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How can we prove this Integral relation? $$\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{\pi }{e}\cdot \frac{e^{\sqrt{1-a^2}}}{\sqrt{1-a^2}}$$ where $\text{ }a\in(-1,1)$.

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2 Answers 2

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$$I(a)=\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{1}{2}\Re\int _{-\pi}^{\pi }\frac{e^{i \left(a\sin x\right)}}{1+a\cos x}e^{a\cos x}dx=\frac{1}{2}\Re\int _{-\pi}^{\pi }\frac{e^{ ae^{ix}}}{1+a\cos x}dx$$ $$=\Re\int _{-\pi}^{\pi }\frac{e^{ ae^{ix}}e^{ix}}{ae^{2ix}+2e^{ix}+a}dx=\Re\,(-i)\oint_{|z|=1}\frac{e^{az}}{az^2+2z+a}dz$$ The zeros of the denominator are $z_{1,2}=-\frac{1}{a}\pm\sqrt{\frac{1}{a^2}-1}$; only one pole ($\,z_1=-\frac{1}{a}+\sqrt{\frac{1}{a^2}-1}\,\,$) lies inside the closed contour $|z|=1$.

Therefore, $$I(a)=\Re \operatorname {Rez}_{z=z_1}\,2\pi i\,(-i)\frac{e^{az}}{az^2+2z+a}=\pi\frac{e^{a\big(-\frac{1}{a}+\sqrt{\frac{1}{a^2}-1}\big)}}{a\sqrt{\frac{1}{a^2}-1}}=\frac{\pi}{e}\frac{e^\sqrt{1-a^2}}{\sqrt{1-a^2}}$$

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Per the generalized result \begin{equation*} \int_0^\pi\frac{f(e^{ix})+f(e^{-ix})}{1+2p\cos x+p^2}\mathrm dx=\frac{2\pi}{1-p^2}f(-p) , \end{equation*} set $f(t)=\frac12e^{at}$ and $p = \frac{1-\sqrt{1-a^2}}a$ to obtain $$\int _0^{\pi }\frac{\cos \left(a\sin x\right)}{1+a\cos x}e^{a\cos x}dx=\frac{\pi e^{\sqrt{1-a^2}-1}}{\sqrt{1-a^2}}$$

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  • $\begingroup$ Really nice general approach! (+1) $\endgroup$
    – Svyatoslav
    Apr 2 at 14:38
  • $\begingroup$ @Svyatoslav - tkx. you could generalize your contour to derive it $\endgroup$
    – Quanto
    Apr 2 at 15:28
  • $\begingroup$ Yes, you are right: the general expression can be obtained via the complex integration. Thank you. $\endgroup$
    – Svyatoslav
    Apr 2 at 16:58

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