- #1
wel
Gold Member
- 36
- 0
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 [itex]x\rightarrow0[/itex].
=> [itex] sin(x.sint)= x.sint - \frac{(x-sint)^3}{3!}+...[/itex]
and integrate term by term should give
[itex] - x.sint - \frac{1}{12}(cos3t-9cost)+...[/itex]
when substituting $t=\pi$ and $t=0$ something else comes up.
please help me.
\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 [itex]x\rightarrow0[/itex].
=> [itex] sin(x.sint)= x.sint - \frac{(x-sint)^3}{3!}+...[/itex]
and integrate term by term should give
[itex] - x.sint - \frac{1}{12}(cos3t-9cost)+...[/itex]
when substituting $t=\pi$ and $t=0$ something else comes up.
please help me.