Just 2 and 3 closed form?

Calculus Level 3

$\frac 1 \pi \int_0^1\dfrac{\ln^4x}{\sqrt{1-x^2}} \,dx= \dfrac{\ln^a(b) }{b} +\dfrac{c\ln^b(b)}{b} \zeta(b)+c\ln(b)\zeta(c) +\dfrac{d}{e}\zeta(a)$

The equation above holds true for positive integers $a$ , $b$ , $c$ , $d$ , and $e$ with $b$ and $c$ being primes, and $d$ and $e$ being coprime integers. Find $a+b+c+d+e$ .

Inspiration

The answer is 82.

1 solution

Aareyan Manzoor
Jul 8, 2019

making a substitution $x = \sin(u)$ makes this integral $\dfrac{1}{\pi} \int_0^{\pi/2} \ln^4{\sin(u)} \text{du} = \dfrac{1}{32 \pi} \dfrac{d^4}{\text{dn}^4} \Beta(n,1/2) \big|_{n=1/2} = \boxed{\dfrac{\ln^4 (2)}{2} +\dfrac{3}{2}\ln^2 (2) \zeta(2) + 3 \zeta(3)\ln(2)+\dfrac{57}{16} \zeta(4)}$ and yes, this was VERY painful to compute

I agree........VERY VERY Painful......!!!!

Aaghaz Mahajan - 1 year, 11 months ago

Just 2 and 3 closed form?

The answer is 82.

1 solution

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