Finding $\int^1_0 \frac{\log(1+x)}{x}dx$ without series expansion 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.
 A: 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.
Chris.
A: With $\int^1_0 \frac{\ln(1+x)}{x}dx
= 2\int^1_0 \frac{\ln(1+x^2)}{x}dx 
= 3\int^1_0 \frac{\ln(1+x^3)}{x}dx $
\begin{align} 
&\int^1_0 \frac{\ln(1+x)}{x}dx\\= &\ \frac67\int^1_0 {\frac{\ln\frac{(1+x^2)(1+x)} {(1+x^3)}}{x} } dx
=\frac67\int^1_0 \frac{\ln\frac{1+x^2}{1-x+x^2}}{x}dx \\= &\ \frac67\int_0^1\int_0^{\frac\pi6}
\frac{2\cos t }{1-2x\sin t + x^2}dt\ dx
=\frac67\int_0^{\frac\pi6}\left(\frac\pi2+t\right)dt
=\frac{\pi^2}{12}
\end{align}
A: How about a contour integral approach?
Let $f(z)= \dfrac{\operatorname{Log}\left(z+1\right)}{z}$ such that $\operatorname{Log}$ is the principal log. Let $C = [0,1] \cup \Gamma \cup [i,0]$ where $\Gamma$ is a quarter-circle on the first quadrant with a radius of $1$. Notice there is a removable discontinuity at $z=0$. By Cauchy's Residue Theorem, we get
$$
\eqalign{
0&=\int_{0}^{1}f\left(z\right)dz+\int_{\Gamma}f\left(z\right)dz+\int_{i}^{0}f\left(z\right)dz \cr
\int_{0}^{1}f\left(z\right)dz &= \int_{0}^{i}f\left(z\right)dz - \int_{\Gamma}f\left(z\right)dz.
}
$$
Using the Dilogarithm definition, we get
$$\int_{0}^{i}\frac{\log\left(z+1\right)}{z}dz = -\operatorname{Li}_2(-i) = \frac{\pi^{2}}{48}+iC,$$
where $C$ denotes Catalan's Constant.
Parameterizing $\Gamma$, let $z=e^{i\theta}$ for $\theta \in [0,\pi/2]$ so that
$$
\eqalign{
\int_{\Gamma}\frac{\operatorname{Log}\left(z+1\right)}{z}dz &= \int_{0}^{\pi/2}\frac{\operatorname{Log}\left(e^{i\theta}+1\right)}{e^{i\theta}}de^{i\theta} \cr
&= i\int_{0}^{\pi/2}\operatorname{Log}\left(e^{i\theta}+1\right)d\theta \cr
&= i\int_{0}^{\pi/2}\operatorname{Log}\left(2e^{i\theta/2}\cdot\frac{e^{i\theta/2}+e^{-i\theta/2}}{2}\right)d\theta \cr
&= i\int_{0}^{\pi/2}\operatorname{Log}\left(2e^{i\theta/2}\cos\left(\frac{\theta}{2}\right)\right)d\theta.
}
$$
From here, we can split up the logarithm and do some computational work using this result and the fact that
$$\operatorname{Log}\left(e^{i\theta/2}\right) = \ln\left|e^{i\theta/2}\right|+i\operatorname{Arg}\left(e^{i\theta/2}\right) = \ln(1)+\dfrac{i\theta}{2} = \dfrac{i\theta}{2}.$$
After some grunt work,
$$\int_{\Gamma}\frac{\operatorname{Log}\left(z+1\right)}{z}dz = iC-\frac{\pi^{2}}{16}.$$
Combining these two results yields the final answer:
$$\int_{0}^{1}\frac{\log\left(x+1\right)}{x}dx = \frac{\pi^{2}}{12}.$$
A: 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).
A: I just want to share this because this is interesting, it uses a definition which is a series.
What you are looking at is
$$\operatorname{Li}_2(-1) = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$$
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 $\operatorname{Li}_2(-1) = -\eta(2)$ using the series definition. It isn't 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:
$$ \sum_{n=-\infty}^{\infty} (-1)^n f(n) = -\sum \operatorname{Res}(f,c) \cdot\pi\csc(\pi z)f(z)$$ at the poles of $f(z)$.
Because $z=0$ is a singularity twice for $z$ (because of $z^2$) and once for $\csc(\pi z)$ it is an order $k=3$ so the residue will be according to the third derivative of $f(z)$
$$-\sum \operatorname{Res}(f,z=0)\cdot\pi\csc(\pi z)f(z) = -\frac{\pi^2}{6}$$
But notice because $H(n) = (-1)^n f(n)$ is even:
$$ \sum_{n=-\infty}^{\infty} (-1)^n f(n) = 2\sum_{n=1}^{\infty} (-1)^n f(n) = -\frac{\pi^2}{6}$$
Finally,
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}$$
It is interesting how so many functions are tied to that one integral.
