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In THIS ANSWER, I showed that

$$2\sum_{s=1}^{\infty}\frac{1-\beta(2s+1)}{2s+1}=\ln\left(\frac{\pi}{2}\right)-2+\frac{\pi}{2}$$

where $\beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet Beta Function.

In the development, it was noted that

$$\begin{align} \sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)&=\log\left(\frac21\cdot \frac23\cdot \frac43\cdot \frac45\cdots\right)\\\\ &=\log\left(\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}\right)\\\\ &=\log\left(\frac{\pi}{2}\right) \tag 1 \end{align}$$

where I used Wallis's Product for $\pi/2$.

If instead of that approach, I had used the Taylor series for the logarithm function, then the analysis would have led to

$$\sum_{n=1}^\infty(-1)^{n-1}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} \tag 2$$

where $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$ is the Dirichlet eta function.

Given the series on the right-hand side of $(2)$ as a starting point, it is evident that we could simply reverse steps and arrive at $(1)$.

But, what are some other distinct ways that one can take to evaluate the right-hand side of $(2)$?

For example, one might try to use the integral representation

$$\eta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{1+e^x}\,dx$$

and arrive at

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}\eta(n)}{n} =\int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx =\int_1^\infty \frac{x-1}{x^2(x+1)\log(x)}\,dx \tag 3$$

Yet, neither of these integrals is trivial to evaluate (without reversing the preceding steps).

And what are some other ways to handle the integrals in $(3)$?

Community
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Mark Viola
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  • I am having trouble with the exact meaning of "distinct ways". Are you, for example, forbidding the Weierstarss product formula for the sine/cosine, of which Wallis' product is a special case? Here are 2 ways for example to evaluate the integrals: 1. expand into series of logarithms by using Frullani's integral. 2. consider a parametric form of the integral, differentiate and obtain a combination of digammas, and then integrate back. Are these ways considered distinct? – Noam Shalev - nospoon May 12 '16 at 19:43
  • @nospoon I invoked the first suggestion and recovered $(1)$ from the OP. The second suggestion was posted by Olivier Oloa roughly 10 minutes before your comment posted. ;-)) ... But, I do sincerely appreciate your comments! Your one of the "good-people" on this site. -Mark – Mark Viola May 12 '16 at 20:07
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    I am grinning ear to ear reading these kind words, especially when they come from you, Mark. I enjoy reading your insightful answers on this site, and you are most certainly of the kind of contributors that made me stick with math. – Noam Shalev - nospoon May 13 '16 at 11:53
  • @nospoon Noam, I enjoy, and often learn from, reading your insightful posts also. By the way ... I wrote "your" instead of "you're" in my previous comment, thereby violating one of my own "don't say that" things. – Mark Viola May 13 '16 at 16:12
  • What was the rationale for the down vote? – Mark Viola Jul 14 '18 at 18:06

3 Answers3

4

What are some other ways to handle the following integral? $$ \int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx \tag 1 $$

One may set $$ I(s):=\int_0^\infty \frac{1-e^{-sx}}{x(e^x+1)}dx, \quad s>0. \tag2 $$ We may differentiate under the integral sign, in order to get rid of the factor $x$ in the denominator, obtaining $$ \begin{align} I'(s)&=\int_0^\infty \frac{e^{-sx}}{e^x+1}dx \\\\I'(s)&=\int_0^\infty e^{-(s+1)x}\sum_{n=0}^\infty(-1)^n e^{-nx} dx \\\\I'(s)&=\sum_{n=0}^\infty(-1)^n\int_0^\infty e^{-(n+s+1)x} dx \\\\I'(s)&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+s} \\\\I'(s)&=\frac12\psi\left(1+\frac{s}2\right)-\frac12\psi\left(\frac{1+s}2\right) \tag3 \end{align} $$ where $\displaystyle \psi : = \Gamma'/\Gamma$.

Integrating $(3)$, with the fact that, as $s \to 0$, $I(s) \to 0$, we get

$$ \int_0^\infty \frac{1-e^{-sx}}{x(e^x+1)}dx=\frac12\log(\pi)+\log \Gamma\left(1+\frac{s}2\right)-\log \Gamma\left(\frac{1+s}2\right), \quad s>0. \tag4 $$

By putting $s:=1$ in $(4)$, one obtains $$ \int_0^\infty \frac{1-e^{-x}}{x(e^x+1)}dx=\log \left(\frac{\pi}2\right).\tag5 $$

Olivier Oloa
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Another way to handle $(2)$ is using the identity $$\eta\left(s\right)=\left(1-\frac{1}{2^{s-1}}\right)\zeta\left(s\right) $$ hence, since $\eta\left(1\right)=\log\left(2\right) $, $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+\sum_{n\geq2}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right) $$ $$=\log\left(2\right)+\sum_{n\geq2}\frac{\left(-1\right)^{n-1}}{n}\zeta\left(n\right)-\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\left(-\frac{1}{2}\right)^{n-1} $$ and now we can use the identity $$\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\left(-x\right)^{n}=x\gamma+\log\left(\Gamma\left(x+1\right)\right),\,-1<x\leq1 $$ which can be proved taking the log of the Weierstrass product of Gamma. So $$\sum_{n\geq1}\frac{\left(-1\right)^{n-1}}{n}\eta\left(n\right)=\log\left(2\right)+2\log\left(\frac{\sqrt{\pi}}{2}\right)=\log\left(\frac{\pi}{2}\right).$$

Marco Cantarini
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  • @Dr.MV In our case we have to take $x=1$, which is admissible, not $x=-1$. – Marco Cantarini May 13 '16 at 16:16
  • Yes, sorry; it's early and I was parallel processing my reading this and speaking on a conference call. On another observation, using the Weierstass product for log(Gamma), we obtain $$\log\left(\Gamma(x+1)\right)+x\gamma = -\gamma -\log(x+1)+\sum_{n=1}^\infty \left(\frac{x+1}{n}-\log\left( 1+\frac{x+1}{n}\right) \right)$$Evidently, there is a bit more work (expanding the logarithm is a Taylor series and interchanging the order of summation) to arrive at the identity in your post. ;-)) Solid way forward! +1 ... -Mark – Mark Viola May 13 '16 at 16:49
  • @Dr.MV Yes, my explanation of the identity isn't too detailed :-D – Marco Cantarini May 13 '16 at 16:58
2

Observation $1$:

A suggestion made in a comment from @nospoon was to expand one of the integrals in a series and exploit Frullani's Integral. Proceeding accordingly, we find that

$$\begin{align} \int_0^\infty \frac{1-e^{-x}}{x(1+e^x)}\,dx&=\int_0^\infty \left(\frac{(e^{-x}-e^{-2x})}{x}\right)\left(\sum_{n=0}^\infty (-1)^{n}e^{-nx}\right)\,dx\\\\ &=\sum_{n=0}^\infty (-1)^{n} \int_0^\infty \frac{e^{-(n+1)x}-e^{-(n+2)x}}{x}\,dx\\\\ &=\sum_{n=0}^\infty (-1)^{n} \log\left(1+\frac{1}{n+1}\right)\\\\ &=\sum_{n=1}^\infty (-1)^{n-1} \log\left(1+\frac{1}{n}\right) \end{align}$$

thereby recovering the left-hand side of Equation $(1)$ in the OP.

Observation $2$:

In the answer posted by Olivier Oloa, note the intermediate relationship

$$I'(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n+s}$$

Upon integrating $I'(s)$, as $\int_0^1 I'(s)\,ds$, we find that

$$I(1)=\sum_{n=1}^\infty (-1)^{n-1}\log\left(1+\frac1n\right)$$

thereby recovering again the left-hand side of Equation $(1)$ in the OP.

Mark Viola
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