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Calculus Level 3

Evalute :

0 x 3 e x 1 d x = a π b c \displaystyle \int_0^\infty \dfrac {x^3}{e^x - 1} dx = \dfrac {aπ^b}{c}

a , c a, c are coprime positive integers and b b is also an integer.

Find 2 ( a + c ) + b . 2(a + c) + b.


The answer is 36.

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1 solution

Using the property, Γ ( s ) ζ ( s ) = 0 x s 1 e x 1 d x \Gamma (s) \zeta (s) = \displaystyle \int_0^\infty \dfrac {x^{s-1}}{e^x - 1} dx

Where Γ ( ) \Gamma (•) and ζ ( ) \zeta (•) denote Gamma Function and Reimann Zeta Function respectively.

Put s = 4 , s= 4, Γ ( 4 ) ζ ( 4 ) = 0 x 4 1 e x 1 d x = S \Gamma (4) \zeta (4) = \displaystyle \int_0^\infty \dfrac {x^{4-1}}{e^x - 1} dx = S

Γ ( s ) = ( s 1 ) ! \Gamma (s) = (s-1)! where ( ! ) (!) denote factorial and ζ ( s ) = k = 1 1 k s \zeta (s) = \displaystyle \sum_{k=1}^\infty \dfrac {1}{k^s}

S = Γ ( 4 ) ζ ( 4 ) = ( 4 1 ) ! × π 4 90 = ( 1 ) π 4 15 = a π b c S = {\color{#20A900}{\Gamma (4)}}{\color{#D61F06}{ \zeta (4)}} = {\color{#20A900}{(4-1)!}} \times {\color{#D61F06}{\dfrac {π^4}{90} }} = \dfrac {(1)π^4}{15} = \dfrac {aπ^b}{c}

Therefore, 2 ( a + c ) + b = 2 ( 1 + 15 ) + 4 = 36 2(a +c) + b = 2( 1 + 15) + 4 = \boxed{36}

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