Please note: Harsh is not harsh

Calculus Level 4

$\large\int_{1}^{\infty} \dfrac{\ln^3(x)}{x^2-x} \, dx$

If the above integral can be expressed as $\dfrac{A\pi^{B}}{C}$ for some coprime integers $A,B,C$ , compute $A+B+C$ .

This problem is original and is dedicated to Harsh S

The answer is 20.

2 solutions

Nihar Mahajan
Feb 23, 2016

We substitute $x=e^y$ , then $dx=e^y \ dy$ and $\ln(x)=y$ . As $x=1 \ , \ y=0$ and as $x\rightarrow \infty \ , \ y\rightarrow \infty$ as well and the integral converts into:

$\int_{1}^{\infty} \dfrac{\ln^3(x)}{x(x-1)} \, dx =\int_{0}^{\infty} \dfrac{y^3}{e^y(e^y-1)} \times e^y \, dy =\int_{0}^{\infty} \dfrac{y^3}{(e^y-1)} \, dy$

Now recall that $\displaystyle\zeta(s)\Gamma(s)=\int_{0}^{\infty} \dfrac{y^{s-1}}{e^y-1} \, dy$ and compare it with above and we get that $s-1=3\Rightarrow s=4$ . Hence we have:

$\int_{0}^{\infty} \dfrac{y^3}{(e^y-1)} \, dy = \zeta(4)\Gamma(4)=\dfrac{\pi^4}{90} \times 6 = \boxed{\dfrac{\pi^4}{15}}$

Hence , $A=1 \ , \ B=4 \ , \ C=15 \Rightarrow A+B+C=1+4+15=\boxed{20}$ .

Moderator note:

That's a nice approach. Sometimes if $x$ is the denominator, we can use the substitution $x = e^y$ to hopefully simplify the integration further.

Proof that $\displaystyle\zeta(s)\Gamma(s)=\int_{0}^{\infty} \dfrac{y^{s-1}}{e^y-1} \, dy$

Using maclaurin series expansion of $\displaystyle e^y-1 = -\sum_{i=0}^{\infty} e^{iy}$

Thus we get $\displaystyle\sum_{i=0}^{\infty} \int_{0}^{\infty} y^{s-1} e^{iy}\, dy$

Which is equal to $\displaystyle\sum_{i=0}^{\infty} \dfrac{\Gamma(s)}{i^{s}}$

Which is again equal to $\zeta (s) \Gamma (s)$

Hence proved.

Thanks @Nihar Mahajan for dedicating a prob to me!

Harsh Shrivastava - 5 years, 3 months ago

Welcome , and thanks for the proof :)

Nihar Mahajan - 5 years, 3 months ago

See my proof in the wiki on Riemann zeta function. I've proved it using RMT. Talking about this problem you could approach it through the relation of polygamma function with Hurwitz zeta function. That'll shorten the problem for just 2 steps.

Aditya Kumar - 5 years, 3 months ago

@Nihar Mahajan Same way!!!

Aaghaz Mahajan - 3 years, 1 month ago

A Former Brilliant Member
Feb 23, 2016

$I =\displaystyle \int_{1}^{\infty} \dfrac{\ln^{3}(x)}{x^{2}-x}dx$
Put $x = \dfrac{1}{t}$ the integral transforms into,
$I = \displaystyle -\int_{0}^{1}\dfrac{\ln^{3}(t)}{1-t}dt$
Using the McLaurin series for $\dfrac{1}{1-t} = \displaystyle \sum_{k=0}^{\infty}t^{k}$
$I = -\displaystyle \sum_{k=0}^{\infty} \int_{0}^{1} \ln^{3}(x)t^{k}dt$

Excercise to the reader :
Prove :
$\displaystyle \int_{0}^{1} x^{m}\ln^{n}(x)dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}}$
$\therefore I = - \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^{3} 3!}{(k+1)^{4}} = 6 \cdot \zeta(4)$

$I = 6 \cdot \dfrac{\pi^{4}}{90} = \dfrac{\pi^{4}}{15}$
$A+B+C = 1 + 4 + 15 = 20$

$\large \mathfrak{I}(m,n)=\displaystyle \int_{0}^{1} x^{m}\ln^{n}(x)dx$ Using Integration by parts : $\mathfrak{I}(m,n)=\dfrac{x^{m+1}(\ln^n x)}{m+1}|_0^1-\dfrac{n}{m+1}\int_0^1 x^m \ln^{n-1}x dx$ $\large =-\dfrac{n}{m+1}\mathfrak{I}(m,n-1)$ $\large =\dfrac{n(n-1)}{(m+1)^2}\mathfrak{I}(m,n-2)$ $\cdots\cdots$ $\large =(-1)^n\dfrac{n!}{(m+1)^n}\mathfrak{I}(m,0)$ $\Large = \dfrac{(-1)^n n!}{(m+1)^{n+1}}$ $\small{(Since~~ \mathfrak{I}(m,0)=\int_0^1x^mdx=\dfrac{1}{m+1})}$

Rishabh Jain - 5 years, 3 months ago

Nicely done.

A Former Brilliant Member - 5 years, 3 months ago

Thanks!! :-}