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Compute

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

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  • $\begingroup$ wolfram gives the answer as 0.272198 $\endgroup$ – user9413 Jun 9 '12 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 '12 at 8:25
  • $\begingroup$ @Chris: Even, i thought of that only :) $\endgroup$ – user9413 Jun 9 '12 at 8:26
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    $\begingroup$ 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 '13 at 17:01
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    $\begingroup$ Why don't you try to show some of your work.... $\endgroup$ – user210387 Jun 12 '15 at 10:29
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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 $$2I = \int_{0}^{\pi/4} \log(2) \ d\theta\Rightarrow I= \log(2) \cdot \frac{\pi}{8}$$

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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.

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  • $\begingroup$ Yes. It's very excellent and different method. $\endgroup$ – Prasad G Jun 9 '12 at 9:11
  • $\begingroup$ Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :) $\endgroup$ – qoqosz Jun 9 '12 at 9:13
  • $\begingroup$ @qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1. $\endgroup$ – user9413 Jun 9 '12 at 9:25
  • $\begingroup$ @Mark Hurd: right. $\endgroup$ – user 1357113 Jun 9 '12 at 11:49
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    $\begingroup$ @Astrobleme in case you are still wondering, it is partial fractions! $\endgroup$ – Vincent Nov 27 '17 at 14:01
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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.

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    $\begingroup$ That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1) $\endgroup$ – user 1357113 Jun 4 '13 at 21:09
  • $\begingroup$ How is the last expression any easier to computre than what we started with ? $\endgroup$ – CivilSigma Sep 6 '15 at 19:58
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    $\begingroup$ @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. $\endgroup$ – Barry Cipra Sep 13 '15 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. $\endgroup$ – Gaurang Tandon Mar 10 '18 at 9:50
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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 $$ \begin{align} \int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x &=\frac12\int_0^1\frac{\log(2)}{1+y^2}\mathrm{d}y\\ &=\frac\pi8\log(2)\tag{4} \end{align} $$

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  • $\begingroup$ I excuse Mr robjohn. I did'nt see your answer. $\endgroup$ – Boulid Jun 4 '13 at 21:09
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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

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    $\begingroup$ @Unoqualunque: The link was really helpful. thanks $\endgroup$ – user9413 Jun 9 '12 at 10:01
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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 :$$\boxed{\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi\ln(2)}{8}}$$

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  • $\begingroup$ How did you see this magical decomposition: $\frac{x}{(1+xy)(1+x^2)} = \frac{1}{1+y^2} \left( \frac{-y}{1+xy} + \frac{x+y}{1+x^2} \right)$ ? $\endgroup$ – Sandeep Silwal Mar 17 '17 at 20:37
  • $\begingroup$ @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 '18 at 9:54
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you can use the residu theory to calculate the integral of ln|1+x|/(1+x^2) between - inf and + inf by putting f(z)=ln|1+z|/(1+z^2) amd we obtain (Pi/2)*ln(2) then int(ln|1+x|/(1+x^2),x=-inf..inf)= int(same function,x=-inf..-1)+int(same function,x=-1..0)+int(same function,x=0..1)+int(same function,x=1..inf) then by putting t=-x in the first and second integrals we obtain: int(ln|1-x^2|/(1+x^2),x=0..1)+int(ln|1-x^2|/(1+x^2),x=1..inf)=(Pi/2)*ln(2) then by putting t=1/x in the second integral we obtain: int(ln|1-x^2|/(1+x^2),x=0..1)-int(ln(x)/(1+x^2),x=0..1)=(Pi/4)*ln(2) and we put x=(1-t)/(1+t) in the first integral we obtain after calculus : int(ln(1+x)/(1+x^2),x=0..1)=(Pi/8)*ln(2)

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    $\begingroup$ It's almost impossible to read this. You should find and read an introduction to MathJax, like this. $\endgroup$ – Henrik Jul 9 '16 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 '16 at 19:48
  • $\begingroup$ why it's impossible ? 😳 $\endgroup$ – Michel Jul 30 '16 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 '18 at 4:18

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