# Derivative of $F(x)=\int^{1}_{0}\frac{\sin(xt)}{t}dt$

The question is: Let $$F(x)=\int^{1}_{0}\frac{\sin(xt)}{t}\,dt$$. Then for $$x\neq0$$, $$F'(x)=\frac{\sin(x)}{x}$$. I have to determine if it is true or false (and give justification). First of all, I saw that $$\frac{\sin(xt)}{t}$$ is not defined for $$t=0$$ and the integration is for $$t=0$$ to $$1$$. This leads me to believe the statement is false, but I am not sure. I tried to integrate $$F$$ with integration by parts, but it did not help. Any advise is appreciated!

## 3 Answers

The function $$\sin(xt)/t$$ is not defined at $$t=0$$ but has a finite limit there: $$\lim_{t\rightarrow 0} \frac{\sin(xt)}{t} = x$$ so there's no problem with the integral.

As for how to find $$F'(x)$$ there are two simple ways:

1. Leibniz integral rule $$\frac{d}{dx} \int_a^b f(x,t) dt = \int_a^b \frac{\partial}{\partial x}f(x,t) dt$$ $$F'(x) = \int_0^1 \frac{\partial}{\partial x}\left(\frac{\sin(xt)}{t}\right) dt = \int_0^1 \cos (xt) dt$$
2. Reparametrization $$u=xt$$, $$t=u/x$$, $$dt =du/x$$: $$F(x) = \int_0^x \frac{\sin(u)}{u/x} \frac{du}{x} = \int_0^x \frac{\sin u}{u}du$$ and fundamental theorem of calculus $$\frac{d}{dx} \int_a^x f(u) du = f(x)$$

Proof without Leibniz rule. Suppose $$x>0$$. Put $$y=tx$$. We get $$F(x)=\int_0^{x} \frac {\sin\,y} y dy$$. By Fundmanetal Theorem of Calculus we gate $$F'(x)=\frac {\sin\,x} x$$. For $$x<0$$ use the fact that $$F(-x)=-F(x)$$.