Where did digamma come from?

Calculus Level 5

∫ 0 1 ( 1 − x 2 ) ln ⁡ x 1 − x 3 d x = π a b − ψ ( 1 ) ( 1 / c ) d \large \int _{ 0 }^{ 1 }{ \dfrac { (1-{ x }^{ 2 })\ln { x } }{ 1-{ x }^{ 3 } } \, dx } =\dfrac { { \pi }^{ a } }{ b } -\dfrac { \psi ^{ (1) }(1/c) }{ d }

If the above equation holds true for positive integers a , b , c a,b,c and d d , find a + b + c + d a+b+c+d .

Notation : ψ ( 1 ) ( ⋅ ) \psi^{(1)}(\cdot) denotes the first derivative of the digamma function .


The answer is 68.

Hamza A by Hamza A , 19, USA

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1 solution

Let the Integral be denoted by S.

S = ∫ 0 1 l n x 1 − x 3 d x ⏟ S 1 − ∫ 0 1 x 2 l n x 1 − x 3 d x ⏟ S 2 \large \displaystyle S=\underbrace{\int_{0}^{1} \frac{lnx}{1-x^3}dx}_{S_1}-\underbrace{\int_{0}^{1}\frac{x^2lnx}{1-x^3}dx}_{S_2} , so that S = S 1 − S 2 S=S_1-S_2 .

S 1 = ∫ 0 1 l n x 1 − x 3 d x \large \displaystyle S_1=\int_{0}^{1}\frac{lnx}{1-x^3}dx . Put x 3 = t x^3=t and the integral changes to ,

S 1 = ∫ 0 1 1 9 l n t 1 − t t − 2 3 d t \large \displaystyle S_1=\int_{0}^{1} \frac{1}{9} \frac{lnt}{1-t}t^{-\frac{2}{3}}dt

By Integral Representation of Trigamma Function we have,

S 1 = ∫ 0 1 1 9 l n t 1 − t t 1 3 − 1 d t \large \displaystyle S_1 = \int_{0}^{1} \frac{1}{9} \frac{lnt}{1-t}t^{\frac{1}{3}-1}dt

S 1 = − ψ ( 1 ) ( 1 3 ) 9 \large \displaystyle \boxed{S_1 =\frac{ -\psi^{(1)}(\frac{1}{3})}{9}}

S 2 = ∫ 0 1 x 2 l n x 1 − x 3 d x \large \displaystyle S_2=\int_{0}^{1} \frac{x^2lnx}{1-x^3}dx , Put 1 − x 3 = t 1-x^3=t we have then ,

S 2 = ∫ 0 1 1 9 l n ( 1 − t ) t d t \large \displaystyle S_2=\int_{0}^{1}\frac{1}{9} \frac{ln(1-t)}{t}dt

S 2 = ∫ 0 1 − 1 9 ∑ k ≥ 1 t k − 1 k d t \large \displaystyle S_2=\int_{0}^{1} -\frac{1}{9}\sum_{k\ge1}\frac{t^{k-1}}{k}dt

S 2 = − 1 9 k ∑ k ≥ 1 [ t k k ] 0 1 = − 1 9 ∑ k ≥ 1 1 k 2 = − ζ ( 2 ) 9 \large \displaystyle S_2=-\frac{1}{9k}\sum_{k\ge1}[\frac{t^k}{k}]_{0}^{1} = -\frac{1}{9} \sum_{k\ge1}\frac{1}{k^2} = -\frac{\zeta(2)}{9}

S 2 = − π 2 54 \large \displaystyle \boxed{S_2 = -\frac{\pi^2}{54}}

So we have , S = S 1 − S 2 = π 2 54 − ψ ( 1 ) ( 1 3 ) 9 \large \displaystyle S=S_1-S_2=\frac{\pi^\color{#D61F06}{2}}{\color{#3D99F6}{54}} - \frac{\psi^{(1)}(\frac{1}{\color{#456461}{3}})}{\color{#624F41}{9}}

Answer : 2 + 54 + 3 + 9 = 68 \boxed{2+54+3+9=68}

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