Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to evaluate this integral: $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$.

I've tried $t=\log(x+1)$, $t=x+1$, but to no avail. I've noticed that:

$\int_0^1 \frac{\log(x+1)}{1+x^2} dx = \int_0^1\log(x+1) \arctan'(x)dx =\left. \log(x+1)\arctan(x) \right|_{x=0}^{x=1} - \int_0^1\frac{\arctan(x)}{x+1}dx$

But can't get further than this.

Any help is appreciated, thank you.

share|improve this question
A related problem. –  Mhenni Benghorbal Mar 12 '13 at 3:43

6 Answers 6

up vote 17 down vote accepted

Going a little round-about way. Consider, for $ s \geqslant 0$, a parametric modification of the integral at hand: $$ \mathcal{I}(s) = \int_0^1 \frac{\log(1+s x)}{1+x^2} \mathrm{d} x $$ The goal is to determine $\mathcal{I}(1)$. Now: $$ \begin{eqnarray} \mathcal{I}(1) &=& \int_0^1 \mathcal{I}^\prime(s) \mathrm{d} s = \int_0^1 \left( \int_0^1 \frac{x}{1+s x} \frac{\mathrm{d} x}{1+x^2} \right) \mathrm{d} s \\ &=& \int_0^1 \left.\left( - \frac{1}{1+s^2} \log(1+s x) + \frac{s}{1+s^2} \arctan(x) + \frac{1}{2} \frac{\log(1+x^2)}{1+x^2} \right) \right|_{x=0}^{x=1} \mathrm{d} s \\ &=& \int_0^1 \left( \color\green{ -\frac{\log(1+s)}{1+s^2}} + \frac{1}{4} \frac{\pi s+\log(4)}{1+s^2}\right) \mathrm{d} s = - \mathcal{I}(1) + \frac{1}{4} \pi \log(2) \end{eqnarray} $$ Hence $$ \mathcal{I}(1) = \frac{\pi}{8} \log(2) $$

share|improve this answer
This seems like a great solution, but I'll need to take some time to properly "digest" it (I've never seen this technique so far, and I want to make sure I properly understand what happens) –  Gabi Purcaru Jul 18 '12 at 8:58

Here's a solution that uses simpler tools (or at least tools that I'm more familiar with):

$I =\int_0^1\frac{\log(1+x)}{1+x^2}$. We change $x$ into $x=\tan(t)$. Then $t=\arctan{x}$, $dt=\frac{1}{1+x^2}dx$, and the integral becomes:

$I = \int_0^{\frac{\pi}{4}}\log(1+\tan(t))dt$. Now $s = \frac{\pi}{4}-t$, $ds=-dt$, and the integral becomes:

$I = -\int_{\frac{\pi}{4}}^0 \log(1+\tan(\frac{\pi}{4}-s))ds = \int_0^{\frac{\pi}{4}} \log(1+\tan(\frac{\pi}{4}-s))ds$

Using $\tan(a+b) = \frac{\tan a - \tan b}{1 + \tan(a)\tan(b)}$, we have

$I = \int_0^{\frac{\pi}{4}} \log(1+\frac{1 - \tan s}{1 + \tan s}) ds = \int_0^{\frac{\pi}{4}}\log(\frac{2}{1+\tan s}) ds = \int_0^{\frac{\pi}{4}} (\log(2) - \log(1+\tan s)) ds = \int_0^{\frac{\pi}{4}}\log(2)ds - I = \frac{\pi}{4}\log(2) - I$.

So $I = \frac{\pi}{4}\log(2) - I$, hence $I = \frac{\pi}{8}\log(2)$.

share|improve this answer
+1 No reason to make it overly complicated ;) –  Pedro Tamaroff Jul 18 '12 at 11:42


$$I(a)=\int_0^1 \frac{\log(1+ax)}{1+x^2}dx$$

Differentiate it, to get

$$I'(a)=\int_0^1 \frac{x}{(1+ax)(1+x^2)}dx$$

Integrate that rational function, then integrate w.r.t. $a$ and find $I(a=1)$.

As Theorem suggested, you can also do the following:

$$\int_0^1 \frac{\log(1+x)}{1+x^2}dx$$

$$\int_0^1 \left(\int_0^x \frac{1}{1+y}dy\right)\frac{1}{1+x^2}dx$$

$$\int_0^1 \int_0^x \frac{1}{1+y}dy\frac{1}{1+x^2}dx$$

Now make a change of variables $y=ux$ in the inner integral:

$$\tag 1 \int_0^1 \int_0^1 \frac{x}{1+ux}\frac{1}{1+x^2}dudx$$

Now partial fractions:

$${x \over {1 + xu}}{1 \over {1 + {x^2}}} = {1 \over {1 + {u^2}}}{x \over {1 + {x^2}}} + {u \over {1 + {u^2}}}{1 \over {1 + {x^2}}} - {u \over {1 + {u^2}}}{1 \over {1 + xu}}$$

Now, integrating the first two terms, which account to the same$^1$, gives that you integral is

$$\mathcal I=\frac \pi 4\log 2-\int_0^1\int_0^1 \frac u {1+u^2}\frac{1}{1+xu}dxdu$$

Now, the latter integral is just our original integral, due to symmetry, as you see in $(1)$

This means that $$\mathcal I =\frac \pi 8 \log 2$$

as desired. $1$: symmetry, once again.

See here for a similar integral and its solution with double integrals.

Some insight about the two methods considered:

Note that as Sasha is showing

$$I(1)=\int_0^1 I'(a)da=\int_0^1 \int_0^1\frac{x}{(1+ax)(1+x^2)}dxda$$ which is exaclty what we got in the second option

$$I=\int_0^1\int_0^1 \frac{1 }{1+mx}\frac{x}{1+x^2}dxdm$$

This means any way you find to solve any of the two will indeed solve the other.

share|improve this answer
For other examples of the technique, please see this. –  André Nicolas Jul 17 '12 at 20:31
@André Nicolas : Is it also possible to convert it into double integral and solve it ? –  Theorem Jul 17 '12 at 20:38
@Theorem That should also work, and I also reccommend it. It should be an eigth of $\pi \log 2$. –  Pedro Tamaroff Jul 17 '12 at 20:38
@PeterTamaroff : nice . –  Theorem Jul 17 '12 at 20:48

I solved this integral a couple of years ago and I had this solution typed out in $\LaTeX$ already. The solution is not conventional, so I think it's worth sharing!

First, substitute the series $$\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}.$$

This series is uniformly convergent on $[0,1]$, so we can interchange the sum and the integral. We get

$$I=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 \frac{x^n}{1+x^2}\: \mbox{d}x. = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}C_n$$

where $$C_n=\int_0^1 \frac{x^n}{1+x^2} \mbox{d}x.$$

Now since $$x^{n-2}-\frac{x^{n-2}}{1+x^2}=\frac{x^n}{1+x^2},$$

we have, integrating this equation on $[0,1]$, $$\frac{1}{n-1}-C_{n-2}=C_n.$$ Hence we have a recurrence relation for the $C_n$'s. Let's see what this gives. We have $$C_0=\int_0^1 \frac{ \mbox{d}x}{1+x^2} = \arctan(1) = \frac{\pi}{4},$$ and

$$C_1=\int_0^1 \frac{x\ \mbox{d}x}{1+x^2} = \frac{1}{2}\log2.$$

Now using the recurrence we find





$C_4 = -1+\frac{1}{3} +\frac{\pi}{4}$

$C_5=-\frac{1}{2}+\frac {1}{4} +\frac{1}{2}\log 2$

$C_6=1-\frac{1}{3}+\frac{1}{5} - \frac{\pi}{4}$


Now if we define

$$A_n = \sum_{k=1}^n\frac{(-1)^{k-1}}{2k},$$ $$B_n = \sum_{k=1}^n\frac{(-1)^{k-1}}{2k-1},$$ it is easy to see by induction that $$C_{2n} = (-1)^n\left(\frac{\pi}{4}-B_n\right)$$ and

$$C_{2n-1} = (-1)^{n-1}\left(\frac{\log 2}{2} -A_{n-1}\right).$$

Notice that $A_n \rightarrow \frac{1}{2}\log2$ as $n\rightarrow \infty$ [recall the series expansion of $\log(1+x)$], and that $B_n \rightarrow \frac{\pi}{4}$ as $n\to \infty$ [recall the series expansion of $\arctan x$, which you can get by integrating $(1+x^2)^{-1}$].

Let us examine the partial sums

$$I_{2N}=\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n}C_n.$$

We can split this into the odd-labeled terms and the even-labeled terms, as

$$I_{2N}= \sum_{n=1}^{N} \frac{C_{2n-1}}{2n-1} - \sum_{n=1}^{N} \frac{C_{2n}}{2n}.$$

Now we can substitute the values of $C_{2n}$ and $C_{2n-1}$ obtained before. First, let us look at the sum of the even-labeled terms. We have

$$\sum_{n=1}^{N} \frac{C_{2n}}{2n} =\sum_{n=1}^{N} \frac{ (-1)^{n}}{2n}\left(\frac{\pi}{4}-B_n\right)=-\frac{\pi}{4}A_N +\sum_{n=1}^N\frac{(-1)^{n-1}}{2n}B_n $$

Now let us recall Cauchy's partial summation formula. For sequences $\{a_n\}, \{b_n\}$, we denote $\{\Delta a_n\}$ the sequence of forward differences $\Delta a_n = a_{n+1}-a_n$. Then we have

$$\sum_{n=1}^N b_n \Delta a_{n-1} = b_Na_N - b_1a_0 -\sum_{n=1}^{N-1} \Delta b_n a_n.$$

Remark that $\Delta A_{n-1} = \frac{(-1)^{n-1}}{2n}$. Also, $A_0=0$. Hence,

$$-\frac{\pi}{4}A_N +\sum_{n=1}^N\Delta A_{n-1}B_n = -\frac{\pi}{4}A_N +B_NA_N -\sum_{n=1}^{N-1}\Delta B_n A_n.$$

(Call this sum (1)).

Now we go back to the sum of the odd-labeled terms. We have, in a similar fashion,

$$\sum_{n=1}^{N} \frac{C_{2n-1}}{2n-1} = \sum_{n=1}^{N} \frac{(-1)^{n-1}}{2n-1}\left(\frac{\log 2}{2} - A_{n-1}\right) = \frac{\log 2}{2} B_N - \sum_{n=1}^N \Delta B_{n-1}A_{n-1}$$ $$= \frac{\log 2}{2} B_N - \sum_{n=0}^{N-1} \Delta B_{n}A_{n}.$$ (Call this sum (2)).

Now, subtracting (1) from (2), we get

$$I_{2N} = \frac{\log 2}{2}B_N + \frac{\pi}{4}A_N - B_NA_N +\Delta B_0 A_0$$ $$= \frac{\log 2}{2}B_N + \frac{\pi}{4}A_N - B_NA_N$$

Now as $N\to \infty$, since $A_N \to \frac{1}{2}\log 2$ and $B_N \to \frac{\pi}{4}$, we have

$$I = \int_0^1\frac{\log(1+x)}{1+x^2}\: \mbox{d}x = \lim_{N\to \infty} I_{2N} = \frac{\pi \log 2}{8}.$$

share|improve this answer

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#00f}{\large\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x}& =\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta \\[3mm]&=\half\bracks{% \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta + \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{{\pi \over 4} - \theta}}\,\dd\theta} \\[3mm]&=\half\bracks{% \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta + \int_{0}^{\pi/4}\ln\pars{% 1 + {1 - \tan\pars{\theta} \over 1 + \tan\pars{\theta}}}\,\dd\theta} \\[3mm]&=\half\bracks{% \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta + \int_{0}^{\pi/4}\ln\pars{2\over 1 + \tan\pars{\theta}}\,\dd\theta} \\[3mm]&=\half\int_{0}^{\pi/4}\ln\pars{2}\,\dd\theta =\color{#00f}{\large{1 \over 8}\,\pi\ln\pars{2}} \end{align}

share|improve this answer
+1 For using simple methods! –  Integrals Apr 30 at 13:05

For these integrals are very useful substitutions homograph type.

To note $\displaystyle I(a) = \int^{a}_{0}\frac{\ln(x+a)}{x^{2}+a^2}dx.$

Using the substitution $x=\dfrac{-at+a^2}{t+a} = u(t)$ with $u'(t)=-\dfrac{2a^2}{(t+a)^2} $ we find $$\begin{align*}I(a) &= \int^{0}_{a}\frac{\ln\left(\frac{-at+a^2}{t+a}+a\right)}{\left(\frac{-at+a^2}{t+a}\right)^2+a^2}\left(- \frac{2a^2}{(t+a)^2}\right)dt=\\ &= \int^{a}_{0}\frac{\ln 2a^2 - \ln(t+a)}{t^2+a^2}dt = \\ &=\int^{a}_{0}\frac{\ln 2a^2}{t^2+a^2}dt -\int^{a}_{0}\frac{\ln(t+a)}{t^2+a^2}dt=\\ & =\frac{\ln2a^{2}}{a} \arctan a- I(a).\end{align*}$$

And finally $$I(a) = \frac{\ln2a^{2}}{2a}\arctan a.$$ For $ a= 1$ we obtain $I(1) = \dfrac{\pi}{8}\ln2.$

See also: http://www.recreatiimatematice.ro/arhiva/articole/RM12011DICU.pdf

share|improve this answer
I edited your answer. If you are interesting in $TeX$ code you can see it clicking on edit button. For some basic information about writing math at this site see e.g. here, here, here and here. And nice answer! (+1) –  Cortizol Jun 29 '13 at 20:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.