nice-calulus-evaluate-0-x-3-e-x-2-sinh-x-2-

Question Number 123720 by mnjuly1970 last updated on 27/Nov/20

\dots n i c e c a l u l u s \dots

e v a l u a t e ::

Φ = \int_{0}^{\infty} \frac{x^{3} e^{\frac{- x}{2}}}{s i n h (\frac{x}{2})} = ? ? ?

Answered by Dwaipayan Shikari last updated on 27/Nov/20

2 \int_{0}^{\infty} \frac{x^{3} e^{- \frac{x}{2}}}{e^{\frac{x}{2}} - e^{- \frac{x}{2}}} d x

= 2 \int_{0}^{\infty} \frac{x^{3}}{e^{x} - 1} d x

= 2 \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} x^{3} e^{- n x} = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{4}} \int_{0}^{\infty} u^{3} e^{- u} d u n x = u

= 2 Γ (4) \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{4}} = \frac{2 π^{4}}{15}

Commented by mnjuly1970 last updated on 27/Nov/20

t h a n k y o u m r D w a i p a y a n \dots

Commented by Dwaipayan Shikari last updated on 27/Nov/20

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Answered by mathmax by abdo last updated on 28/Nov/20

Φ = \int_{0}^{\infty} \frac{x^{3} e^{- \frac{x}{2}}}{sh (\frac{x}{2})} dx changement \frac{x}{2} = t give

Φ = \int_{0}^{\infty} \frac{{8 t}^{3} e^{- t}}{sh (t)} (2) dt = 32 \int_{0}^{\infty} \frac{t^{3} e^{- t}}{e^{t} - e^{- t}} dt

= 32 \int_{0}^{\infty} \frac{t^{3} e^{- 2 t}}{1 - e^{- 2 t}} dt = 32 \int_{0}^{\infty} t^{3} e^{- 2 t} \sum_{n = 0}^{\infty} e^{- 2 nt} dt

= 32 \sum_{n = 0}^{\infty} \int_{0}^{\infty} t^{3} e^{- 2 (n + 1) t} dt =_{2 (n + 1) t = z} 32 \sum_{n = 0}^{\infty} \int_{0}^{\infty} \frac{z^{3}}{8 {(n + 1)}^{3}} e^{- z} \frac{dz}{2 (n + 1)}

= 2 \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{4}} \int_{0}^{\infty} z^{3} e^{- z} dz = 2 Γ (4) \sum_{n = 1}^{\infty} \frac{1}{n^{4}}

= 2 Γ (4) ξ (4) = 2.3! \frac{π^{4}}{90} = \frac{12 π^{4}}{90} = \frac{6.2 π^{4}}{6.15} \Rightarrow Φ = \frac{2 π^{4}}{15}