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I hope there is no problem asking this. I am struggling to find a closed form/pattern for this integral $$I(n)=\int_0^\pi \exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right) \cos(nt )dt$$ while trying to solve another integral found here: https://math.stackexchange.com/a/2912878/515527 .I started by finding small values for $I(n)$. Il just write them for $\frac{\pi}{I(n)}$ \begin{array}{|c|c|} \hline n & \frac{\pi}{I(n)} \\ \hline 1 & 2^4 \\ \hline 2 & 2^8 \\ \hline 3 & 2^{11}\cdot3 \\ \hline 4 & 2^{16}\cdot3 \\ \hline 5 & 2^{19}\cdot3\cdot5 \\ \hline 6 & 2^{23} \cdot 3^2 \cdot5 \\ \hline 9 & 2^{35}\cdot3^4\cdot5\cdot7 \\ \hline 10 & 2^{39}\cdot3^{4}\cdot5^2\cdot 7 \\ \hline 20 & 2^{79}\cdot3^8\cdot5^4\cdot7^2\cdot11\cdot13\cdot17\cdot19 \\ \hline 25 & 2^{98}\cdot3^{10}\cdot5^6\cdot7^3\cdot11^2\cdot13\cdot17\cdot19\cdot23\\ \hline \end{array} It seems to me, that if $n$ is not simultaneously a multiple of $2$ and a perfect square then the first term is of the form $2^{4n-1}$, howeverer this fails for $n=25$. Also the power counts how many primes (and multiples of that primes) are found untill $n$ and adds $1$ (to the power) if $n$ was already a perfect square, for example for $n=25$. $13,17,19,23$ are found once. $11$ is found twice as it can be written as $11 $and $2\cdot11$. $5$ is found $5$ times between $1$ and $25$ but as $25=5^2$ adds $1$ more to its power. Unfortunately I cant see a proper pattern. Could you perhaps share some help with this integral?

It might be helpful to mention that $$\sum_{n=1}^\infty \frac{1}{3^n} I(n) =\frac{\pi}{2} \left(\sqrt[24]{e}-1\right)$$

Szeto
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Zacky
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    Have you tried differentiating it with respect to $n$? If you differentiate it twice you should be able to obtain a differential equation for $I$. – MSDG Sep 11 '18 at 11:30
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    @Sobi That doesn't work. You get $$I''(n)=-\int_0^\pi t^2\exp\left(\frac{\cos t}8\right)\cos\left(\frac{\sin t}8\right)\cos(nt),dt$$ with an extra $t^2$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 11 '18 at 11:35
  • @SimpliFire Ah yes, of course, I forgot about the chain rule. Sorry. – MSDG Sep 11 '18 at 11:37

1 Answers1

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(There was a typo in the OP and I corrected it.)

The integral can be evaluated in closed form as $$I(n)=\frac{\pi}{2^{3n+1}n!}$$

$$ \begin{align} I(n) &= \frac12 \int^\pi_{-\pi}\exp\left(\frac{\cos t}{8}\right)\cos\left(\frac{\sin t}{8}\right) \cos(nt )dt \\ &= \frac12 \Re\left[\int^\pi_{-\pi}\exp\left(\frac{\cos t}{8}\right)\exp\left(\frac{i\sin t}{8}\right) \cos(nt )dt \right]\\ &=\frac12 \Re\left[\int^\pi_{-\pi}\exp\left(\frac{\cos t+i\sin t}{8}\right)\cos(nt )dt \right]\\ &=\frac12 \Re\left[\int^\pi_{-\pi}\exp\left(\frac{e^{it}}{8}\right)\left(\frac{e^{int}+e^{-int}}2\right)dt \right]\\ &=\frac14 \Re\left[\oint_{|z|=1} e^{z/8}\left(z^n+\frac1{z^n}\right)\frac{dz}{iz}\right] \qquad{(1)}\\ &=\frac14 \Re\left[ 2\pi i\operatorname*{Res}_{z=0}\frac{e^{z/8}}{iz^{n+1}} \right]\\ &=\frac14\cdot2\pi\cdot\frac1{8^nn!} \\ &=\frac{\pi}{2^{3n+1}n!} \end{align} $$

$(1)$: let $z=e^{it}$.

The observation of pattern related to prime directly follows.

Szeto
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