Question-163954

Question Number 163954 by mnjuly1970 last updated on 12/Jan/22

Answered by mathmax by abdo last updated on 13/Jan/22

Ψ = 2 \int_{0}^{\infty} \frac{x^{2} e^{x}}{sh (2 x)} dx =_{2 x = t} 2 \int_{0}^{\infty} \frac{t^{2} e^{\frac{t}{2}}}{4 sh (t)} \frac{dt}{2} = \frac{1}{4} \int_{0}^{\infty} \frac{t^{2} e^{\frac{t}{2}}}{\frac{e^{t} - e^{- t}}{2}} dt

= \frac{1}{2} \int_{0}^{\infty} \frac{t^{2} e^{\frac{t}{2}}}{e^{t} - e^{- t}} dt = \frac{1}{2} \int_{0}^{\infty} \frac{t^{2} e^{- t + \frac{t}{2}}}{1 - e^{- 2 t}} dt

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

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

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

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

Ψ = \frac{2!}{2} \sum_{n = 0}^{\infty} \frac{1}{{(2 n + \frac{1}{2})}^{3}} = 8 \sum_{n = 0}^{\infty} \frac{1}{{(4 n + 1)}^{3}} = \frac{8}{4^{3}} \sum_{n = 0}^{\infty} \frac{1}{{(n + \frac{1}{4})}^{3}}

rest to find this sum using digamma function \dots .

Answered by Lordose last updated on 13/Jan/22

I = \int_{0}^{\infty} \frac{x^{2} e^{x}}{\sinh (x) \cosh (x) dx} = 2 \int_{0}^{\infty} x^{2} e^{x} csch (2 x) dx

I \overset{x = \frac{x}{2}}{=} \frac{1}{2} \int_{0}^{\infty} \frac{x^{2} e^{\frac{x}{2}}}{e^{x} (1 - e^{- 2 x})} dx \overset{x = - \ln (x)}{=} \frac{1}{2} \int_{0}^{1} \frac{x^{- \frac{1}{2}} \ln^{2} (x)}{1 - x^{2}} dx

I = \frac{1}{2} \int_{0}^{1} \frac{x^{- \frac{1}{2}} \ln^{2} (x)}{1 - x^{2}} dx \overset{x = x^{\frac{1}{2}}}{=} \frac{1}{16} \int_{0}^{1} \frac{x^{\frac{1}{4} - 1} \ln^{2} (x)}{1 - x} dx

- \int_{0}^{1} \frac{t^{x - 1} \ln^{2} (x)}{1 - t} dx = ψ^{(2)} (x)

I = - \frac{1}{16} ψ^{(2)} (\frac{1}{4}) = \frac{7}{2} ζ (3) + \frac{π^{3}}{8}

α = \frac{7}{2}, β = \frac{1}{8}

\emptyset sE

Commented by mnjuly1970 last updated on 13/Jan/22

m e r c r y s i r l o r d o s

Answered by mindispower last updated on 13/Jan/22

= \int_{0}^{\infty} \frac{4 x^{2} e^{- x}}{(1 + e^{- 2 x}) (1 - e^{- 2 x})} d x

= 2 (\int_{0}^{\infty} \frac{x^{2} e^{- x}}{1 + e^{- 2 x}} d x + \int_{0}^{\infty} \frac{x^{2} e^{- x}}{1 - e^{- 2 x}})

= 2 \sum_{n ⩾ 0} (\int_{0}^{\infty} x^{2} {(- 1)}^{n} e^{- (1 + 2 n) x} d x + \int_{0}^{\infty} x^{2} e^{- (1 + 2 n) d x}

= 2 \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} Γ (3) + 2 \sum_{n ⩾ 0} \frac{1}{{(1 + 2 n)}^{3}} Γ (3)

= 4 β (3) + 4 (ζ (3) - \frac{ζ (3)}{8})

= 4 β (3) + \frac{7}{2} ζ (3) = \frac{π^{3}}{8} + \frac{7}{2} ζ (3)

a + b = \frac{29}{8}

Commented by mnjuly1970 last updated on 13/Jan/22

v e r y n i c e s o l u t i o n a s a l w a y s

s i r p o w e r \dots

Commented by mindispower last updated on 14/Jan/22

w i t h e P l e a s u r n i c e d a y

Commented by mnjuly1970 last updated on 14/Jan/22

Answered by mnjuly1970 last updated on 13/Jan/22