Computation of $\int _{-\pi} ^\pi \frac {e^{in\theta} - e^{i(n-1)\theta}} {\mid \sin {\theta} \mid} d\theta .$ I need to compute
$$\int \limits _{-\pi} ^\pi \frac {e^{in\theta} - e^{i(n-1)\theta}} {\mid \sin {\theta} \mid} d\theta .$$
Does anyone see any good strategy? Thanks.
 A: It seems to me that
$$\lim_{\theta\rightarrow\pi^-}\left(e^{in\theta}-e^{i(n-1)\theta}\right)=\lim_{\theta\rightarrow\pi^-}e^{in\theta}(1-e^{-i\theta})=2e^{in\pi}\ne0$$
So the integral diverges because
$$\lim_{\theta\rightarrow\pi^-}\frac{|\sin\theta|}{(\pi-\theta)}=1$$
So the integrand looks like
$$\frac{2e^{in\theta}}{\pi-\theta}$$
As $\theta\rightarrow\pi^-$.
A: A possible integration strategy could be derived from putting
$$
\eqalign{
  & e^{\,i\,n\,\theta }  - e^{\,i\,\left( {n - 1} \right)\,\theta }  = e^{\,i\,\left( {n - {1 \over 2} + {1 \over 2}} \right)\,\theta }  - e^{\,i\,\left( {n - {1 \over 2} - {1 \over 2}} \right)\,\theta }  =   \cr 
  &  = e^{\,i\,\left( {n - {1 \over 2}} \right)\,\theta } \left( {e^{\,i\,{1 \over 2}\,\theta }  - e^{\, - i\,{1 \over 2}\,\theta } } \right) = 2\,i\,e^{\,i\,\left( {2n - 1} \right){\theta  \over 2}\,} \sin \left( {{\theta  \over 2}} \right) \cr} 
$$
Thereby reaching to:
$$
\eqalign{
  & I = \int_{ - \pi }^\pi  {{{e^{\,i\,n\,\theta }  - e^{\,i\,\left( {n - 1} \right)\,\theta } } \over {\left| {\sin \theta } \right|}}\,d\theta }  = \int_{ - \pi }^\pi  {{{2\,i\,e^{\,i\,\left( {2n - 1} \right){\theta  \over 2}\,} \sin \left( {{\theta  \over 2}} \right)} \over {\left| {2\sin \left( {{\theta  \over 2}} \right)\cos \left( {{\theta  \over 2}} \right)} \right|}}\,d\theta }  =   \cr 
  &  = 2\,i\,\int_{ - \pi /2}^{\pi /2} {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} \sin \beta } \over {\left| {\sin \beta \cos \beta } \right|}}\,d\beta }  = 2\,i\,\int_{ - \pi /2}^{\pi /2} {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} {\rm sign}\left( {\sin \beta } \right)} \over {\cos \beta }}\,d\beta }  =   \cr 
  &  = 2\,i\,\left( {\int_0^{\pi /2} {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} } \over {\cos \beta }}\,d\beta }  - \int_{ - \pi /2}^0 {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} } \over {\cos \beta }}\,d\beta } } \right) =   \cr 
  &  = 2\,i\,\left( {\int_0^{\pi /2} {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} } \over {\cos \beta }}\,d\beta }  + \int_0^{ - \pi /2} {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} } \over {\cos \beta }}\,d\beta } } \right) =   \cr 
  &  = 2\,i\,\left( {\int_0^{\pi /2} {{{e^{\,i\,\left( {2n - 1} \right)\beta \,} } \over {\cos \beta }}\,d\beta }  - \int_0^{\pi /2} {{{e^{\, - i\,\left( {2n - 1} \right)\beta \,} } \over {\cos \beta }}\,d\beta } } \right) =   \cr 
  &  =  - 4\int_0^{\pi /2} {{{\sin \left( {\left( {2n - 1} \right)\beta } \right)} \over {\cos \beta }}\,d\beta }  \cr} 
$$
From here various methods can be applied, e.g. :
$$
\eqalign{
  & I =  - 4\int_0^{\pi /2} {{{\sin \left( {\left( {2n - 1} \right)\beta } \right)} \over {\cos \beta }}\,d\beta }  =   \cr 
  &  =  - 4\int_0^{\pi /2} {{{\sin \left( {2n\beta } \right)\cos \left( \beta  \right) - \cos \left( {2n\beta } \right)\sin \left( \beta  \right)} \over {\cos \beta }}\,d\beta }  =   \cr 
  &  =  - 4\int_0^{\pi /2} {\sin \left( {2n\beta } \right)\,d\beta }  - 4\int_{\beta  = 0}^{\pi /2} {{{\cos \left( {2n\beta } \right)} \over {\cos \beta }}\,d\cos \beta }  \cr} 
$$
from which you see that you end with expressions that contain the term 
$
\ln \left( {\cos \beta } \right)
$
thus confirming the answer above that the integral does not converge.
