piarepm

2022-01-03

Finding integral:
${\int }_{0}^{1}\frac{\mathrm{ln}\left(1+x\right)}{x}dx$

Paul Mitchell

Expert

Here is an elementary integration of $I={\int }_{0}^{1}\frac{\mathrm{ln}\left(1+x\right)}{x}dx$
$I=-\frac{1}{2}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+x\right)}{x}dx+\frac{3}{2}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+x\right)}{x}dx=-\frac{3}{2}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1-x+{x}^{2}\right)}{x}dx$
Let $J\left(a\right)={\int }_{0}^{1}\frac{\mathrm{ln}\left(1-2x\mathrm{sin}a+{x}^{2}\right)}{x}dx$
${J}^{\prime }\left(a\right)=-{\int }_{0}^{1}\frac{2\mathrm{cos}a}{{\left(x-\mathrm{sin}a\right)}^{2}+{\mathrm{cos}}^{2}a}dx=-\left(\frac{\pi }{2}+a\right)$
Then, width $J\left(0\right)={\int }_{0}^{1}\frac{\mathrm{ln}\left(1+{x}^{2}\right)}{x}dx=\frac{12}{I}$
$I=-\frac{32}{J}\left(\frac{\pi }{6}\right)=-\frac{32}{J\left(0\right)+{\int }_{0}^{\frac{\pi }{6}}{J}^{\prime }\left(a\right)da}=-\frac{34}{I}+{\frac{32}{\int }}_{0}^{\frac{\pi }{6}}\left(\frac{\pi }{2}+a\right)da$
$I=\frac{{\pi }^{2}}{12}$

sonorous9n

Expert

Hint. One may recall that

one may then divide by x and one is allowed to integrate termwise obtaining
${\int }_{0}^{1}\frac{\mathrm{ln}\left(1+x\right)}{x}dx=\sum _{n=1}^{\mathrm{\infty }}{\left(-1\right)}^{n-1}{\int }_{0}^{1}\frac{{x}^{n-1}}{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n-1}}{{n}^{2}}$
Can you take it from here?

user_27qwe

Expert

By integration by parts:
${\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)dx}{x}=-{\int }_{0}^{1}\frac{\mathrm{log}x}{1+x}dx=2{\int }_{0}^{1}\frac{-\mathrm{log}x}{1-{x}^{2}}dx-{\int }_{0}^{1}$
but since ${\int }_{0}^{1}\left(-\mathrm{log}x\right){x}^{k}dx=\frac{1}{\left(k+1{\right)}^{2}}$, by expanding $\frac{1}{1-{x}^{2}}$ and $\frac{1}{1-x}$ as geometric series we get:
${\int }_{0}^{1}\frac{\mathrm{log}\left(1+x\right)dx}{x}=2\sum _{kodd}\frac{1}{{k}^{2}}-\sum _{k\ge 1}\frac{1}{{k}^{2}}=\sum _{k\ge 1}\frac{1}{{k}^{2}}$
$=\frac{{\pi }^{2}}{12}$

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