# Show that there exists an $\alpha \in (0,1)$ such that $\int_{0}^{\pi} x^{\alpha}\sin x dx=3$

Show that there exists an $\alpha \in (0,1)$ such that $\int_{0}^{\pi} x^{\alpha}\sin x dx=3$

By using Mean value theorem for integrals, I got that $\exists c\in(0,\pi) \text{such that}\quad c^{\alpha}\sin c=\pi/3$

After that I am not able to proceed. Is there any other easy method for this problem? Help is needed. Thanks in advance.

Let $f(x,t)=x^t\sin x$, it is continuous with respect to both $x$ and $t$ over $[0,\pi]$ and $[0,1]$ respectively and is dominated by $\phi :t \mapsto \pi$. Therefore, $g: t \mapsto \int_0^\pi f(x,t) dx$ is continuous over $[0,1]$ and since $g(0)=2$ and $g(1)=\pi$, you may conclude using the intermediate value theorem.
• Have a look at this post, domination is one of the hypotheses of the Dominated Convergence Theorem that yields continuity of $g$. – Bill O'Haran May 29 '18 at 11:45