Show $ \int_{0}^{1}\frac{\sin (ax)}{x}dx $ is an entire function.

Question

I am struggling in evaluating the following integral:

$$\int_{0}^{1}\frac{\sin (ax)}{x}dx$$

I know that if the integral is from $0$ to infinity, it will be a constant of $\pi/2$ which is analytic, but not sure when it's from 0 to 1.

Answer 1

Notice that\begin{align}\int_0^1\frac{\sin(ax)}x\,\mathrm dx&=\int_0^1\sum_{n=0}^\infty\frac{(-1)^na^{2n+1}x^{2n}}{(2n+1)!}\,\mathrm dx\\&=\sum_{n=0}^\infty\left(\int_0^1\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx\right)a^{2n+1}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)\times(2n+1)!}a^{2n+1}.\end{align}

Answer 2

You may simply notice that $$ F(a)=\int_{0}^{1}\frac{\sin(ax)}{x}\,dx \stackrel{x\mapsto\frac{z}{a}}{=}\int_{0}^{a}\frac{\sin z}{z}\,dz $$ so $F'(a)=\frac{\sin a}{a}$ is an entire function. This implies that $F(z)$ is an entire function too (the space of entire functions is closed with respect to $\frac{d}{dz}$ and $\int(\ldots)\,dz$, since these operators do not alter the radius of convergence of a power series).