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An integral representation of nth Harmonic number is

$$H_n = \displaystyle \int_0^1 \frac{1 - x^n}{1 - x}\,dx$$

Wikipedia states that for every x > 0, integer or not, we have: $$H_{n} = n \displaystyle\sum_{k=1}^\infty \frac{1}{k(n+k)}$$

How can I get to this result from integral representation?

Karlis Olte
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3 Answers3

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We have $$H_n = \sum_{k=1}^n \dfrac1k = \sum_{k=1}^n \int_0^1 x^{k-1} dx = \int_0^1 \left(\sum_{k=1}^n x^{k-1} \right) dx = \int_0^1 \left(\dfrac{1-x^n}{1-x} \right)dx$$

You may also want to look at these answers [1] and [2] and , where I derived a similar/same result. The second part of your question is answered in [1].

Community
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  • I have 2 questions about your proof. Q1) Does it work for all fixed z? (It was not stated for what z does it work.). Q2) How do you substantiate interchange of sum and integral signs in 2nd row? Thank you. – Karlis Olte Apr 07 '13 at 09:50
  • The integral is linear, so integral of (finite) sums is the (finite) sum of the integrals. – copper.hat Apr 07 '13 at 16:09
  • @copper.hat there is infinity sign above summation sign, doesn't it mean it is infinite sum? – Karlis Olte Apr 07 '13 at 16:59
  • I don't see any $\infty$ in the above answer? – copper.hat Apr 07 '13 at 17:06
  • @copper.hat the proof as Marvis stated is in reference [1] and there is no "z" in above answer as well :) – Karlis Olte Apr 07 '13 at 17:09
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    @user64985 It might be better if you had asked this question in the comment section of reference [1], sicne this is related to that. Anyway, the idea is as follows. Since the convergence of the series is uniform in any closed interval of the form $[0,1-\epsilon]$, we are allowed to interchange the integral and summation for $\int_0^{1-\epsilon} f(x) dx$. Now argue that the function you get is continuous and hence the answer you can let $\epsilon \to 0$. –  Apr 07 '13 at 17:27
  • You could also combine the two answers to this question which would avoid dealing with convergence issues. (You still would need to justify that the expression in the other answer is indeed $H_n$, so you are just replacing one converges issue with another.) – copper.hat Apr 07 '13 at 17:51
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For the second, $$n \sum\limits_{k=1}^{+ \infty} \frac{1}{k(n+k)}= \sum\limits_{k=1}^{+ \infty} \left( \frac{1}{k}-\frac{1}{n+k} \right)$$

Seirios
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"How can I get to this result from integral representation?"

Here's how, taking your request literally. The trick is integrating by parts

$$ \begin{align} &\int_0^1 \frac{1-x^n}{1-x} \, dx\tag{1}\\ &=-\int_0^1 \left(1-x^n\right) \frac{d}{dx} \log (1-x) \, dx\tag{2a}\\ &=-\left(1-x^n\right) \log (1-x)\Big|^{1}_{0}-\int_0^1 n\; x^{n-1} \log (1-x) \, dx\tag{2b}\\ &=\int_0^1 n\; x^{n-1} \sum _{k=1}^{\infty } \frac{x^k}{k} \, dx\tag{2c}\\ &= \sum _{k=1}^{\infty } \frac{n}{k}\int_0^1 x^{n+k-1} \, dx\tag{2d}\\ &=\sum _{k=1}^{\infty } \frac{n}{k (n+k)}\tag{2e}\\ \end{align} $$

Explanation
(2a): rewrite the integrand to prepare integration by parts
(2b): do integration by parts. Notice that the integrated part vanishes at both ends of the interval
(2c): expand $\log (1-x)$ into a series
(2d): Exchange summation and integration
(2e): do the integral
Derivation finished

Dr. Wolfgang Hintze
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