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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!

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Hint: Leibniz rule: $$\frac{d}{dt}\int_a^b f(t,x)\,dx = \int_a^b \frac{\partial f}{dt}(t,x)\,dx$$

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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 $$ \frac{d}{dx} \int_a^x f(u) du = f(x)$$
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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 $F(-x)=-F(x)$.

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