1. ## Integrations

Consider the integral

I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt

show that

I(x)= \frac{2x}{\pi} +O(x^{3})

as $x\rightarrow0$.
=> I Have used the expansion of McLaurin series of $I(x)$ but did not work.

2. ## Re: Integrations

it does work.

What is your Maclaurin expansion of $\sin(x \sin(t))$ ?

3. ## Re: Integrations

Originally Posted by romsek
it does work.

What is your Maclaurin expansion of $\sin(x \sin(t))$ ?
$sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$,
then I guess I have to integrate term by term.
$I(0)=0$
integrate $I(x)= x sint$ at $t =\pi$ and $t=0$ gives the answer but I really don't know why and how?

4. ## Re: Integrations

Originally Posted by grandy
$sin(x\cdot sint) = x\cdot sint - \dfrac{(x\cdot sint)^3}{3!} + ...$,
then I guess I have to integrate term by term.
$I(0)=0$
integrate $I(x)= x sint$ at $t =\pi$ and $t=0$ gives the answer but I really don't know why and how?
basically as $x \to 0, \sin(x \sin(t))$ can be closely approximated by it's Maclaurin series in x

which in this case is simply $\sin(t) x+O(x^3)$

we drop the $O(x^3)$ part and integrate

$\dfrac 1 \pi \displaystyle{\int_0^{\pi}} \sin(t)x~dt=\dfrac x \pi \cos(t)|_{\pi}^0=\dfrac {2x}{\pi}$