I'm working through a complex analysis problem and I'm wondering whether there is a straightforward way to simplify these integrals any further (or compute them):
$$ \int_0^1 \frac{e^{2{\pi}it}}{e^{e^{2{\pi}it}} - 1}\,dt,$$
and $$\int_0^1 \sin\left(\frac{1}{e^{2{\pi}it}}\right)\,dt.$$
Any help would be appreciated.
Let $z=e^{i2\pi t}$ so that $dt=\frac{1}{i2\pi z} dz$. Then,
$$\int_0^{1}\frac{e^{i2\pi t}}{e^{e^{i2\pi t}}-1}=\frac{1}{i2\pi}\oint_{|z|=1}\frac{1}{e^z-1}\,dz=1$$
Let $z=e^{-i2\pi t}$ so that $dt=-\frac{1}{i2\pi z} dz$
$$\int_0^1\sin\left(\frac{1}{e^{i2\pi t}}\right)\,dt=\frac{1}{i2\pi}\oint_{|z|=1}\sin(z)\,\frac{1}{z} dz=0$$
where the integral in traversed in the counter-clockwise direction thereby absorbing the minus sign.