98
$\begingroup$

Compute

$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$

$\endgroup$
7
  • $\begingroup$ wolfram gives the answer as 0.272198 $\endgroup$
    – user9413
    Jun 9, 2012 at 7:57
  • $\begingroup$ Mathematica gives the answer $\frac{\pi}{8} \log 2$. But I do not know how it computed this number... $\endgroup$
    – Siminore
    Jun 9, 2012 at 8:25
  • $\begingroup$ @Chris: Even, i thought of that only :) $\endgroup$
    – user9413
    Jun 9, 2012 at 8:26
  • 14
    $\begingroup$ Why dont you try and solve these Integrals yourself. Browsing through your most recent questions, you have had this type of question almost exclusively. Other users get downvoted for this. I see no reason not to hint the same to you and at least show some effort. -1 $\endgroup$
    – CBenni
    Mar 20, 2013 at 17:01
  • 6
    $\begingroup$ Why don't you try to show some of your work.... $\endgroup$
    – user210387
    Jun 12, 2015 at 10:29

8 Answers 8

122
$\begingroup$

Put $x = \tan\theta$, then your integral transforms to $$I= \int_{0}^{\pi/4} \log(1+\tan\theta) \ d\theta. \tag{1}$$

Now using the property that $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx,$$ we have $$I = \int_{0}^{\pi/4} \log\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} \log\biggl(\frac{2}{1+\tan\theta} \biggr) \ d\theta.\tag{2}$$

Adding $(1)$ and $(2)$ we get $$2I = \int_{0}^{\pi/4} \log(2) \ d\theta\Rightarrow I= \log(2) \cdot \frac{\pi}{8}.$$

$\endgroup$
0
65
$\begingroup$

Consider: $$I(a) = \int_0^1 \frac{\ln (1+ax)}{1+x^2} \, dx$$ than, the derivative $I'$ is equal: $$I'(a) = \int_0^1 \frac{x}{(1+ax)(1+x^2)} \, dx = \frac{2 a \arctan x - 2\ln (1+a x) + \ln (1+x^2)}{2(1+a^2)} \Big|_0^1\\ = \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2}$$ Hence: $$I(1) = \int_0^1 \left( \frac{\pi a + 2 \ln 2}{4(1+a^2)} - \frac{\ln (1+a)}{1+a^2} \right) \, da \\ 2 I(1) = \int_0^1 \frac{\pi a + 2 \ln 2}{4(1+a^2)} \, da = \frac{\pi}{4} \ln 2$$ Divide both sides by $2$ and you're done.

$\endgroup$
6
  • $\begingroup$ Yes. It's very excellent and different method. $\endgroup$
    – Prasad G
    Jun 9, 2012 at 9:11
  • $\begingroup$ Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :) $\endgroup$
    – qoqosz
    Jun 9, 2012 at 9:13
  • $\begingroup$ @qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1. $\endgroup$
    – user9413
    Jun 9, 2012 at 9:25
  • $\begingroup$ @Mark Hurd: right. $\endgroup$ Jun 9, 2012 at 11:49
  • 1
    $\begingroup$ @Astrobleme in case you are still wondering, it is partial fractions! $\endgroup$
    – Vincent
    Nov 27, 2017 at 14:01
52
$\begingroup$

Good evening, I've got another method by putting $x=(1-t)/(1+t)$, we obtain $$\int_0^1\frac{\ln (x+1)}{x^2+1}dx=\int_1^0\frac{\ln\frac{2}{1+t}}{\left(\frac{1-t}{1+t}\right)^2+1}\cdot\left\{-\frac{2}{(1+t)^2}\right\}dt =\int_0^1\frac{\ln 2-\ln (1+t)}{t^2+1}\ dt.$$ You can finish easily.

$\endgroup$
4
  • 8
    $\begingroup$ That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1) $\endgroup$ Jun 4, 2013 at 21:09
  • $\begingroup$ How is the last expression any easier to computre than what we started with ? $\endgroup$
    – CivilSigma
    Sep 6, 2015 at 19:58
  • 2
    $\begingroup$ @CivilSigma, if you look at the first and last expressions, you see you have an equation of the form $A=B-A$, where $A$ is the integral you're trying to evaluate and $B$ is an integral that's easy to evaluate. $\endgroup$ Sep 13, 2015 at 15:10
  • $\begingroup$ Great solution (+1) but how does one even think of that substitution in such situation? I mean, this is a clearly a "let's try it out and hope it works method". The accepted solution is more straightforward. $\endgroup$ Mar 10, 2018 at 9:50
39
$\begingroup$

If $(1+x)(1+y)=2$, then $$\begin{align} x&=\frac{1-y}{1+y}\\ 1+x^2&=2\frac{1+y^2}{(1+y)^2}\\ \frac{1+x^2}{1+x}&=\frac{1+y^2}{1+y} \end{align}\tag{1} $$ and since $(1+y)\,\mathrm{d}x+(1+x)\,\mathrm{d}y=0$ we get $$ \frac{\mathrm{d}x}{1+x^2}=-\frac{\mathrm{d}y}{1+y^2}\tag{2} $$ Therefore, $$ \begin{align} \int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x &=\int_0^1\frac{\log(2)-\log(1+y)}{1+y^2}\mathrm{d}y\tag{3} \end{align} $$ Adding the left side to both sides and dividing by $2$ yields $$ \begin{align} \int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x &=\frac12\int_0^1\frac{\log(2)}{1+y^2}\mathrm{d}y\\ &=\frac\pi8\log(2)\tag{4} \end{align} $$

$\endgroup$
4
  • $\begingroup$ I excuse Mr robjohn. I did'nt see your answer. $\endgroup$
    – Boulid
    Jun 4, 2013 at 21:09
  • $\begingroup$ I love your answers! $\endgroup$
    – user730361
    Jun 27, 2021 at 15:08
  • $\begingroup$ $(1+x)(1+y)=2$ mysteriously appears in the beginning. Did you do reverse-engineering? $\endgroup$
    – user271232
    Sep 6, 2021 at 15:12
  • $\begingroup$ Actually, I have used $(1+x)(1+y)=2$ many times. It means that $\tan^{-1}(x)+\tan^{-1}(y)=\frac\pi4$ and so, is useful to try in situations with $\frac1{1+x^2}$ in the denominator. The arctangent relation leads immediately to $(2)$. $\endgroup$
    – robjohn
    Sep 6, 2021 at 16:36
25
$\begingroup$

Start with $$\begin{align*} \int_0^{\pi/4} \ln(1+\tan x)dx &= \int_0^{\pi/4} \ln(\sin x+\cos x)dx - \int_0^{\pi/4} \ln(\cos x)dx \\ &= \int_0^{\pi/4} \ln\left(\cos(x-\frac{\pi}4)\right)dx +\int_0^{\pi/4} \ln(\sqrt 2)dx - \int_0^{\pi/4} \ln(\cos x)dx. \end{align*}$$ Now change $\pi/4-x=t$ in the first integral: $$=\int_0^{\pi/4} \ln(\cos t) dt +\int_0^{\pi/4} \ln(\sqrt 2)dx - \int_0^{\pi/4} \ln(\cos x)dx$$ and the result follows. Changing $x=\tan u$ in the first integral yields your integral. As far as I know these are said Bertrand's integrals.

@Chandrasehkar: see here http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf

$\endgroup$
1
  • 1
    $\begingroup$ @Unoqualunque: The link was really helpful. thanks $\endgroup$
    – user9413
    Jun 9, 2012 at 10:01
14
$\begingroup$

Let us consider $$A=\iint_{[0,1]^2}\frac{x}{(1+xy)(1+x^2)}dx dy$$ By Fubini's theorem, we have : $$A=\int_0^1\left[\frac{1}{1+x^2}\int_0^1\frac{x\,dy}{1+xy}\right]dx=\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$ and$$A=\int_0^1\left[\int_0^1\frac{x}{(1+xy)(1+x^2)}dx\right]dy$$But$$\frac{x}{(1+xy)(1+x^2)}=\frac{1}{1+y^2}\left(\frac{-y}{1+xy}+\frac{x+y}{1+x^2}\right)$$and therefore$$A=\int_0^1\frac{1}{1+y^2}\left(-\ln(1+y)+\frac{\ln(2)}{2}+\frac{\pi y}{y}\right)dy=-A+\frac{\pi\ln(2)}{4}$$Finally :$$\boxed{\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi\ln(2)}{8}}$$

$\endgroup$
2
  • $\begingroup$ How did you see this magical decomposition: $\frac{x}{(1+xy)(1+x^2)} = \frac{1}{1+y^2} \left( \frac{-y}{1+xy} + \frac{x+y}{1+x^2} \right)$ ? $\endgroup$ Mar 17, 2017 at 20:37
  • 1
    $\begingroup$ @SandeepSilwal : it is merely a partial fraction decomposition in the real domain (where the main variable is considered to be $x$). No magic at all in it ;-) The interesting part is that this transformation leads to an explicit calculation. $\endgroup$
    – Adren
    Sep 22, 2018 at 9:54
1
$\begingroup$

Note \begin{align} &\int_0^1\frac{\ln (x+1)}{x^2+1}dx =\int_0^1\underset{x\to\frac{1-x}{1+x}}{\frac{\ln \frac{x+1}{\sqrt{x^2+1}}}{x^2+1}}dx + \int_0^1\frac{\ln \sqrt{x^2+1}}{x^2+1}dx\\ &= \int_0^1\frac{\ln \frac{\sqrt2}{\sqrt{x^2+1}}}{x^2+1}dx + \int_0^1\frac{\ln \sqrt{x^2+1}}{x^2+1}dx = \frac{\ln2}2\int_0^1\frac{dx}{x^2+1}=\frac\pi8\ln2 \end{align}

$\endgroup$
1
  • $\begingroup$ This is the same method posted by both @boulid and Robert Johnson more than 7 years before your post herein. $\endgroup$
    – Mark Viola
    May 20, 2021 at 20:21
-1
$\begingroup$

You can use the residue theory to calculate
$$J=\int_{-\infty}^\infty \frac{\ln|1+x|}{(1+x^2)} \, dx $$
by putting $$f(z)=\frac{\ln(1+z)}{(1+z^2)}$$ and we obtain $$J = 2\pi i\frac{\ln(1+i)}{(i+i)}=\pi\ln\left(\sqrt2e^{\frac14\pi i}\right)=\pi\left(\frac12\ln2+\frac14\pi i\right)=\frac{\pi}{2}\ln 2+\frac{\pi^2i}4$$

$$J=\int_{-\infty}^{-1}\frac{\ln(1+x)}{1+x^2} \, dx +\int_{-1}^0 \frac{\ln(1+x)}{1+x^2} \, dx +\int_0^1\frac{\ln(1+x)}{1+x^2} \, dx +\int_1^\infty \frac{\ln(1+x)}{1+x^2} \, dx$$

Then by putting $(t-1)e^{\pi i}=1+x$ in the first integral we obtain: $$\int_{-\infty}^{-1}\frac{\ln(1+x)}{1+x^2}dx=\int_1^{\infty}\frac{\pi i+\ln(t-1)}{t^2+1}dt=\frac{\pi^2i}4+\int_1^{\infty}\frac{\ln(t-1)}{t^2+1}dt$$

Then by putting $t=-x$ in the second integral we obtain: $$\int_0^1\frac{\ln(1-x^2)}{1+x^2}dx+\int_1^{\infty}\frac{\ln(x^2-1)}{1+x^2}dx=\frac{\pi}2\ln2$$

then by putting $t=1/x$ in the second integral we obtain: $$\int_0^1\frac{\ln(1-x^2)}{1+x^2}dx-\int_0^1\frac{\ln x}{1+x^2}dx=\frac{\pi}4\ln2$$

and we put $x=(1-t)/(1+t)$ in the first integral we obtain after calculus : $$\int_0^1 \frac{\ln \, (1+x)}{1+x^2}= \frac{\pi}{8} \ln 2$$

$\endgroup$
11
  • 12
    $\begingroup$ It's almost impossible to read this. You should find and read an introduction to MathJax, like this. $\endgroup$ Jul 9, 2016 at 17:39
  • $\begingroup$ In posting an Answer to a (by now four year old) Question having already an Accepted Answer, it is helpful to your Readers to highlight what new information you provide. $\endgroup$
    – hardmath
    Jul 9, 2016 at 19:48
  • $\begingroup$ why it's impossible ? 😳 $\endgroup$
    – Michel
    Jul 30, 2016 at 9:15
  • $\begingroup$ @Michel I mean, compared to the accepted answer, which one is much easier to read? Also, you spelled "residue" wrong. $\endgroup$
    – Frank W
    May 20, 2018 at 4:18
  • 1
    $\begingroup$ @mjw For $-1< x<0$, $0<(1+x)<1$ so the absolute value bars do nothing. Then he has $$\begin{align}\int_{-1}^0\frac{\ln(1+x)}{1+x^2}dx+\int_0^1\frac{\ln(1+x)}{1+x^2}dx&=\int_0^1\frac{\ln(1-t)}{1+t^2}dt+\int_0^1\frac{\ln(1+t)}{1+t^2}dt\\ &=\int_0^1\frac{\ln(1-x^2)}{1+x^2}\end{align}$$ Where he let $x=-t$ in the first integral and $x=t$ in the second. $\endgroup$ Mar 13, 2020 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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