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Problem: Evaluate:

$$\displaystyle I=\int _{ 0 }^{ 1 }{ \ln\bigg(\frac { 1+x }{ 1-x } \bigg)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }$$

On Lucian Sir's advice, I substituted $x=\cos(\theta)$. Thus, the Integral becomes $$\int_0^{\pi/2} \ln\bigg(\dfrac{1+\cos(\theta)}{1-\cos(\theta)}\bigg)\dfrac{1}{\cos(\theta)}d\theta$$ $$=\int_0^{\pi/2} \ln\bigg(\cot^2\dfrac{\theta}{2}\bigg)\dfrac{1}{\cos(\theta)}d\theta$$ Unfortunately I'm stuck now. I would be indeed grateful if somebody could assist me in solving this Integral. Thanks very much in advance! $$$$Note: My trigonometry is quite poor and so I may have overlooked some glaring facts. Also, for the same reason, if possible, I would prefer a solution using Complex Numbers instead of Trigonometry (as long as the method with Complex Numbers is shorter). Many thanks once again!

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  • $\begingroup$ You should have tried $x=\cos2t$, since there are formulas $\dfrac{1\pm\cos2t}2$ $\endgroup$ – Lucian Jun 22 '15 at 6:08
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Let $\frac { 1+x }{ 1-x }=e^u$ to have \begin{align} I&=\int _{ 0 }^{ 1 }{ \ln\bigg(\frac { 1+x }{ 1-x } \bigg)\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }\\ &=\int _{ 0 }^{ \infty }\frac{u}{2\sinh\frac u2}du\\ &=4\int _{ 0 }^{ \infty }\frac{u}{e^u-e^{-u}}du\\ &=-4\int _{ 0 }^{ 1 }\frac{\log u}{1-u^2}du\\ \end{align}

The rest (binomial stuff) is as Lucian mentioned.


Addendum:

Note that $\frac{1}{1-u^2}=\sum_{j=0}^{\infty}u^{2j}$, hence we need to find $I_j=\int_0^1u^{2j}\log u \,d u$: \begin{align} I_j&=\int_0^1u^{2j}\log u \,d u\\ &=\frac{2j\log u+\log u-1}{(1+2j)^2}u^{2j+1}\Big|_{0^{+}}^1\\ &=-\frac{1}{(2j+1)^2} \end{align}

Therefore \begin{align} \int _{ 0 }^{ 1 }\frac{\log u}{1-u^2}du&=\sum_{j=0}^{\infty}I_j\\ &=-\sum_{j=0}^{\infty}\frac{1}{(2j+1)^2}\\ &=\sum_{j=1}^{\infty}\frac{1}{(2j)^2}-\sum_{j=1}^{\infty}\frac{1}{j^2}\\ &=\frac14\frac{\pi^2}{6}-\frac{\pi^2}{6}\\ &=-\frac{\pi^2}{8} \end{align}

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  • $\begingroup$ Sir, could you please explain your last step? $\endgroup$ – Ishan Jun 22 '15 at 7:34
  • $\begingroup$ Sure, in the last step I used $e^{-u} \to u$. $\endgroup$ – Math-fun Jun 22 '15 at 7:43
  • $\begingroup$ Sir, please could you tell me how to integrate $$4\int^0_1 \dfrac{log(u)}{1-u^2} du?$$ I couldn't get the binomial series you were referring to. $\endgroup$ – Ishan Jun 22 '15 at 11:31
  • $\begingroup$ sure, see the addendum. $\endgroup$ – Math-fun Jun 22 '15 at 12:01
  • $\begingroup$ Thanks Sir. Sir, could you please tell me how you integrated $\int_0^1u^{2j}\log u \,d u$, and how $$-\sum_{j=0}^{\infty}\frac{1}{(2j+1)^2}$$ $$=\sum_{j=0}^{\infty}\frac{1}{(2j)^2}-\sum_{j=0}^{\infty}\frac{1}{j^2}?$$ $\endgroup$ – Ishan Jun 22 '15 at 12:17
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Hint: Let $x=\cos2t$ and use the formulas for $\dfrac{1+\cos2t}2=\cos^2t$ and $\dfrac{1-\cos2t}2=\sin^2t$,

then let $u=\tan t$ in conjunction with $\cos2t=2\cos^2t-1=\dfrac2{1+\tan^2t}-1$. Lastly, expand

the integrand into its own binomial series, and reverse the order of summation and integration.

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  • $\begingroup$ Sir I was trying to understand (and have so far failed) when we can change the order of summation and Integration.$$$$Sir, [over here](math.stackexchange.com/questions/83721/…) it is said that if $f_n(x) \ge 0,$$$\sum \int f_n(x) dx = \int \sum f_n(x) dx$$Also, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$, then$$\int \sum f_n =\sum \int f_n$$Sir, could you please help me understand the meaning of $f_n(x)?$ Could you also please tell me how to evaluate $ \int |f_n|?$ I've never Integrated the $modulus$ of a function before. $\endgroup$ – Ishan Jun 22 '15 at 12:45
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It is straightforward to perform a simple sub of $u=(1-x)/(1+x) \implies x=(u-1)/(u+1)$ and $dx = 2/(u+1)^2 du$ to get

$$\int_1^{\infty} du \, u^{-1/2} \frac{\log{u}}{u-1} = 2 \int_0^{\infty} dv \, \frac{\log{v}}{v^2-1}$$

The latter integral may be computed using the residue theorem. To do this, consider

$$\oint_C dz \frac{\log^2{z}}{z^2-1} $$

where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$, except we deform about the removable singularity at $z=1$ above and below the axis with semicircular detours of radius $\epsilon$. In the limit as $R \to \infty$ and $\epsilon \to 0$ we get that the contour integral is, by the residue theorem,

$$-i 4 \pi \int_0^{\infty} dx \, \frac{\log{x}}{x^2-1} + 4 \pi^2 PV \int_0^{\infty} \frac{dx}{x^2-1} +i 2 \pi^3 = i 2 \pi \frac{\log^2{e^{i \pi}}}{e^{i \pi}-1} = i \pi^3$$

Using the fact that the second integral on the left is zero (this can be verified by taking the limit in the Cauchy PV definition directly), we find that the integral we seek is

$$2 \int_0^{\infty} dv \, \frac{\log{v}}{v^2-1} = \frac{\pi^2}{2} $$

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  • $\begingroup$ Sir I was trying to understand (and have so far failed) when we can change the order of summation and Integration.$$$$Sir, [over here](math.stackexchange.com/questions/83721/…) it is said that if $f_n(x) \ge 0,$$$\sum \int f_n(x) dx=\int \sum f_n(x) dx$$And, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$, then$$\int \sum f_n =\sum \int f_n$$Sir, could you please help me understand the meaning of $f_n(x)?$ Could you also please tell me how to evaluate $ \int |f_n|?$ I've never Integrated the $modulus$ of a function before. $\endgroup$ – Ishan Jun 22 '15 at 12:49
  • $\begingroup$ @BetterWorld: maybe you should ask that as a question. It seems off-topic to ask that within a comment to my answer to a completely different question. $\endgroup$ – Ron Gordon Jun 22 '15 at 12:59
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Here is how I finally worked it out: $$\displaystyle I=\int _{ 0 }^{ 1 }{ \ln\bigg(\dfrac { 1+x }{ 1-x } \bigg)\dfrac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } } $$

Put $x=\cos(\theta)$ to get our integral as : $$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln\bigg({ \cot }^{ 2 }\bigg(\frac { \theta }{ 2 }\bigg) \bigg)\dfrac { d\theta }{ \cos\theta } } $$

$$\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln(\tan\frac { \theta }{ 2 } )\dfrac { d\theta }{ cos\theta } } $$

Put $\tan\bigg(\dfrac{\theta}{2}\bigg)=x$ to get our integral as :

$$\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } $$

Using the result $$\displaystyle \int _{ 0 }^{ 1 }{ \dfrac { \ln(t)dt }{ 1-{ t }^{ 2 } } } =\dfrac { -{ \pi }^{ 2 } }{ 8 } $$

$$I=\frac{{\pi}^{2}}{2}$$

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