I know that the integral $\int_0^1 \frac{(x+1)^n-1}{x} dx,$ for $n \in \mathbb{Z}^+$, can be evaluated by expanding the numerator with the binomial theorem and integrating term by term. You get the nice expression $$\int_0^1 \frac{(x+1)^n-1}{x} dx = \sum_{k=1}^n \binom{n}{k} \frac{1}{k}.$$ My question is this:

Is there some other "nice" expression for $$\int_0^1 \frac{(x+1)^n-1}{x} dx,$$ when $n$ is a positive integer?

Mathematica and Wolfram Alpha both tell me that $$\int_0^1 \frac{(x+1)^n-1}{x} dx = n \text{HypergeometricPFQ}[\{1, 1, 1 - n\}, \{2, 2\}, -1],$$ but the hypergeometric function is just another way of writing $\sum \limits_{k=1}^n \binom{n}{k} \frac{1}{k}.$

I'm led to think that there might be such a nice expression because we do get one for the same integrand but slightly different bounds: $$\int_{-1}^0 \frac{(x+1)^n-1}{x} dx = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^{k+1}}{k} = H_n,$$ where $H_n$ is the $n$th harmonic number. On the other hand, alternating binomial sums often evaluate to simpler expressions than the corresponding sums that don't alternate, so perhaps the existence of the simpler expression here doesn't mean much for my question.

I would accept a known special function as an answer (other than the hypergeometric one I already mention), or a simple expression in terms of well-known numbers, like the harmonic numbers.

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    $\begingroup$ Hmm, heuristically for n=0.5 I get $$\small \int_0^b {(x+1)^{0.5}-1\over x} dx = 2 \cdot \left( \log(1+1)+\sqrt{1+b}-1-\log(1+\sqrt{1+b}) \right) $$. But I don't see how this could possibly be generalized to other n ... $\endgroup$ Jan 4, 2012 at 7:28
  • $\begingroup$ @GottfriedHelms: Interesting. I didn't even consider the case where $n$ is not a positive integer $\endgroup$ Jan 4, 2012 at 16:35

2 Answers 2


Substitute $x = u-1,$ so $u = x+1,$ I get $$ \int_1^2 \; \; \frac{u^n -1}{u-1} \; du \; = \; \int_1^2 \; \; u^{n-1} + u^{n-2} + \cdots + u^2 + u^2 + u + 1 \; \; du $$ or $$ \left. \frac{u^n}{n} + \frac{u^{n-1}}{n-1} + \frac{u^{n-2}}{n-2} + \cdots + \frac{u^3}{3} + \frac{u^2}{2} + u \; \; \right|_{u=1}^{u=2} $$ The value at $u=1$ is $H_n.$ The value at $u=2$ is, well, just one of those things.

  • $\begingroup$ Thanks, Will. Your substitution makes it clear why I shouldn't expect something "nice" for my integral but why we get $H_n$ for the other problem. $\endgroup$ Jan 4, 2012 at 16:31

Let $J_n$ be your integral, and define $J_0 = 0$. The generating function for this sequence is $$G(z) = \sum_{n=1}^\infty J_n z^n = \frac{\ln((1-2t)/(1-t))}{t(t-1)} $$ The generating function satisfies the differential equation $$(1-2z)(1-z)^2 \frac{dy}{dz} - (1-2z)(1-z) y = 1 $$ and the coefficients satisfy the recurrence $$ (2n+2) J_n - (3n+4) J_{n+1} + (n+2) J_{n+2} = 1 $$ A "closed-form" formula obtained by having Maple solve this recurrence is $$J_n =-i\pi -{\frac {{2}^{n} \left( {\it LerchPhi} \left( 2,1,n \right) n-1 \right) }{n}}-\Psi \left( n+1 \right) -\gamma $$

  • $\begingroup$ Thanks, Robert. This is interesting to know. $\endgroup$ Jan 4, 2012 at 16:32
  • $\begingroup$ +1, I wouldn't thought of using this method to derive a recurrence relation, that is really interesting. $\endgroup$ Jan 4, 2012 at 16:43
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    $\begingroup$ Just a quick thing to mention about Maples "closed-form" answer: We have that $\psi(n+1)=-\gamma+H_n$, and the $i\pi $ along with the LerchPhi function is simply an elaborate encoding of the finite series $\sum_{k=1}^n \frac{2^k}{k},$ so the final result is really the same as that which was mentioned in Will Jagy's answer. $\endgroup$ Jan 4, 2012 at 16:43

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