (1 - e^{ - x/c} )^{M-1}dx}.$$ So you want to know why $${\rm E}[Y_M] = c\sum\limits_{k = 1}^M {\frac{1}{k}}.$$ Now, $Y_M$ is equal in distribution to $E_1 + \cdots + E_M$ where the $E_k$ are independent exponentials and $E_k$ has mean $c/k$; for this fact, see James Martin's answer to this MathOverflow question (where $c=1$). The result is thus established.

EDIT: As an additional reference, see Example 4.22 on p. 157 in the book Probability, stochastic processes, and queueing theory by Randolph Nelson.

EDIT: It is interesting to note that $$\int_0^\infty {x\frac{M}{c}e^{ - x/c} (1 - e^{ - x/c} )^{M - 1} dx} = c\int_0^1 { - \log (1 - x^{1/M} )dx} .$$ This follows by using a change of variable $x \mapsto x/c$ and then $x \mapsto (1-e^{-x})^M$. So, this gives the following integral representation for the $M$-th harmonic number $H_M := \sum\nolimits_{k = 1}^M {\frac{1}{k}}$: $$H_M = \int_0^1 { - \log (1 - x^{1/M} )dx}.$$ Finally, it is both interesting and useful to note note $$H_M = \int_0^1 {\frac{{1 - x^M }}{{1 - x}}dx} = \sum\limits_{k = 1}^M {( - 1)^{k - 1} \frac{1}{k}{M \choose k}},$$ see Harmonic number. With the above notation, the right-hand side corresponds to $${\rm E}[Y_M] = \int_0^\infty {{\rm P}(Y_M > x)dx} = \int_0^\infty {[1 - F_M (x)]dx} .$$

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