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Evaluation of $$\int^{1}_{0}\frac{\ln^2(1+x)}{x}dx$$

$\bf{My\; Try::}$ Let $$I = \int^{1}_{0}\frac{\ln^2(1+x)}{x}dx = \int^{1}_{0}\ln^2(1+x)\cdot\frac{1}{x}dx$$

Using By parts, We get

$$I = \left[\ln^2(1+x)\cdot \ln(x)\right]^{1}_{0}-2\int^{1}_{0}\frac{\ln(1+x)\cdot \ln x}{1+x}dx$$

So we get $$I = -2\int^{1}_{0}\frac{\ln(1+x)\cdot \ln x}{1+x}dx$$

Now how can i solve above Integral , Help required, Thanks

juantheron
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  • According to Wolfram, its $\zeta(3) \over 4$ where $\zeta$ is the Riemann Zeta Function. Strange :/ – gowrath Nov 27 '16 at 10:06
  • Also sorry if this is a noob question but isn't $\ln(0)$ undefined? How did you get rid of it in the left term of your by parts? I haven't really verified it, just looking at your working. – gowrath Nov 27 '16 at 10:10
  • @juantheron Have a look here http://math.stackexchange.com/questions/795867/evaluation-of-int-01-frac-log21xx-dx – Olivier Oloa Nov 27 '16 at 10:14

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