Calculus!

For a function to have a discontinuity at a point, its derivative must not exist on that point. Here, we are given $f(x)=\int_{0}^{x}tsin\frac{1}{t}dt$ We compute its derivative by using the First Fundamental Theorem of Calculus $\Rightarrow f'(x)=\frac{d}{dx}\int_{0}^{x}tsin\frac{1}{t}dt$ $\Rightarrow f'(x)=xsin\frac{1}{x}$ $xsin\frac{1}{x}$ is defined $\forall x\in (0,\pi)$ $\therefore f(x)$ doesn't have any discontinuity at $(0,\pi)$ .