# 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!

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

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 $F(b) = F(0)=-1$.

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

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.

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

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

$$\cos(be^{i\theta})=\cos(b\cos\theta+ib\sin\theta)=-\cos (b\cos \theta)\cosh(b\sin \theta)+i\sin(b\cos\theta)\sinh(b\sin \theta)$$ 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.

• 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. Feb 24, 2015 at 13:15