I want to evaluate:

$$\int_{0}^{1} \frac{\ln x}{x+1} dx$$

If I was asked I would to evaluate:

$$\int_{0}^{1} \frac{\ln x}{x-1} dx$$

That would be easy because if I use the Taylor series for $\ln x$ centered at $1$ then things will cancel out and leave me with a easy integral.

So how about this integral, I'm guessed to use the Taylor series centered around $-1$. But even with that thought in mind, it does not take me anywhere because $\ln (-1)$ is not defined. Can someone help.

  • $\begingroup$ Is this a normal calculus class? If not I'd suggest looking up Polylogarithms, most specifically dilogarithms $\endgroup$ – Triatticus May 30 '16 at 3:24
  • $\begingroup$ I only know about basic techniques but thanks anyway. $\endgroup$ – Ahmed S. Attaalla May 30 '16 at 3:26

We know that $$\int_0^1 x^a~dx=\frac{1}{a+1}$$

Then we can differentiate with respect to $a$:

$$\int_0^1 x^a \ln x ~dx=-\frac{1}{(a+1)^2}$$

Now we can use the geometric series:

$$\sum_{a=0}^\infty (-1)^a x^a=\frac{1}{1+x},~~~|x|<1$$

$$\sum_{a=0}^\infty (-1)^a \int_0^1 x^a \ln x ~dx=- \sum_{a=0}^\infty (-1)^a \frac{1}{(a+1)^2}$$

(Now we interchange integration and summation on the left hand side)

$$\int_0^1 \frac{\ln x}{1+x} ~dx=-\sum_{a=0}^\infty \frac{(-1)^a}{(a+1)^2}=\sum_{a=1}^\infty \frac{(-1)^a}{a^2}=-\frac{\pi^2}{12}$$

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  • $\begingroup$ I understand everything until you get to the .actual integral, what did you do? Integration by parts? It would really help. Thanks $\endgroup$ – Ahmed S. Attaalla May 31 '16 at 23:00
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    $\begingroup$ @AhmedS.Attaalla, I edited in the middle. No, I don't use integration by parts here $\endgroup$ – Yuriy S May 31 '16 at 23:23
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    $\begingroup$ @AhmedS.Attaalla : he started from $\int_0^1 x^a dx$ and made the $\ln x$ appear by differentiating with respect to $a$ $\endgroup$ – reuns May 31 '16 at 23:32
  • $\begingroup$ I understand it now thanks $\endgroup$ – Ahmed S. Attaalla May 31 '16 at 23:34

Make the change of variables, $x = 1 - u$ to get

$$-\int_0^1{\log(1-u)\over u}$$

Now you can do the Taylor series easily

$$\sum_{k=1}^\infty \int_0^1{u^{k-1}\over k}\,du$$

This gives us

$$\sum_{k=1}^\infty {1\over k^2}={\pi^2\over 6}.$$

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  • $\begingroup$ About where did you center the Taylor series? $u=0$? $\endgroup$ – Ahmed S. Attaalla May 30 '16 at 3:28
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    $\begingroup$ @AhmedS.Attaalla note that the powers of $u$ are $(u-0)^k$ so it must be centered at $0$. $\endgroup$ – Adam Hughes May 30 '16 at 3:29
  • $\begingroup$ Can you help me with another thing, I'm trying to get a technique down that will help me to solve integrals of these kind and possible others, I will edit in another question in the body of this question if you will help. $\endgroup$ – Ahmed S. Attaalla May 30 '16 at 3:36
  • $\begingroup$ @AhmedS.Attaalla it is standard MSE procedure to put new questions in new topics. $\endgroup$ – Adam Hughes May 30 '16 at 3:37
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    $\begingroup$ I'm sorry but I'm forced to accept another answer. Your substitution idea gives $2-u$ in the denominator. $\endgroup$ – Ahmed S. Attaalla Aug 8 '16 at 19:26


$$I=\int_{0}^{1}\frac{\ln(x)}{1+x}dx= \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$$ You can show this result by noticing since $0<x<1$ that $$\frac{1}{1+x}= \sum_{n=0}^{\infty}(-x)^n$$ and performing term by term integration. You will need integration by parts to do that.

Now you can also show that:

$$-I=\int_{0}^{1}\int_{0}^{1}\frac{1}{1+xy}dydx$$ by expanding the integrand into a geometric series and doing term by term integration with respect to $y$ and the with respect to $x$.

Now, in the double integral, let $x=u+v,y=-u+v$

Then $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\begin{vmatrix} 1 && 1\\ -1 && 1 \\ \end{vmatrix}=2$

The change of variables causes the double integral to become:

$$\iint_{T} \frac{2}{1+v^2-u^2}dvdu$$ where $T$ is the square with vertices $(0,0),(\frac{-1}{2},\frac{1}{2}),(0,1),(\frac{1}{2},\frac{1}{2})$

Now, we need to compute two double integrals (over two different subregions of the square) and sum them to get the final answer.

First let $\frac{-1}{2}\leq u \leq 0 , -u \leq v \leq 1+u$.

$$\int_{\frac{-1}{2}}^{0}\int_{-u}^{1+u} \frac{2}{1+v^2-u^2}dvdu=\int_{\frac{-1}{2}}^{0}\frac{2\tan^{-1}\left(\frac{1+u}{\sqrt{1-u^2}}\right)+2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$$

For the first term on the right hand side, let : $z=2\tan^{-1}\left(\frac{1+u}{\sqrt{1-u^2}}\right),dz=\frac{1}{\sqrt{1-u^2}}du$. For the second term, if you draw a right triangle, notice $2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=2\sin^{-1}\left(u\right)$ so let $z=\sin^{-1}\left(u\right),dz=\frac{1}{\sqrt{1-u^2}}du$ for the second term.When evaluating these two terms with these substitutions, the sum of these terms should be $\frac{\pi^2}{24}$.

On the other hand, let $0\leq u \leq \frac{1}{2} , u \leq v \leq 1-u$.

$$\int_{0}^{\frac{1}{2}}\int_{u}^{1-u} \frac{2}{1+v^2-u^2}dvdu=\int_{0}^{\frac{1}{2}}\frac{2\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)-2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$$

For the first term on the right hand side, let $z=2\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right),dz=\frac{-1}{\sqrt{1-u^2}}$ and for the second term, use $2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=2\sin^{-1}\left(u\right)$ and let $z=\sin^{-1}\left(u\right),dz=\frac{1}{\sqrt{1-u^2}}du$ for this.When evaluating these two terms with these substitutions, the sum of these terms should be $\frac{\pi^2}{24}$.


$$\iint_{T} \frac{2}{1+v^2-u^2}dvdu=\frac{2\pi^2}{24}=\frac{\pi^2}{12}$$ thus, $$I=\int_{0}^{1}\frac{\ln(x)}{1+x}dx=\frac{-\pi^2}{12}$$

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Convert the integral to the easier one via

\begin{align} \int_{0}^{1} \frac{\ln x}{x+1} dx &= \int_{0}^{1} \frac{\ln x}{1-x} dx - \int_{0}^{1} \frac{2x\ln x}{1-x^2} dx, \>\>\>\>\>t=x^2\\ &= \int_{0}^{1} \frac{\ln x}{1-x} dx -\frac12 \int_{0}^{1} \frac{\ln t}{1-t} dt \\ &= \frac12 \int_{0}^{1} \frac{\ln t}{1-t} dt=\frac12\cdot \frac{\pi^2}{6}=\frac{\pi^2}{12} \end{align}

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