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Taken from the post: The Integral that Stumped Feynman?

I want to know if the integral:

$$\int_0^{2\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx = 2\pi e^{2/3}$$

can be evaluated using strictly real methods.

I've tried series of $e^x$ and $\cos x$ but to no avail. I tried differentiating under the integral, but nothing seemed to come out of it. Is there any wizardry that can conjure up this answer without complex analysis.

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  • $\begingroup$ Would Weierstrass substitution help? $\endgroup$
    – Henry Lee
    Sep 10, 2018 at 11:34
  • $\begingroup$ Just gave it a try, didn't seem to get anywhere. $\endgroup$
    – Tom Himler
    Sep 10, 2018 at 14:53
  • $\begingroup$ Hmmm. Is there no way you could use complex analysis rather than strictly real? $\endgroup$
    – Henry Lee
    Sep 10, 2018 at 14:54
  • $\begingroup$ Since exp(a+b)=exp(a)exp(b) then the equality is equivalent to something like: $\int$ exponential() cosinus()=$2\pi$ $\endgroup$
    – FDP
    Sep 10, 2018 at 15:47
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    $\begingroup$ Does this solution not count? $\endgroup$
    – Caddyshack
    Sep 11, 2018 at 19:52

1 Answer 1

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Start with $\tan \left(\frac{x}{2} \right)$$=2\tan \left( \frac{t}{2} \right)$ $$I=2\int_0^{\pi} \exp\left(\frac{7+5 \cos x}{10+6\cos x}\right) \cos \left( \frac{\sin x}{10+6 \cos x} \right) dx=8e^{5/8}\int_0^{\pi}\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)\frac{dt}{5-3\cos t} $$ Using $$ \sum_{n=1}^{\infty} a^{n} \cos(nx) = \frac12\left(\frac{1-a^{2}}{1-2 a \cos x + a^{2}}-1\right)$$ we can rewrite $$\frac{1}{5-3\cos x} =\frac14 +\frac12 \sum_{n=1}^\infty \frac{1}{3^n} \cos (nx) $$ thus we have $$I=2e^{5/8} \int_0^\pi \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right) dt +4e^{5/8} \sum_{n=1}^\infty \frac{1}{3^n} \int_0^\pi \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right) \cos(nt )dt $$ $$=2\pi e^{5/8}+4e^{5/8}\sum_{n=1}^\infty \frac{1}{3^n} I(n)$$ I dont know how to evaluate $I(n)$, but maybe someone can help. $$I(0)=\pi,I(1)=\frac{\pi}{2^4}, I(2)=\frac{\pi}{2^8}, I(3)=\frac{\pi}{3\cdot 2^{11}}, I(4)=\frac{\pi}{3\cdot2^{16}},I(5)=\frac{\pi}{3\cdot 5 \cdot 2^{19}}$$$$ I(6)=\frac{\pi}{3^2\cdot5\cdot 2^{23}}, I(10)=\frac{\pi}{3^4\cdot5^2\cdot 7 \cdot 2^{39}}, I(20)=\frac{\pi}{3^8\cdot5^4 \cdot 7^2\cdot11\cdot 13\cdot 17\cdot 19\cdot 2^{79}}$$

Edit: As seen here: https://math.stackexchange.com/a/2913057/515527 $ I(n) =\frac{\pi} {2^{3n+1}n!}$, plugging this into the sum and using the series for $e^x$ will give the result immediately..

Another approach to evaluate $I(n)$ is to use that $$\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right)=\sum_{n=0}^{\infty} \frac{\cos(nt)}{8^nn!}$$ Since $$\int_0^\pi \cos(nx) \cos(mx) dx=\begin{cases} \frac{\pi} {2} & n=m \\ 0 & n \neq m\end{cases}$$ We get that $I(n) =\frac{\pi} {2} \frac{1} {8^n n!} $ and the result follows.

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  • $\begingroup$ We might conjecture that $I(n) = \pi/2^{4n}$ $\endgroup$
    – user370967
    Sep 11, 2018 at 8:40
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    $\begingroup$ Well, this way we get $I=\frac{98}{47}\pi e^{5/8}$, Unfortunately $ \frac{98}{47}\pi e^{5/8}-2\pi e^{2/3}= 0.000073...$ :( $\endgroup$
    – Zacky
    Sep 11, 2018 at 8:54
  • $\begingroup$ Yes, the patterns unfortunately breaks down. $\endgroup$
    – user370967
    Sep 11, 2018 at 10:07
  • $\begingroup$ By the way, I'm not sure you can prove the series identity used by real methods. $\endgroup$
    – FDP
    Sep 11, 2018 at 16:31
  • $\begingroup$ look here: math.stackexchange.com/questions/685278/… $\endgroup$
    – Zacky
    Sep 11, 2018 at 18:40

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