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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Ishan Banerjee
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4 Answers4


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).

L. F.
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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.


user 1591719
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Note that the value of $\int^1_0 \frac{n\ln(1+x^n)}{x}dx$ is invariant of $n$. Then \begin{align} \int^1_0 \frac{\ln(1+x)}{x}dx= -\frac67\int^1_0 {\frac{\ln\frac{(1+x^3)}{(1+x^2)(1+x)}}{x} } dx =-\frac67\int^1_0 \frac{\ln\left( 1-\frac x{1+x^2}\right)}{x}dx \\ \end{align} Let $J(t)=\int^1_0 \frac{\ln\left( 1-\frac {2x\sin t}{1+x^2}\right)}{x}dx$ and $$J’(t)=-\int^1_0 \frac{2\cos t\ dx }{(x-\sin t)^2 +\cos^2t}=-\left(\frac\pi2+t\right)$$ Thus $$\int^1_0 \frac{\ln(1+x)}{x}dx = -\frac67 J(\frac\pi6)= -\frac67\int_0^{\frac\pi6}J’(t)dt=\frac67\int_0^{\frac\pi6}\left(\frac\pi2+t\right)dt = \frac{\pi^2}{12}$$

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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 $Li_2(-1) = -\eta(2)$ 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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