# Number of ways to evaluate an integral

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• February 5th 2011, 07:50 AM
wonderboy1953
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.
• February 5th 2011, 08:44 AM
TheEmptySet
Here is another way using the Laplace transform.

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

Then via the laplace transform we get

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

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

Now taking the transform above gives

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

This gives $\displaystyle \lim_{s \to 0^+}\tan^{-1}\left( \frac{1}{s}\right)=\frac{\pi}{2}$
• February 5th 2011, 09:32 AM
Drexel28
Quote:

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

This is actually something proof worthy, but you are right of course.
• February 16th 2011, 12:35 PM
uniquesailor
Solution Details By use of Laplace
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