Find the limit and derivative of integral function. $\psi_m(x)$ is defined as $$\int_0^{\ln|x|}e^{mt}\sin(t)^m\mathop{dt}$$ with $m$ a natural number greater then zero. Now the question is, does $\lim\limits_{x\to 0}\psi_m(x)$ exist. I've tried using the squeeze theorem using $-1\leq \sin(t)^m\leq 1$ and this resulted in $\frac{-1}{m}\leq\psi_m(x)\leq\frac{1}{m}$, which isn't useful.
Another part of the question was to determine the derivative. But I have no idea how, since the variable $x$, is in the limit of integration. I tried doing integration by parts, in hope of finding recursion, but this didn't really work (integrated by parts twice).
Any ideas how to solve this?
 A: Since $e^{mt}\sin(t)^m$ is a $L^1(\mathbb{R}^-)$ function, we just have to compute:
$$I_m=\int_{0}^{+\infty}\sin(t)^m e^{-mt}dt.$$
Notice that we have:
$$\int_{0}^{+\infty}e^{i\eta t}\,e^{-mt}\,dt=\frac{1}{m-i\eta}\tag{1}$$
while the binomial theorem gives:
$$\sin(t)^m = \frac{1}{(2i)^m}\sum_{j=0}^{m}\binom{m}{j}(-1)^j e^{(m-2j)it}\tag{2}$$
so it follows that:
$$ I_m = \frac{1}{(2i)^m}\sum_{j=0}^{m}\binom{m}{j}\frac{(-1)^j}{m-(m-2j)i}$$
or:

$$ I_{2m} = \frac{1}{4^m}\left(\frac{1}{2m}\binom{2m}{m}+(-1)^m\sum_{j=0}^{m-1}\binom{2m}{j}\frac{(-1)^j m}{m^2+(m-j)^2}\right),$$
$$ I_{2m+1} = \frac{(-1)^m}{2\cdot 4^m}\sum_{j=0}^{m}\binom{2m+1}{j}\frac{(-1)^j(2m-2j+1)}{j^2+(2m+1-j)^2}.\tag{3}$$

So for any $m\in\mathbb{N}$ we have that $\lim_{x\to 0^+}\psi_m(x)$ exists and equals a rational number:
$$ \lim_{x\to 0^+}\psi_m(x) = (-1)^m\, I_m.$$
A: As a hint:
$$\left|\int\limits_0^{\log|x|}e^{mt}\sin^mt\,dt\right|\le\int\limits_0^{\log |x|}e^{mt}dt=\left.\frac1me^{mt}\right|_0^{\log|x|}=\frac1m\left(|x|^m-1\right)\xrightarrow[x\to 0]{}...?$$
A: Basically you are looking for evaluating the integral
$$I = \int_{0}^{-\infty} e^{mt}\sin(t)^m\, dt. $$
A possible closed form is
$$I= {\frac{\left( -1/2-i/2 \right) \Gamma( m ) \Gamma\left( -m/2 + 1-im/2\right)  }{( -2\,i )^{m}\,\Gamma( -im/2+m/2+1 ) }}, \quad i=\sqrt{-1}. $$
Here some special values for $m=1,2,3,4,5$
$$ \left\{ \frac{1}{2},-\frac{1}{8},\frac{1}{30},-{\frac {3}{320}},{\frac {3}{1105}} \right\} $$
