Laplace transform of $\frac{sin(t)}{t}$

Question

How to compute the Laplace transform $\mathcal{L}\frac{sin(t)}{t}$? I can't use the shifting theorem because of the $t^{-1}$?

Thank you.

Answer 1

HINT:

Note that if $F(s)=\int_0^\infty \frac{\sin(t)}{t}e^{-st}\,dt$, then

$$F'(s)=-\int_0^\infty \sin(t) e^{-st}\,dt=-\frac{1}{s^2+1}$$

Now integrate $F'(s)$ and apply $\lim_{s\to \infty}F(s)=0$.

Answer 2

Another way:

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

So, we get:

$$\text{F}\left(\text{s}\right)=\mathcal{L}_t\left[\frac{\sin\left(t\right)}{t}\right]_{\left(\text{s}\right)}:=\int_0^\infty e^{-\text{s}t}\times\frac{\sin\left(t\right)}{t}\space\text{d}t=$$ $$\int_0^\infty e^{-\text{s}t}\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}t^{1+2\text{n}}}{t\left(1+2\text{n}\right)!}\space\text{d}t=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(1+2\text{n}\right)!}\int_0^\infty e^{-\text{s}t}\times\frac{t^{1+2\text{n}}}{t}\space\text{d}t=$$ $$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(1+2\text{n}\right)!}\int_0^\infty e^{-\text{s}t}t^{2\text{n}}\space\text{d}t=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}\Gamma\left(1+2\text{n}\right)}{\text{s}^{1+2\text{n}}\left(1+2\text{n}\right)!}=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{s}^{1+2\text{n}}\left(1+2\text{n}\right)}$$

Now, when $\left|\text{s}\right|=\sqrt{\Re^2\left[\text{s}\right]+\Im^2\left[\text{s}\right]}>1$:

$$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{s}^{1+2\text{n}}\left(1+2\text{n}\right)}=\arctan\left(\frac{1}{\text{s}}\right)$$

For the integral we used:

$$\int_0^\infty e^{-\text{s}t}t^{2\text{n}}\space\text{d}t=\frac{\Gamma\left(1+2\text{n}\right)}{\text{s}^{1+2\text{n}}}$$

When $\Re\left[\text{s}\right]>0\space\wedge\space\Re\left[\text{n}\right]>-\frac{1}{2}$