The problem is as follows,
Show that
$$f(x)=\sin\left(\dfrac{1}{x}\right)-\left(\dfrac{1}{x}\right)\cos\left(\dfrac{1}{x}\right)$$
is improper integrable on $(0,1]$ and find the value of
$$\int_0^1 f(x) \,dx$$
How can i solve? I think the substitution of $\frac{1}{x}$ as $t$. Please give me a detailed explanation.
On
The main idea is already laid down by Ty above.
For a few details, and in the spirit of Riemann improper integrals, the function $f:x\mapsto \sin(x^{-1}) -x^{-1}\cos(x^{-1})$ is not continuous at zero, so a limiting process is used to define $\int^1_0f = \lim_{\varepsilon\rightarrow0^+}\int^1_\varepsilon f$. As noted by Ty, an application of the fundamental theorem of Calculus gives $$ \int^1_\varepsilon \sin(x^{-1}) -x^{-1}\cos(x^{-1})\,dx = x\sin(x^{-1})\Big|^1_\varepsilon=\sin1 -\varepsilon\sin(\varepsilon^{-1})\xrightarrow{\varepsilon\rightarrow0}\sin 1$$
since $|\varepsilon\sin(\varepsilon^{-1})|\leq\varepsilon$.
$$ \frac{d}{dx} \left( x\sin{\left(\frac{1}{x}\right)} \right)= \sin{\left(\frac{1}{x}\right)}-\frac{1}{x}\cos{\left(\frac{1}{x}\right)}$$ Therefore, $$\int_0^1 \sin{\left(\frac{1}{x}\right)}-\frac{1}{x}\cos{\left(\frac{1}{x}\right) \; dx = x\sin{\left(\frac{1}{x}\right)} \bigg \rvert_0^1=\boxed{\sin{1}}}$$