Is McLaurin Expansion the Key to Solving Integration by Expansion?

[ra_forever8]

Consider the integral
\begin{equation}
I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
\end{equation}
show that
\begin{equation}
I(x)= \frac{2x}{\pi} +O(x^{3})
\end{equation}
as $x\rightarrow0$.
=> I Have used the expansion of McLaurin series of $I(x)$ but did not work.
please help me.

 

[stainburg]

I'm not sure if the expansion correct. However, we can do it as follow.

By McLaurin expansion, we have

\(\displaystyle f(x,t)=\sin(x\sin t)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}(\sin t)^{2n-1}=\sum_{n=1}^{\infty} f_n(x,t)\).

Since we integrate \(\displaystyle f(x,t)\) on \(\displaystyle E=[0,\pi]\ni t\), we should test uniform convergence of the series, \(\displaystyle \sum_{n=1}^{\infty} f_n(x,t)\), at first.

Clearly, \(\displaystyle |f_n(x,t)|\leq \frac{|x|^{2n-1}}{(2n-1)!}\) on \(\displaystyle E\ni t\), and \(\displaystyle \sum_{n=1}^{\infty} \frac{|x|^{2n-1}}{(2n-1)!}\) converges uniformly on \(\displaystyle \mathbb{R}\ni x\). Thus, \(\displaystyle \sum_{n=1}^{\infty} f_n(x,t)\) converges uniformly on \(\displaystyle E\ni t\).Hence \(\displaystyle \frac{1}{\pi}\int_E f(x,t)dt=\frac{1}{\pi}\int_E \sum_{n=1}^{\infty}f_n(x,t)dt=\frac{1}{\pi}\sum_{n=1}^{\infty}\int_E f_n(x,t)dt=\frac{1}{\pi}\int_E x\sin tdt+O(x^3)\), as \(\displaystyle x\) goes to \(\displaystyle 0\).

I think you can complete the rest.