Integral $\int_{0}^{i}z\sin{z} \operatorname dz$ Find this complex integral $$I=\int_{0}^{i}z\sin{z}dz=\dfrac{5\pi^2}{96}?$$ where $i^2=-1$ My try: use $$\sin{z}=\dfrac{e^{iz}-e^{-iz}}{2i}$$ so $$I=\dfrac{1}{2i}\int_{0}^{i}z(e^{iz}-e^{-iz})dz$$ My idea is true? Thank you Thank you Christian Blatter help.and I have $$I=\int_{0}^{i}z\sin{z}dz=\int_{0}^{i}zd(-\cos{z})=z(-\cos{z})|_{0}^{i}+\int_{0}^{i}\cos{z}dz=-i\cos{i}+\sin{i}$$ so $$I=-i\cos{i}+\sin{i}$$ can have simple form?

A hint: Find a primitive $F$ of the function $f(z):=z\sin z$. To this end find a primitive of the real function $f(x):=x\sin x$ in the usual way and check whether its "complexified version" works. Your integral is then simply $=F(i)-F(0)$.