1-pi-0-pi-e-2cos-d-

Question Number 114526 by bobhans last updated on 19/Sep/20

\frac{1}{π} \underset{0}{\int^{π}} e^{2 \cos θ} d θ ?

Commented by JDamian last updated on 19/Sep/20

https://youtu.be/-UhFu0g9740

Commented by mathmax by abdo last updated on 19/Sep/20

the problem how to calculate Σ \frac{1}{{(n!)}^{2}} \dots!

Answered by mathmax by abdo last updated on 19/Sep/20

I = \frac{1}{π} \int_{0}^{π} e^{2 \cos θ} d θ \Rightarrow π I = \int_{0}^{\frac{π}{2}} e^{2 \cos θ} d θ + \int_{\frac{π}{2}}^{π} e^{2 \cos θ} d θ (\to θ = \frac{π}{2} + u)

= \int_{0}^{\frac{π}{2}} e^{2 \cos θ} d θ + \int_{0}^{\frac{π}{2}} e^{- 2 sinu} du = \int_{0}^{\frac{π}{2}} \sum_{n = 0}^{\infty} \frac{{(2 \cos θ)}^{n}}{n!} d θ

+ \int_{0}^{\frac{π}{2}} \sum_{n = 0}^{\infty} \frac{{(- 2 \sin θ)}^{n}}{n!} d θ = \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} \int_{0}^{\frac{π}{2}} \cos^{n} θ d θ

+ \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n!} \int_{0}^{\frac{π}{2}} \sin^{n} θ d θ changement θ = \frac{π}{2} - t give

\int_{0}^{\frac{π}{2}} \sin^{n} θ d θ = \int_{0}^{\frac{π}{2}} \cos^{n} t dt \Rightarrow

π I = \sum_{n = 0}^{\infty} {\frac{2^{n}}{n!} + \frac{{(- 2)}^{n}}{n!}} \int_{0}^{\frac{π}{2}} \cos^{n} θ d θ let w_{n} = \int_{0}^{\frac{π}{2}} \cos^{n} θ d θ (wallis integral)

π I = \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} (1 + {(- 1)}^{n}) W_{n} = \sum_{n = 0}^{\infty} \frac{2^{2 n}}{(2 n)!} (2) W_{2 n} \Rightarrow

I = \frac{1}{π} \sum_{n = 0}^{\infty} \frac{2^{2 n + 1}}{(2 n)!} W_{2 n} (w_{2 n} is known by recurrence)