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I’d like help with computing the following integral:

$$\int_{0}^{\pi} e^{\cos t}\,dt$$

(This is a problem in complex analysis [supposedly].)

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2 Answers 2

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(the complex analysis way)

$\cos t$ is even hence $$\int_{-\pi}^\pi e^{\cos t} dt = 2 \int_0^{\pi} e^{\cos t} dt$$

then apply the (complex) change of variable $z = e^{it}, dz = i e^{it} dt$, $$\int_{-\pi}^{\pi} e^{\cos t} dt = \int_{-\pi}^{\pi} e^{\frac{e^{it}+e^{-it}}{2}} (-i e^{-it}) i e^{it} dt = -i\int_{\gamma} e^{\frac{z+z^{-1}}{2}} z^{-1} dz$$

with $\gamma$ the unit circle traversed counter-clockwise.

the only one singularity inside the contour is at $z=0$, where $$e^{\frac{z+z^{-1}}{2}} z^{-1} = z^{-1}\left(\sum_{k=0}^\infty \frac{z^k}{2^k k!}\right)\left(\sum_{k=0}^\infty \frac{z^{-k}}{2^k k!}\right) = z^{-1}\sum_{n=-\infty}^\infty z^n \sum_{k=\max(0,n)}^\infty \frac{1}{2^k k!}\frac{1}{2^{k-n} (k-n)!} = \sum_{n=-\infty}^\infty c_n z^n$$

with $$c_{-1} = \sum_{k=0}^\infty \frac{1}{2^{2k} (k!)^2}= Res\left(e^{\frac{z+z^{-1}}{2}} z^{-1},0\right)$$

hence

$$\int_{\gamma} e^{\frac{z+z^{-1}}{2}} z^{-1} dz = 2 i \pi c_{-1}$$

$$\int_0^{\pi} e^{\cos t} dt = \frac{-i}{2} \int_{\gamma} e^{\frac{z+z^{-1}}{2}} z^{-1} dz = \pi \sum_{k=0}^\infty \frac{1}{2^{2k} (k!)^2}$$

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Hint: expand the exponential in a power series, and express the cosine in terms of exponential functions. The answer turns out to involve a modified Bessel function. In fact, one of the formulas on that page contains the answer.

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  • $\begingroup$ This was an exam question. The answer is presumably not supposed to be in terms of special functions (other than maybe sinusoidal functions). $\endgroup$ May 25, 2016 at 1:21
  • $\begingroup$ can you check if the result agrees with mine ? $\endgroup$
    – reuns
    May 25, 2016 at 2:18
  • $\begingroup$ Yes, it does. $$\pi\sum_{k=0}^\infty \dfrac{1}{2^{2k} (k!)^2} = \pi I_0(1)$$ $\endgroup$ May 25, 2016 at 7:28
  • $\begingroup$ Presumably, if you have not studied modified Bessel functions, the intended answer on the exam was the series representation. You won't get a "closed form" answer that doesn't involve special functions. $\endgroup$ May 25, 2016 at 7:32

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