Question:
Find the number of points of discontinuity of $f(x) = \int_0^x{t\sin{\frac1t}}dt$; for $x \in (0,\pi)$
I tried using By Parts assuming $t$ as first function,
$$\begin{align} \\
\int_0^x{t\sin{\frac1t}}dt &= \frac12x^2\sin\frac1x + \int_0^x{\frac12\cos{\frac1t}}dt \\
&=\frac12x^2\sin\frac1x + \frac x2\cos\frac1x – \frac12\int_0^x \frac1t\sin{\frac1t}dt
\end{align}$$
Now, it seems to be unsolvable from here. I couldn't think of Newton-Leibniz, as it is helpful in case of differentiability. Any ideas?
The correct answer is:
0
Best Answer
The function $$t \mapsto t\sin\left(\frac{1}{t}\right)$$ is continuous on $[0,\pi]$ so by the fundamental theorem of calculus $f$ is differentiable on $[0,\pi]$ and so continuous as well.