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Evaluate the integral : $$\int_0^1 \frac{\log(1+x)}{x}dx$$ It is an improper integral & I tried it by substituting $\log(1+x)=z$ . But it does not open any way to evaluate it.

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    $\begingroup$ Taylor series for $\ln (1+x)$ around $x=0$ is the way to go $\endgroup$
    – Yuriy S
    Nov 25, 2016 at 8:35
  • $\begingroup$ The integral can be reduced to gamma functions by substitutions.Contour integral is also another choice $\endgroup$
    – vidyarthi
    Nov 25, 2016 at 8:40
  • $\begingroup$ @vidyarthi By which substitution it reduce to gamma function? $\endgroup$
    – Empty
    Nov 25, 2016 at 8:43
  • $\begingroup$ the result should be $$\frac{\pi^2}{12}$$ $\endgroup$ Nov 25, 2016 at 8:44
  • $\begingroup$ this must be duplicate here ... $\endgroup$
    – Math-fun
    Nov 25, 2016 at 8:48

2 Answers 2

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Since $\log(1+x)=\sum_{j=1}^{\infty}\frac{(-1)^{j+1}x^j}{j}$ we have $$I=\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{j}\int_0^1x^{j-1}=\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{j^2}=\frac{\pi^2}{12}$$

where the last summation can be calcualted using the well known summation: $\sum_{j=1}^{\infty}\frac{1}{j^2}=\frac{\pi^2}{6}$.

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Let $$I = \int^{1}_{0}\frac{\ln(1+x)}{x}dx = \int^{1}_{0}\ln(1+x)\cdot \frac{1}{x}dx$$

Using By parts, We get

$$I = \left[\ln(1+x)\cdot \ln x\right]^{1}_{0}-\int^{1}_{0}\frac{\ln x}{1+x}dx$$

So $$I=0-\int^{1}_{0}\sum^{\infty}_{n=0}(-x)^{n}\ln xdx = \sum^{\infty}_{n=0}(-1)^n\int^{1}_{0}\ln x \cdot x^ndx$$

Again, Using By parts, We get

$$I = -\sum^{\infty}_{n=0}\bigg[(-1)^n\bigg(\ln x \cdot \frac{x^{n+1}}{n+1}\bigg)^{1}_{0}-(-1)^n\int^{1}_{0}\frac{x^{n}}{(n+1)}dx\bigg]$$

So $$I = -\sum^{\infty}_{n=0}\frac{(-1)^n}{(n+1)^2} = \zeta(2)-\frac{\zeta(2)}{4} = \frac{3}{4}\zeta(2)$$

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