I was trying to evaluate $$\int^1_0 \frac{\log(1+x)}{x}dx.$$

I expanded $\log(1+x) $ as $x -\frac{x^2}{2}... $ and got the answer. I would like to know if there is any way to do it without series expanding.

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    $\begingroup$ You then get another integral, which isn't easy to solve. $\endgroup$ – Ishan Banerjee Jan 28 '13 at 10:28
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    $\begingroup$ $$ \int_0^1 \frac{\log(1+x)}{x} dx =\frac{\pi^2}{12} $$ $\endgroup$ – Rustyn Jan 28 '13 at 10:40
  • $\begingroup$ @Rustyn Yazdanpour I also got the same answer.I solved it in terms of $\zeta (2)$. But as I want to avoid series expansion or using the fact that $\zeta (2)= \frac{\pi^2}{6}$ I want a different method. $\endgroup$ – Ishan Banerjee Jan 28 '13 at 10:45
  • $\begingroup$ Ok. I'm not sure of any other way. Hopefully somebody has some insight for you, sorry. $\endgroup$ – Rustyn Jan 28 '13 at 10:48

Step I
Integrating by part we get that

$$\int^1_0 \frac{\log(1+x)}{x}dx=-\int^1_0 \frac{\log(x)}{x+1}dx$$

Step II
Letting $x=e^{-u}$, we have $$\int_0^{\infty}\frac{u}{e^u+1}du$$
Step III $$\int_0^{\infty}\frac{u^{s-1}}{e^u+1}du=\Gamma(s)\cdot\eta(s)\tag1$$ that is the product between gamma function and Dirichlet eta function

Step IV
Let $s=2$ in $(1)$ and we're done.


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    $\begingroup$ I did go up till step 2 in my attempts. Unfortunately, I've never even heard of the Dirichlet eta function. Is there any other way(using only elementary functions)? $\endgroup$ – Ishan Banerjee Jan 28 '13 at 13:39
  • $\begingroup$ @IshanBanerjee: you may express $1/(e^u+1)$ as a geometrical progression sum that is easy to do. $\endgroup$ – user 1357113 Jan 28 '13 at 13:41
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    $\begingroup$ @ Chris's sister Yes, but then you again need to know the value of $\eta(2)$ from it's series, which is what I wanted to avoid. $\endgroup$ – Ishan Banerjee Jan 28 '13 at 13:46
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    $\begingroup$ My point is that proving $(1)$ is more or the same work (and the general solution) to this problem. $\endgroup$ – Pedro Tamaroff Jan 28 '13 at 17:58
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    $\begingroup$ ...and one usually uses series. $\endgroup$ – Pedro Tamaroff Jan 28 '13 at 17:59

You might be interested in this: noticing that

$$\int_0^1 \frac{1}{1+xy}dy=\frac{\ln (x+1)}{x}$$

We can rewrite the integral as:

$$\int_0^1\frac{\ln (x+1)}{x}\;dx=\int_0^1\int_0^1 \frac{1}{1+xy}\;dy\;dx$$

Now read page 11 of this article (you'll have to slightly adapt the above of course).


I just want to share this because this is interesting, it uses a definition which is a series.

What you are looking at is

$$Li_2(-1) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$$

Which is the most "important" polylogarithm.

What is even cooler is this is $-\eta(2)$ which is the dirchlet-eta function.

I will show you how to compute $Li_2(-1) = -\eta(2)$ using the series definition. It isnt what you asked for but I cant resist to share this as it is really cool. Let $S$ represent the required sum.

$$S = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$$

The residue theorem states:

$\displaystyle \sum_{n=-\infty}^{\infty} (-1)^n f(n) = (-)\sum$ Res(f,c)$\cdot\pi\csc(\pi z)f(z)$ at the poles of $f(z)$

Because $z=0$ is a singularity twice for $z$ (because of $z^2$) and once for $\csc(z)$ it is an order $k=3$ so the residue will be according to the third derivative of $f(z)$

$$(-)\sum \space \text{Res}(f,z=0)\cdot\pi\csc(\pi z)f(z) = -\frac{\pi^2}{6}$$

But notice because $H(n) = (-1)^n f(n)$ is even:

$\displaystyle \sum_{n=-\infty}^{\infty} (-1)^n f(n) = 2\sum_{n=1}^{\infty} (-1)^n f(n) = -\frac{\pi^2}{6}$


$$\sum_{n=1}^{\infty} \frac{(1-)^n}{n^2} = -\frac{\pi^2}{12}$$

It is interesting how so many functions are tied to that one integral.


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