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

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wolfram gives the answer as 0.272198 –  user9413 Jun 9 '12 at 7:57
Mathematica gives the answer $\frac{\pi}{8} \log 2$. But I do not know how it computed this number... –  Siminore Jun 9 '12 at 8:25
@Chris: Even, i thought of that only :) –  user9413 Jun 9 '12 at 8:26
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 –  CBenni Mar 20 '13 at 17:01
@CBenni: are you sure you saw all my questions? How about this? math.stackexchange.com/questions/299801/… or math.stackexchange.com/questions/289172/… –  Chris's sis Mar 20 '13 at 17:29

5 Answers 5

up vote 57 down vote accepted

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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Yes. It's very excellent and different method. –  Prasad G Jun 9 '12 at 9:11
Thanks, but it needs much more calculations than the soulution proposed by @Chandrasekhar :) –  qoqosz Jun 9 '12 at 9:13
@qoqosz Nice way of seeing how differentiation under the integral sign works. Thanks +1. –  user9413 Jun 9 '12 at 9:25
@Mark Hurd: right. –  Chris's sis Jun 9 '12 at 11:49

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

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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That substitution is very powerful, and used it when I met ugly integrals. I didn't use it here (unfortunately). Great solution! (+1) –  Chris's sis Jun 4 '13 at 21:09

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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@Unoqualunque: The link was really helpful. thanks –  user9413 Jun 9 '12 at 10:01

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