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I am trying to find the value of $$ -\frac{1}{\pi}\int_{-\pi/2}^{\pi/2} \cos\left(be^{i\theta}\right) \mathrm{d}\theta,$$ where $b$ is a real number.

Any helps will be appreciated!

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  • $\begingroup$ Do you know any complex analysis yet? $\endgroup$ Feb 24, 2015 at 12:45
  • $\begingroup$ I know just a little bit about complex analysis. $\endgroup$
    – Firman
    Feb 24, 2015 at 13:14

5 Answers 5

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Let $F(b)$ denote your integral. Then by differentiating under the $\int$ sign, $$ F'(b) = \frac1\pi \int_{-\pi/2}^{\pi/2}e^{i\theta}\sin(be^{i\theta}){\rm d}\theta $$ The integrand is the derivative of $(1-\cos(be^{i\theta}))/(ib)$ wrt. $\theta$ (that's understood to be $0$ if $b=0$), so that $$ F'(b) = \frac{1}{\pi}\Big[\frac{1-\cos(be^{i\theta})}{ib}\Big]_{-\pi/2}^{\pi/2} = 0 $$ because $\cos$ is even, which gives you $F(b) = F(0)=-1$.

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Yet another elementary solution: $$ -\frac{1}{\pi}\int_{-\pi/2}^{\pi/2} \sum_{n=0}^\infty \frac{(be^{i\theta})^{2n}}{(2n)!}d\theta = -\frac{1}{\pi}\sum_{n=0}^\infty \left[\frac{b^{2n}}{(2n)!}\int_{-\pi/2}^{\pi/2} e^{i2n\theta}d\theta\right] = -\frac{1}{\pi} \frac{b^0}{0!}\pi + 0= -1. $$

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Using the Substitution $z=e^{i \theta}$ our intergral can be rewritten as

$$ -\pi I(b)=\frac{1}{i}\int_{SC_{-i\rightarrow i}}\frac{\cos(bz)}{z}dy $$

$SC_{-i\rightarrow i}$ denotes the arc of a semicircle connecting $-i$ and $i$.

By Cauchy integral theorem we can conclude that $$ -\pi I(b)=\frac{1}{i}P\int_{-i}^i\frac{\cos(by)}{y}dz $$

By choosing the standardbranch of the Logartihm, the above integral can be shown to be $\pi $. Therfore we conclude that $$ I(b)=\frac{1}{-\pi}\times\pi=-1 $$

Which is, independent of $b$, which is (at least for me) by no means obvious from the original integral.

This is result is confirmed by some numerical test in Mathematica.

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Because cosine is an even function you may write the integral as

$$-\frac1{2 \pi} \int_{-\pi/2}^{3 \pi/2} d\theta \, \cos{\left ( b e^{i \theta} \right )} = -\frac{1}{i 2 \pi} \oint_{|z|=1} dz \frac{\cos{b z}}{z} $$

which, by the residue theorem or Cauchy's theorem, is

$$-\frac{1}{i 2 \pi} i 2 \pi \cos{0} = -1$$

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$$\cos(be^{i\theta})=\cos(b\cos\theta+ib\sin\theta)=-\cos (b\cos \theta)\cosh(b\sin \theta)+i\sin(b\cos\theta)\sinh(b\sin \theta)$$ Thus your integral reduces to $$I=\frac{2}{\pi}\int_{0}^{\pi/2}\cos (b\cos \theta)\cosh(b\sin \theta)d\theta$$ I think you have to use Bessel's function hereon to simplify this, though I am not quite sure.

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  • $\begingroup$ Actually this is where my integral comes from. I simplified it into the form I wrote above, because I thought it should be easier to do in that form. $\endgroup$
    – Firman
    Feb 24, 2015 at 13:15

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