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# int_0^1 lnx/(1+x)dx ?

Jun 22, 2018

${\int}_{0}^{1} \setminus \ln \frac{x}{1 + x} \setminus \mathrm{dx} = - {\pi}^{2} / 12$

#### Explanation:

We seek:

$I = {\int}_{0}^{1} \setminus \ln \frac{x}{1 + x} \setminus \mathrm{dx}$

We can apply Integration By Parts:

Let  { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/(1+x), => v,=ln(1+x) ) :}

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

We have:

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

$\therefore I = - {\int}_{0}^{1} \setminus \frac{\ln \left(1 + x\right)}{x} \setminus \mathrm{dx}$

The resulting integral, is cannot be expressed in terms of elementary functions, in fact by defining the polylogarithm function, $L {i}_{2} \left(x\right)$ we write:

$L {i}_{2} \left(- x\right) = - \int \setminus \frac{\ln \left(1 + x\right)}{x} \setminus \mathrm{dx}$

So that

$\therefore I = - {\left[L {i}_{2} \left(- x\right)\right]}_{0}^{1} = L {i}_{2} \left(- 1\right) - L {i}_{2} \left(0\right)$

And (fortunately) special values of $L {i}_{2} \left(x\right)$ are:

$L {i}_{2} \left(0\right) = 0$
$L {i}_{2} \left(- 1\right) = - {\pi}^{2} / 12$

$I = - {\pi}^{2} / 12$