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Is there an analytic way to solve this integral?

$$\int_0^{2 \pi } \large e^{\frac{11 \cos (\phi )+29}{14 \cos (\phi )+50}} \cos \left(\frac{3 \sin (\phi )}{14 \cos (\phi )+50}\right) \, d\phi$$

I tried with the bad substitution $z = 50 + 14\cos(\phi)$ (and a successive easy substitution) and I got nothing good.

Mathematica says the result is $11.126232827(...)$

Is this result obtained via numerical integration only, or is there something we can do?

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  • $\begingroup$ Where did you find this integral? Btw the answer is $2\pi e^{4/7}$, I will try to type an answer soon. $\endgroup$ – Zacky Apr 28 '19 at 12:28
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    $\begingroup$ This looks like it can be tackled with the method used to produce "difficult" integrals in this answer. $\endgroup$ – MathIsFun7225 Apr 28 '19 at 12:49
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    $\begingroup$ @mrtaurho well, I spent some time with a similar integral before. math.stackexchange.com/a/2912878/515527 $\endgroup$ – Zacky Apr 28 '19 at 12:50
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    $\begingroup$ @Zacky Ah, I see. I am looking forward towards your solution :) $\endgroup$ – mrtaurho Apr 28 '19 at 12:54
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    $\begingroup$ @Zacky I actually wrote it down a paper sheet long time ago, after having found it online somewhere (probably in some sort of maths forum) and I got really curious about. Just today, cleaning the desk, I found it again and... Here I am! Looking forward for your answer, when you will, to learn something new! $\endgroup$ – Von Neumann Apr 28 '19 at 13:13
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Due to symmetry we can first reduce the bounds from $2\pi$ to $\pi$ first, then take on with Weierstrass substitution.$$I=2\int_0^{ \pi } \exp{\left(\frac{11 \cos (\phi )+29}{14 \cos (\phi )+50}\right)} \cos \left(\frac{3 \sin (\phi )}{14 \cos (\phi )+50}\right) d\phi$$ $$\overset{\large \tan\frac{\phi}{2}=x}=2\int_0^\infty \exp{\left(\frac12+\frac{2}{9\left(x^2+\frac{16}{9}\right)}\right)}\cos\left(\frac32 \frac{x}{9\left(x^2+\frac{16}{9}\right)}\right)\frac{2}{1+x^2}dx$$ $$\overset{\large x=\frac{4}{3}\tan t}=4e^{1/2}\int_0^\frac{\pi}{2} \exp{\left(\frac29 \frac{1}{\frac{16}{9}\left(\tan^2 t +1\right)}\right)}\cos\left(\frac3{2\cdot 9} \frac{\frac43\tan t}{\frac{16}{9}(\tan^2 t+1)}\right)\frac{\frac43\sec^2 t}{1+\frac{16}{9}\tan^2 t}dt$$ $$=4e^{1/2}\int_0^\frac{\pi}{2} \exp{\left(\frac{1}{8\sec^2 t}\right)}\cos \left(\frac{\tan t}{8\sec^2 t}\right)\frac{12}{9+7\sin^2 t}dt$$ $$=96e^{1/2}\int_0^\frac{\pi}{2} \exp\left(\frac{1+\cos(2t)}{16}\right)\cos\left(\frac{\sin(2t)}{16}\right)\frac{dt}{25-7\cos(2t)}$$ $$\overset{2t=x}=48e^{9/16}\int_0^\pi \exp\left(\frac{\cos x}{16}\right)\cos\left(\frac{\sin x}{16}\right)\frac{dx}{25-7\cos x}$$ $$=48e^{9/16}\Re\left(\int_0^\pi \exp\left(\frac{\cos x}{16}+i\frac{\sin x}{16}\right)\frac{dx}{25-7\cos x}\right)$$$$=24e^{9/16}\Re\left(\int_0^{2\pi}\exp\left(\frac{e^{ix}}{16}\right)\frac{dx}{25-\frac{7}{2}\left(e^{ix}+e^{-ix}\right)}\right)$$ Via the substitution $e^{ix}=z\Rightarrow x=\frac{dz}{iz}$, which transforms $0\to 2\pi$ into the unit circle $|z|=1$, we get: $$I=24e^{9/16}\Re\left(\oint_{|z|=1}\exp\left(\frac{z}{16}\right)\frac{1}{25-\frac72\frac{z^2+1}{z}}\frac{dz}{iz}\right)=-\Re\left({\frac{48}{7i}e^{9/16}}\oint_{|z|=1}\exp\left(\frac{z}{16}\right)\frac{dz}{(z-7)(z-1/7)}\right)$$ Now we can see that our only poles are at $z_1=7$ and $z_2=\frac{1}{7}$, but only $z_2$ is found inside our contour $|z|=1$. Thus by the Residue theorem we have: $$I=-\Re\left(\frac{48}{7i}e^{9/16}\cdot 2\pi i \operatorname{Res}(f(z),z_2) \right), \quad f(z)=\frac{e^{z/16}}{(z-7)(z-1/7)}$$ $$\require{cancel}=-\Re\left(\frac{24\pi}{7}e^{9/16} \lim_{z\to z_2} \cancel{(z-z_2)}\frac{e^{z/16}}{(z-7)\cancel{(z-1/7)}}\right)=-\Re\left(\frac{96\pi}{7}e^{9/16} \frac{e^{1/{7\cdot 16}}}{\frac{{1-49}}{7}}\right)=\boxed{{2\pi e^{4/7}}}$$


From this line: $$I=12e^{9/16}\int_0^\pi \exp\left(\frac{\cos x}{16}\right)\cos\left(\frac{\sin x}{16}\right)\frac{dx}{25-7\cos x}$$ One can do the same approach as seen here, but I tried to take a different approach in this answer.

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    $\begingroup$ Ohh that was amazing to read! Thank you so much for this and for the other links too! $\endgroup$ – Von Neumann Apr 28 '19 at 16:45
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    $\begingroup$ (+1) Interesting! The first time I see you utilizing complex analysis ;) $\endgroup$ – mrtaurho Apr 29 '19 at 9:09

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