# Thread: Definite integral - tricky substitutions

1. ## Definite integral - tricky substitutions

Hi.

So I am asked to analyze

$\displaystyle $\int\limits_0^{2\pi } {\frac{{dx}}{{1 + \varepsilon \cdot \cos x}}}$$

However I believe substitution $\displaystyle $u = \tan \frac{x}{2}$$ might be very wrong in this situation because of discontinuity.

Also I guess this may diverge when $\displaystyle $\left| \varepsilon \right| \geqslant 1$$

Unfortunately at this very moment I am not aware of a nice proper solution, maybe you have an idea?

2. Originally Posted by Pranas
Hi.

So I am asked to analyze

$\displaystyle $\int\limits_0^{2\pi } {\frac{{dx}}{{1 + \varepsilon \cdot \cos x}}}$$

However I believe substitution $\displaystyle $u = \tan \frac{x}{2}$$ might be very wrong in this situation because of discontinuity.

Also I guess this may diverge when $\displaystyle $\left| \varepsilon \right| \geqslant 1$$

Unfortunately at this very moment I am not aware of a nice proper solution, maybe you have an idea?
This integral is perfect to use a little complex analysis and the residue theorem.

First note that the substitution $\displaystyle z=e^{ix} \implies dz=ie^{ix}dx \iff \frac{dz}{iz}=dx$ where $i=\sqrt{-1}$ will be used.

$\displaystyle \cos(x)=\frac{e^{ix}+e^{-ix}}{2}=\frac{z+z^{-1}}{2}$

Putting all of this in gives

$\displaystyle \int_{0}^{2\pi}\frac{dx}{1+\epsilon\cos(x)}=\oint_ {|z|=1}\frac{dz}{\left[1+\epsilon(\frac{z+z^{-1}}{2}) \right]iz}=\frac{2}{i\epsilon}\oint_{|z|=1}\frac{dz}{(z+\ frac{1-\sqrt{1-\epsilon^2}}{\epsilon})(z+\frac{1+\sqrt{1-\epsilon^2}}{\epsilon})}$

As you noted the integral only converges when $|\epsilon|<1$

So their is only one pole inside the unit disk and it occurs at

$\displaystyle z=\frac{-1+\sqrt{1-\epsilon^2}}{\epsilon}$

Now by the residue theorem we get

$\displaystyle \frac{2}{i\epsilon}\cdot 2\pi i\left( \frac{1}{\frac{-1+\sqrt{1-\epsilon^2}}{\epsilon}+\frac{1+\sqrt{1-\epsilon^2}}{\epsilon}}\right)=\frac{4\pi}{\epsilo n} \left( \frac{1}{\frac{2\sqrt{1-\epsilon^2}}{\epsilon}}\right)=\frac{2 \pi}{\sqrt{1-\epsilon^2}}$

3. TheEmptySet, thank you very much.