Proving $f(z)=\int_{0}^1\sin(zt)/t\;dt$ is an entire function with Morera's theorem

Question

This is from Bak and Newman exercise 7.11. Let $$f(z)=\int_{0}^1\frac{\sin zt}{t}dt.$$ Prove that $f(z)$ is entire from Morera's theorem.

This question will need to employ LDCT, Fubini's theorem. I will be delighted if someone can guide me step by step.

Thanks in advance for answering.

Answer 1

Let $B_R$ define the open disk of radius $R$ centered at the origin on the complex plane.

Claim $$\left|\frac{\sin w}{w}\right|\le\frac{\sinh(R)}{R}$$ for $w\in B_R$.

Proof

$$\begin{align} \left|\frac{\sin w}{w}\right|&=\left|\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)!}w^{2n}\right| \\ &\le \sum^\infty_{n=0}\frac{|w|^{2n}}{(2n+1)!} \\ &\le \sum^\infty_{n=0}\frac{R^{2n}}{(2n+1)!} \\ &=\frac{\sinh R}{R} \end{align} $$

Therefore, let $\gamma$ be a rectifiable closed curve in $B_R$, then $$\oint_\gamma\int^1_0 |z|\left|\frac{\sin zt}{zt}\right| dt |dz|\le R\cdot\frac{\sinh R}{R}\oint_{\gamma}|dz|<+\infty$$

Hence, by Fubini’s theorem, $$\oint_\gamma\int^1_0 \frac{\sin zt}{t} dt dz =\int^1_0\frac 1t\left(\oint_\gamma\sin zt\, dz\right)dt=0$$

By Monera’s theorem, $\displaystyle\int^1_0\frac{\sin zt}{t}dt$ is holomorphic on $B_R$. Since the choice of $R$ is arbitrary, it immediately follows that such function is entire.