Integral using Laplace Transform: int [0,infty][sin t/t] dt

[kunalshdr]

Find using Laplace transform

integration(0,infinity) (sin t / t ) dt.

What i did to solve the question.

Let I = integration(0,infinity) (sin t / t ) dt
Then i took Laplace transform on both sides and using Laplace transform definition RHS was transformed into a double integral. It was like this,

integration(0,infinity) e^-st (integration(0,infinity) (sin t / t ) dt)dt

then i took 1/t in outer integration sign and e^-st in inner. Inner part became L(sin t).

I put value of L(sin t) in integration part and took the constant out. i was left with only
integration(0,infinity) (1/t)

(Of course constant was still there). How to solve this integral. I hope there is nothing wrong in my procedure as i m feeling that this integral can be evaluated.

 

[royhaas]

Re: Integral using Laplace Transform

Start with the Laplace transform of the sine function itself. Then form the double integral \(\displaystyle \int_0^\infty e^{-st}\int_0^\infty \sin(t) dt ds\).

 

[kunalshdr]

Re: Integral using Laplace Transform

not able to understand wat u told

 

[galactus]

Re: Integral using Laplace Transform

\(\displaystyle f(t)=\frac{sin(t)}{t}\)

\(\displaystyle L(f(t))=tan^{-1}(\frac{1}{s})\)

\(\displaystyle \lim_{s\rightarrow{0}}tan^{-1}(\frac{1}{s})=\frac{\pi}{2}\)