Contour integration of cosine of a complex number 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!
 A: 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 $F(b) = F(0)=-1$.
A: Yet another elementary solution:
$$
-\frac{1}{\pi}\int_{-\pi/2}^{\pi/2} \sum_{n=0}^\infty \frac{(be^{i\theta})^{2n}}{(2n)!}d\theta = -\frac{1}{\pi}\sum_{n=0}^\infty \left[\frac{b^{2n}}{(2n)!}\int_{-\pi/2}^{\pi/2} e^{i2n\theta}d\theta\right] = -\frac{1}{\pi} \frac{b^0}{0!}\pi  + 0= -1.
$$
A: 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.
A: 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 
$$-\frac{1}{i 2 \pi} i 2 \pi \cos{0} = -1$$
A: $$\cos(be^{i\theta})=\cos(b\cos\theta+ib\sin\theta)=-\cos (b\cos \theta)\cosh(b\sin \theta)+i\sin(b\cos\theta)\sinh(b\sin \theta)$$ 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.
