Prove that $\int\limits_0^\pi {\frac{{\sin \left( {xt} \right)}} {t} \mathrm dt} $ is continuous How can I prove that this function is continuous? $$
f\left( x \right) = \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt} 
$$
Some hint? 
Don´t consider the zero in the endpoint of the integration zone, just take it as a limit $$
f\left( x \right) = \mathop {\lim }\limits_{\varepsilon ^ +   \to 0} \int\limits_\varepsilon ^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt} 
$$
How can I do it? DX!
 A: First of all, observe that
$$
\lim_{t\to0}\frac{\sin(x\,t)}{t}=x\ ,
$$
so that the integral exists as a bona fide Riemann integral. Next, given $x,y\in\mathbb{R}$,
$$
|f(x)-f(y)|\le\int_0^{\pi}\frac{|\sin(x\,t)-\sin(y\,t)|}{t}\,dt.
$$
Now use the inequality $|\sin a-\sin b|\le\dots$ to conclude that $f$ is continuous.
A: $$I(x)=\int_0^\pi\frac{\sin(xt)}{t}dt$$
$$I'(x)=\int_0^\pi\cos(xt)dt$$
now $u=xt,du=xdt\Rightarrow dt=\frac{du}{x}$ and so:
$$I'(x)=\int_0^{\pi x}\frac{\cos(u)}{x}du=\left[\frac{\sin(u)}{x}\right]_{u=0}^{\pi x}$$
$$I'(x)=\frac{\sin(\pi x)}{x}$$
and it is clear that $I'(x)$ is continuous for $x>0$ and then try taking the limit:
$$\lim_{x\to0^+}\frac{\sin(\pi x)}{x}=\pi$$
so the derivative exists for all $x\in\mathbb{R}$. One other thing that you have to look at is the value of $I(0)$. we can do this:
$$I(0)=\lim_{x\to 0}\int_0^\pi\frac{\sin(xt)}{t}dt=\int_0^\pi\lim_{x\to0}\frac{\sin(xt)}{t}dt=\int_0^\pi\lim_{x\to0}\frac{xt}{t}dt=0$$
and so $I(x)\in C(\mathbb{R})$
A: $f\left( x \right) 
= \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt}
= \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{xt} x\mathrm dt}
= \int\limits_0^{x\pi}  {\frac{{\sin \left( {t} \right)}}
{t} \mathrm dt}
$
and the integral of a continuous function is continuous.
