Prove convergence of $\int^{\infty}_{0}\ t^{z-1}cos(t)dt$ and $\int^{\infty}_{0}\ t^{z-1}sin(t)dt$

Question

For a complex analysis problem set I am trying to show that the integrals

$$\int^{\infty}_{0}\ t^{z-1}cos(t)dt \quad and \quad \int^{\infty}_{0}\ t^{z-1}sin(t)dt $$

is convergent for $0<Re(z)<1$ and $-1<Re(z)<1$ respectively, but am unsure of how to progress. I've tried expressing $cos(t)$ and $sin(t)$ as a Taylor series, and also expressing the integrals as shown below,

$$\int^{\infty}_{0}\ t^{z-1}cos(t)dt = \int^{\infty}_{0}\frac{e^{(z-1)ln(t)}(e^{-it}+e^{it})}{2}dt$$

$$\int^{\infty}_{0}\ t^{z-1}sin(t)dt = \int^{\infty}_{0}\frac{e^{(z-1)ln(t)}(ie^{-it}-ie^{it})}{2}dt$$

and then trying to bound the real component, but I haven't been able to make much progress. Any help in the right direction would greatly appreciated!

Answer 1

For the first integral,

1) First show that $\displaystyle \int_0^1 t^{z-1}\cos(t)dt$ is convergent (we have $|t^{z-1}\cos(t)|\leq t^{x-1}$, ($ {\rm Re}(z)=x$).

2) Let $A>1$. Integrate by parts $\displaystyle \int_1^A t^{z-1}\cos(t)dt$, (integrate $\cos(t)$), and show that the integrated part has a limit as $A\to +\infty$, and that the remaining integral is (absolutely) convergent on $[1,+\infty[$.

3) Same method for the second integral.