Question

Use this fact proved by the previous problem
\[
\sum _{n=1}^{\infty } \frac{(2 n-1)!}{2^{4 n}
(n!)^2}=\ln \left(-4
\left(\sqrt{3}-2\right)\right)
\]
to solve this problem
http://openstudy.com/study#/updates/54a2cb99e4b054f0c3b448b0

Answer 1

The previous problem was
http://openstudy.com/study#/updates/54a31068e4b0b8a54fb0621d

Answer 2

@ganeshie8

Answer 3

@SithsAndGiggles

Answer 4

One has to show that
\[
\int_0^{\pi } \ln (\cos (x)+2) \, dx=-\pi \ln
\left(4-2 \sqrt{3}\right)
\]

Answer 5

Something that I will use later
\[\int_0^{\pi } \cos ^{2 k+1}(x) \, dx=0 \\
\int_0^{\pi } \cos ^{2 k}(x) \, dx=\frac{\pi (2
k)!}{2^{2 k} (k!)^2}
\]
I will post this as a question. It is an easy and well known recurrence formula

Answer 6

in all posts belo log(x)=ln(x)
\[
\log (t+2)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1}
t^n}{n 2^n}+\log (2)
\]

Answer 7

\[
\log (\cos (x)+2)=\sum _{n=1}^{\infty }
\frac{(-1)^{n-1} \cos ^n(x)}{n 2^n}+\log (2)
\]

Answer 8

\[
\int_0^{\pi } \log (\cos (x)+2) \, dx=\pi \log
(2)-\sum _{k=1}^{\infty } \frac{\int_0^{\pi }
\cos ^{2 k}(x) \, dx}{(2 k) 2^{2 k}}
\]

Answer 9

\[
\int_0^{\pi } \log (\cos (x)+2) \, dx=\\\pi \log
(2)-\sum _{k=1}^{\infty } \frac{\pi (2
k)!}{\left((2 k) 2^{2 k}\right) \left(2^{2 k}
(k!)^2\right)}=\pi \log (2)-\sum _{k=1}^{\infty
} \frac{\pi (2 k-1)!}{2^{4 k} (k!)^2}=\\ \pi \log
(2)-\pi \log \left(-4
\left(\sqrt{3}-2\right)\right)=\\\pi \log (2)-\pi
\left(\log (2)+\log \left(4-2
\sqrt{3}\right)\right)=-\pi \log \left(4-2
\sqrt{3}\right)
\]

Answer 10

@ganeshie8 @SithsAndGiggles

Answer 11

Beautiful! XD In the morning, I was messing with \(\ln(2+\cos x) = \sum\limits^{\infty} \frac{(-1)^{n-1}(1+\cos x)^n}{n}\) and made things worse haha!
\[\int_0^{\pi } \cos ^{2 k+1}(x) \, dx \stackrel{x = \frac{\pi}{2}-u}{=} -\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2} } \sin^{2 k+1}(u) \, du = 0\]

I'll attempt the even powers later when you post the question.. :D

Answer 12

Nice, I'll have to keep this sort of strategy in mind if I come across any more "fun" integrals!

Answer 13

I am glad you liked it.

Answer 14

WOW man...kudos!! @eliassaab