Integrand becomes infinite when length of interval goes to zero So I want to evaluate the following integral
$$ \int_{- \pi }^{\pi } | \theta |^{ \alpha}  e^{x ( \cos \theta   -1) } d \theta$$
when $\alpha < 0$, and look at the limit as $ x \rightarrow +\infty $. 
Since $ \cos \theta -1 < 0 $ when $\theta \in [ -\pi , \pi ] \setminus\{0\}$ we have that $e^{x ( \cos \theta   -1) } \rightarrow 0 $ for all $\theta \in [ -\pi , \pi ] \setminus\{0\}$ and for $\theta = 0 $ we have $e^{x ( \cos \theta   -1) } = 1$ for all $x$. Thus if we split the integral in two parts we have 
$$  \int_{- \pi }^{\pi } | \theta |^{ \alpha}  e^{x ( \cos \theta   -1) } d \theta =  \int_{ | \theta | < \varepsilon  } | \theta |^{ \alpha}  e^{x ( \cos \theta   -1) } d \theta +  \int_{ \varepsilon \leq |\theta | \leq \pi } | \theta |^{ \alpha}  e^{x ( \cos \theta   -1) } d \theta   $$
Now the second integral goes to zero as $ x \rightarrow \infty $, but I am not sure have to handle the first integral $$ \int_{ | \theta | < \varepsilon  } | \theta |^{ \alpha}  e^{x ( \cos \theta   -1) } d \theta $$
As $\varepsilon$ goes to zero the lenght of the interval goes to zero but at the same time the inegrand becomes infinite. My guess is that the limit is zero, or that we approach some kind of delta distribution, but  I'am not sure how to proceed. 
Thanks
 A: $x \to \infty\,,\quad \alpha < 0$.
\begin{align}
\int_{-\pi}^{\pi}\left\vert\theta\right\vert^{\,\alpha}\,
{\rm e}^{x\left[\cos\left(\theta\right) - 1\right]}\,{\rm d}\theta
&=
2\int_{0}^{\pi/2}\theta^{\,\alpha}\,
{\rm e}^{-x\left[1 - \cos\left(\theta\right)\right]}\,{\rm d}\theta
+
2\int_{-\pi/2}^{0}\theta^{\,\alpha}\,
{\rm e}^{-x\left[1 + \cos\left(\theta\right)\right]}\,{\rm d}\theta
\end{align}
$$
\begin{array}{c}\hline\\
\int_{-\pi}^{\pi}\!\!\left\vert\theta\right\vert^{\,\alpha}\,
{\rm e}^{x\left[\cos\left(\theta\right) - 1\right]}\,{\rm d}\theta
=
\overbrace{2\int_{0}^{\pi/2}\!\!\!\theta^{-\left\vert\alpha\right\vert}\,
{\rm e}^{-x\left[1 - \cos\left(\theta\right)\right]}\,{\rm d}\theta}
^{\equiv\ {\cal J}_{-}}\
+\
\overbrace{2\int_{0}^{\pi/2}\!\!\!\theta^{-\left\vert\alpha\right\vert}\,
{\rm e}^{-x\left[1 + \cos\left(\theta\right)\right]}\,{\rm d}\theta}
^{\equiv\ {\cal J}_{+}}
\\ \\ \hline
\end{array}
$$
${\bf\mbox{Let's analyze the first term}}\ \left(~{\cal J}_{-}~\right)$:
When $x \to \infty$, the main contribution to the integral comes from
$\theta \gtrsim 0$ where $cos\left(\theta\right) \approx 1 - \theta^{2}/2$. Then
$$
{\cal J}_{-}
\approx
2\int_{0}^{\infty}\theta^{-\left\vert\alpha\right\vert}\,
{\rm e}^{-x\,\theta^{2}/2}\,{\rm d}\theta
$$
Whith the variable change $\theta \equiv \sqrt{2/x\,}\,t^{1/2}.\quad
{\rm d}\theta = \sqrt{2/x\,}\,\left(t^{-1/2}/2\right)\,{\rm d}t$:
$$
{\cal J}_{-}
\approx
2\int_{0}^{\infty}
\left(2 \over x\right)^{-\left\vert\alpha\right\vert/2}t^{-\left\vert\alpha\right\vert/2}
{\rm e}^{-t}
\left(2 \over x\right)^{1/2}\,{t^{-1/2} \over 2}\,{\rm d}t
=
\left(2\over x\right)^{\left(1 - \left\vert\alpha\right\vert\right)/2}
\int_{0}^{\infty}t^{-\left(1 + \left\vert\alpha\right\vert\right)/2}
{\rm e}^{-t}\,{\rm d}t
$$
$$
\begin{array}{c}\hline\\
{\cal J}_{-}
\approx
\left(2\over x\right)^{\left(1 - \left\vert\alpha\right\vert\right)/2}
\Gamma\left(1 - \left\vert\alpha\right\vert \over 2\right)
\\ \\ \hline
\end{array}
$$
${\bf\mbox{Let's consider the second term}}\ \left(~{\cal J}_{+}~\right)$:
\begin{align}
{\cal J}_{+}
&=
2\int_{-\pi/2}^{0}
\left(\theta + {\pi \over 2}\right)^{-\left\vert\alpha\right\vert}
{\rm e}^{-x\left[1 - \sin\left(\theta\right) \right]}\,{\rm d}\theta
=
2\int^{\pi/2}_{0}
\left(-\theta + {\pi \over 2}\right)^{-\left\vert\alpha\right\vert}
{\rm e}^{-x\left[1 + \sin\left(\theta\right) \right]}\,{\rm d}\theta
\\[3mm]&\approx
2\left(\pi \over 2\right)^{-\left\vert\alpha\right\vert}\,{\rm e}^{-x}
\int_{0}^{\infty}
\exp\left(\vphantom{\LARGE A}
-\left\vert\alpha\right\vert\ln\left(1 - 2\theta/\pi\right)
-
x\theta\right)\,{\rm d}\theta
\\[3mm]&\approx
2\left(\pi \over 2\right)^{-\left\vert\alpha\right\vert}\,{\rm e}^{-x}
\int_{0}^{\infty}
\exp\left(\vphantom{\Large A}
-\left[x - {2\left\vert\alpha\right\vert \over\pi}\right]\theta
\right)\,{\rm d}\theta
=
2\left(\pi \over 2\right)^{-\left\vert\alpha\right\vert}\,{\rm e}^{-x}\,
{1 \over x - 2\left\vert\alpha\right\vert/\pi}
\end{align}
$$
\begin{array}{c}\hline\\
{\cal J}_{+}
\approx
2\left(\pi \over 2\right)^{-\left\vert\alpha\right\vert}\,
{{\rm e}^{-x} \over x}
\\ \\ \hline
\end{array}
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\int_{-\pi}^{\pi}\left\vert\theta\right\vert^{\,\alpha}\,
{\rm e}^{x\left[\cos\left(\theta\right) - 1\right]}\,{\rm d}\theta
\color{#000000}{\ \approx\ }
\left(2\over x\right)^{\left(1 - \left\vert\alpha\right\vert\right)/2}
\Gamma\left(1 - \left\vert\alpha\right\vert \over 2\right)
+
2\left(\pi \over 2\right)^{-\left\vert\alpha\right\vert}\,
{{\rm e}^{-x} \over x}
\quad}
\\[3mm]\color{#0000ff}{\large%
\left\vert\alpha\right\vert < 1\,,
\qquad
x \to \infty}
\\ \\ \hline
\end{array}
$$
In order that the original integral converges, $\alpha < 0$ must satisfy $\left\vert\alpha\right\vert < 1$.
A: Notice that, the integrand is an even function, then you can deal with integral

$$ \int_{- \pi }^{\pi } | \theta |^{ \alpha}  e^{x ( \cos \theta   -1) } d \theta=2\int_{0 }^{\pi } \theta^{ \alpha}  e^{x ( \cos \theta -1) } d \theta. $$

