I-n-0-t-2t-sin-2n-tdt-Prove-that-I-n-1-1-e-2pi-pi-0-e-2t-sin-2n-t-dt-and-I-n-1-2sh-pi-pi-n-

Question Number 195320 by Erico last updated on 30/Jul/23

I_{n} = \int_{0}^{+ \infty} t^{- 2 t} {s i n}^{2 n} t d t

Prove that I_{n} = \frac{1}{1 - e^{- 2 π}} {\int_{0}}^{π} e^{- 2 t} {s i n}^{2 n} t d t

and I_{n} \underset{\infty}{∽} \frac{1}{2 s h (π)} \sqrt{\frac{π}{n}}

Answered by witcher3 last updated on 31/Jul/23

I_{n} = \frac{1}{1 - e^{- 2 π}} \int_{0}^{π} e^{- 2 t} \sin^{2 n} (t) dt

\int_{0}^{\infty} e^{- 2 t} \sin^{2 n} (t) = \underset{k = 0}{\sum^{\infty}} \int_{k π}^{(k + 1) π} e^{- 2 t} \sin^{2 n} (t) dt

t - k π = x \Rightarrow dt = dx

= \sum_{k ⩾ 0} \int_{0}^{π} e^{- 2 x - 2 k π} \sin^{2 n} (x + k π) dx

= \sum_{k ⩾ 0} e^{- 2 k π} \int_{0}^{π} e^{- 2 x} \sin^{2 n} (x) dx

= \frac{1}{1 - e^{- 2 π}} \int_{0}^{π} e^{- 2 x} \sin^{2 n} (x) dx \dots

J_{n} = \int_{0}^{π} e^{- 2 x} \sin^{2 n} (x) dx

By = {[\frac{e^{- 2 x}}{- 2} \sin^{2 n} (x)]}_{0}^{π} + \frac{1}{2} \int_{0}^{π} 2 ncos (t) \sin^{2 n - 1} (t) e^{- 2 t} dt

= n \int_{0}^{π} \cos (t) \sin^{2 n - 1} (t) e^{- 2 t} dt

= n [\frac{e^{- 2 t}}{- 2} \cos (t) \sin^{2 n - 1} (t)] + \frac{n}{2} \int_{0}^{π} e^{- 2 t} [- \sin^{2 n} (t) + (2 n - 1) \cos^{2} (t) \sin^{2 n - 2}] dt

= \frac{n}{2} \int_{0}^{π} [(- 2 n) \sin^{2 n} (t) + (2 n - 1) \sin^{2 n - 2}] e^{- 2 t} dt

= - n^{2} \int_{0}^{π} e^{- 2 t} \sin^{2 n} (t) dt + \frac{n (2 n - 1)}{2} \int_{0}^{π} \sin^{2 n - 2} (t) e^{- 2 t} dt

= - n^{2} J_{n} + \frac{n (2 n - 1)}{2} J_{n - 1}

\Leftrightarrow (1 + n^{2}) J_{n} = \frac{n (2 n - 1)}{2} J_{n - 1} \Leftrightarrow \frac{J_{n}}{J_{n - 1}} = \frac{n (2 n - 1)}{2 (1 + n^{2})}

\Rightarrow \underset{k = 1}{\prod^{n}} \frac{J_{k}}{J_{k - 1}} = \frac{J_{n}}{J_{0}} = \underset{k = 1}{\prod^{n}} \frac{k (2 k - 1)}{2 (k + i) (k - i)}

= \underset{k = 1}{\prod^{n}} \frac{2 k (2 k - 1)}{4 (k + i) (k - i)} = \frac{(2 n)!}{4^{n}} . \frac{Γ (1 - i) Γ (1 + i)}{Γ (n + 1 - i) Γ (n + 1 + i)}

Γ (1 + i) Γ (1 - i) = i Γ (1 - i) Γ (i) = \frac{i π}{\sin (i π)}

= \frac{π}{sh (π)}

Γ (n + 1 \underline{+} i) \sim \sqrt{2 π n} e^{- n} n^{n \underline{+} i}

Γ (2 n + 1) = n^{2 n} . 2^{2 n} e^{- 2 n} 2 \sqrt{2 π n}

\frac{J_{n}}{J_{0}} \sim \frac{2 \sqrt{2 π n} . 4^{n} . n^{2 n} e^{- 2 n}}{2 π n . e^{- 2 n} n^{2 n} . 4^{n}} . \frac{π}{sh (π)} = \frac{2 π}{sh (π) \sqrt{2 π n}}

J_{0} = \int_{0}^{π} e^{- 2 x} = \frac{1 - e^{- 2 π}}{2}

J_{n} \sim \frac{1 - e^{- 2 π}}{2 sh (π)} . \frac{2 π}{\sqrt{2 π n}}

I_{n} = \frac{1}{1 - e^{- 2 π}} J_{n} \sim \frac{1}{sh (π)} . \sqrt{\frac{π}{2 n}}

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