A calculus problem by Aman Dubey

$\begin{aligned} I &= \int_1^\infty \frac {\ln x}{x(x-1)} dx & \small \color{#3D99F6} \text{Let }x=e^u, \ dx = e^u\ du \\&= \int_0^\infty \frac {u e^u}{e^u(e^u-1)} du \\ &= \int_0^\infty \frac {u}{e^u-1} du & \small \color{#3D99F6} \text{Using the identity below} \\ \zeta (s) &= \frac 1{\Gamma(s)} \int_0^\infty \frac {u^{s-1}}{e^u-1} du \\ \implies I &= \zeta(2) = \frac {\pi^2}6 \approx \boxed{1.645} \end{aligned}$

Notations:

• $\zeta (\cdot)$ denotes the Riemann zeta function.
• $\Gamma (\cdot)$ denotes the gamma function. $\Gamma(2)=1!=1$ .

Let $I = \int_{1}^{\infty} \frac{ln(x)}{x(x-1)}$

$\frac{ln(x)}{x(x-1)}=\frac{ln(x)}{x^{2}(1-\frac{1}{x})}$

$\frac{ln(x)}{x^{2}}(1+\frac{1}{x}+\frac{1}{x^2}+...)$ Since $-1< \frac{1}{x} <1$

Using Substitution $x = e^{t }$ and further solving we can show that $\int_{1}^{\infty}x^{-n}{ln(x)} =\frac{1}{(n-1)^2}$

Using this result in above expression we get $I = \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...=\zeta(2)=\frac{\pi^2}{6}$

Simply do it like this...

Making change of variables , x = 1/y , then again make a change of variable as , y = 1-y , then the integral become ,

    int(0 to 1) (- ln(1-x))dx/x

I = (- sum(n=1 to infinity)) (1/n) int(0 to 1) (-x^(n-1)) dx

= (sum(n=1 to infinity))(1/n^2)

I = zeta(2)

(OR)

 I = ((pi)^2)/6

              (Q.E.D)