maclaurin series for this function

[iDoof]

i'm supposed to get the infinite series representation for the INTEGRAL of (sin(x))/x dx.

so i start by finding the maclaurin series for sin(x)/x right? then i need to integrate term by term, but when i try to get the first couple terms, i get undefined values (divide by 0)...does that make sense?

am i doing something wrong? if my specific issue doesn't make sense, could someone just walk through how to get the maclaurin series?

thanks so much

 

[galactus]

You could start with the MacLaurin series for sin(x).

Let's derive it:

\(\displaystyle f(x)=sin(x), f(0)=0\)

\(\displaystyle f'(x)=cos(x), f'(0)=1\)

\(\displaystyle f''(x)=-sin(x), f''(0)=0\)

\(\displaystyle f'''(x)=-cos(x), f'''(0)=-1\)

The pattern 0,1,0,-1 will repeat.

So we have, using the MacLaurin series definition:

\(\displaystyle x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+.....+(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}+.....\)

Divide by x to get:

\(\displaystyle \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k+1)}=

1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!}+....+(-1)^{k}\frac{x^

{2k}}{(2k+1)!}+....\)

Now integrate the series. Is this what you were looking for?.

 

[iDoof]

yes! thanks so much!!