nice-calculus-evaluate-x-2-1-e-x-1-e-x-dx-

Question Number 123060 by mnjuly1970 last updated on 22/Nov/20

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

e v a l u a t e :::

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

Answered by mindispower last updated on 22/Nov/20

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

= 2 {[- \frac{x^{2}}{1 + e^{x}}]}_{0}^{\infty} + 4 \int_{0}^{\infty} \frac{x}{1 + e^{x}} d x

= 4 \int_{0}^{\infty} x Σ {(- 1 e^{- x})}^{k} e^{- x} d x

= 4 \sum_{k ⩾ 0} \int_{0}^{\infty} {(- 1)}^{k} e^{- (1 + k) x} x d x

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

= 4 Σ \frac{{(- 1)}^{k}}{{(1 + k)}^{2}} = 4 (\frac{ζ (2)}{2}) = \frac{π^{2}}{3}

Commented by mnjuly1970 last updated on 22/Nov/20

g r a t e f u l

m r m i n d s p o w e r \dots

Answered by Dwaipayan Shikari last updated on 22/Nov/20

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

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

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

= 4 \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{\infty} {x e}^{- n x} d x

= 4 \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{n^{2}} \int_{0}^{\infty} {u e}^{- u} d u

= 4 \sum^{\infty} \frac{{(- 1)}^{n + 1}}{n^{2}} Γ (2) = \frac{π^{2}}{12} . 4 = \frac{π^{2}}{3}

Commented by mnjuly1970 last updated on 22/Nov/20

t h a n k y o u m r

p a y a n . .

Answered by mathmax by abdo last updated on 22/Nov/20

Ω = \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + e^{x}) (1 + e^{- x})} dx changement e^{x} = t give

Ω = \int_{0}^{\infty} \frac{\ln^{2} t}{(1 + t) (1 + t^{- 1})} \frac{dt}{t} = \int_{0}^{\infty} \frac{\ln^{2} t}{{(1 + t)}^{2}} dt = \int_{0}^{1} \frac{\ln^{2} t}{{(1 + t)}^{2}} dt + \int_{1}^{\infty} \frac{\ln^{2} t}{{(1 + t)}^{2}} dt (\to t = \frac{1}{x})

= 2 \int_{0}^{1} \frac{\ln^{2} x}{{(1 + x)}^{2}} dx we hsve for ∣ x ∣< 1 \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} \Rightarrow

- \frac{1}{{(1 + x)}^{2}} = \sum_{n = 1}^{\infty} n {(- 1)}^{n} x^{n - 1} = \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n + 1} x^{n} \Rightarrow

\frac{1}{{(1 + x)}^{2}} = \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} x^{n} \Rightarrow

Ω = 2 \int_{0}^{1} \ln^{2} x (\sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} x^{n}) dx

= 2 \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} \int_{0}^{1} x^{n} \ln^{2} x dx we have by parts

\int_{0}^{1} x^{n} \ln^{2} xdx = {[\frac{x^{n + 1}}{n + 1} \ln^{2} x]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} \frac{2 lnx}{x} dx

= - \frac{2}{n + 1} \int_{0}^{1} x^{n} lnx dx = - \frac{2}{n + 1} {{[\frac{x^{n + 1}}{n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} \frac{dx}{x}}

= \frac{2}{{(n + 1)}^{3}} \Rightarrow Ω = 2 \sum_{n = 0}^{\infty} (n + 1) {(- 1)}^{n} \frac{2}{{(n + 1)}^{3}}

= 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{2}} = 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = - 4 \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}}

let δ (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{x}} we know δ (x) = (2^{1 - x} - 1) ξ (x) \Rightarrow

δ (2) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n^{2}} = (2^{1 - 2} - 1) ξ (2) = - \frac{1}{2} . \frac{π^{2}}{6} = - \frac{π^{2}}{12} \Rightarrow

Ω = - 4 \times (- \frac{π^{2}}{12}) \Rightarrow Ω = \frac{π^{2}}{3}

Commented by mnjuly1970 last updated on 22/Nov/20

t h a n k y o u s o m u c h s i r m a x .

Commented by mathmax by abdo last updated on 23/Nov/20

you are welcome sir