# Laplace Transform of sin(t)/t

1. May 11, 2010

### Char. Limit

Just out of curiosity, I've been looking into the Laplace transform and noticed that...

$$\mathcal{L}(sin(t))=\frac{1}{s^2+1}=\frac{d}{ds}(tan^{-1}(s))$$

So, I was wondering if the Laplace transform of the sinc function was the inverse tangent function. In LaTEX...

$$\mathcal{L}(sinc(t))=tan^{-1}(s)$$

In short, is the above a true statement?

2. May 11, 2010

### gabbagabbahey

Close, I think you get $\cot^{-1}(s)=\frac{\pi}{2}-\tan^{-1}(s)$ when you do the integration properly.

$$\mathcal{L}\left[\frac{f(t)}{t}\right]=\int_s^\infty F(\sigma)d\sigma$$

3. May 11, 2010

### Cyosis

You're missing a minus sign and the constant of integration. To get the correct result you will have to find that constant.

4. May 11, 2010

### Char. Limit

Wait, constant of integration? I thought that in a definite integral, the constant of integration would cancel out...

And yeah, I see that I got something wrong there gabba. I integrated from 0 to s, instead of from s to infinity. Doing it again, you're right. Here's the true statement, I guess:

$$\mathcal{L}(sinc(t))=cot^{-1}(s)$$

I love it when I put in a trig function and get out a trig function.

5. May 11, 2010

### Cyosis

Yes I was under the impression you did something else than using the identity in post #2 namely.

I thought you had noticed that if

$$f(s)=\int_0^\infty \frac{\sin (t)}{t} e^{-s t} ds$$

then

$$f'(s)=-\int_0^\infty \sin( t )e^{-s t} ds = - \frac{d}{ds} \arctan s$$

from this it follows that

$$f(s)=-\arctan s+c$$

but $f(\infty)= 0$ therefore $c=\pi/2$ and finally $f(s)=\pi/2-\arctan s=\text{arccot} s$

Edit: Whoops this post was a bit premature. I thought you finished in post #4, but seeing as you finished while I was fixing my tex I won't edit this post.

Last edited: May 11, 2010
6. May 11, 2010

### Char. Limit

And then lastly, $$F(s)=\frac{\pi}{2} - tan^{-1}(s)=cot^{-1}(s)$$, right?