# Thread: Number of ways to evaluate an integral

1. I know I'm not being explicit and I may be coinciding with somebody's preceding post (sorry about that if it is), but it seems to me to be the type of problem you can apply polar coordinates to to get a solution.

2. Here is another way using the Laplace transform.

$\displaystyle \displaystyle f(t)=\frac{\sin(t)}{t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(2n+1)!}$

Then via the laplace transform we get

$\displaystyle \displaystyle F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}e^{-st}dt=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int_{0}^{\infty}t^{2n}e^{-st}dt$

Notice that $\displaystyle \displaystyle \lim_{s \to 0^{+}}F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt$

Now taking the transform above gives

$\displaystyle \displaystyle F(s)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left( \frac{(2n)!}{s^{2n+1}}\right)=\sum_{n=0}^{\infty}\ frac{(-1)^n}{(2n+1)}\left( \frac{1}{s^{2n+1}}\right)=\tan^{-1}\left( \frac{1}{s}\right)$

This gives $\displaystyle \displaystyle \lim_{s \to 0^+}\tan^{-1}\left( \frac{1}{s}\right)=\frac{\pi}{2}$

3. Originally Posted by TheEmptySet
Notice that $\displaystyle \displaystyle \lim_{s \to 0^{+}}F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt$
This is actually something proof worthy, but you are right of course.

4. ## Solution Details By use of Laplace

Page 2 of 2 First 12