# Find $\int_0^\infty \int_0^t e^{-t}\frac{\sin(\tau)}{\tau}d\tau dt$ using laplace transform.

Find $$\int_0^\infty \int_0^t e^{-t}\frac{\sin(\tau)}{\tau}d\tau dt$$ using laplace transform

My attempt: I know that $$\int_0^t f(\tau)d\tau=\mathscr{L}^{-1}(\frac{F(s)}{s}).$$ Consider the integral $$\int_0^\infty \int_0^t e^{-t}\frac{\sin(\tau)}{\tau}d\tau dt=\int_0^\infty e^{-t} \int_0^t\frac{\sin(\tau)}{\tau}d\tau dt.$$ Now let $$I=\int_0^t\frac{\sin(\tau)}{\tau}d\tau=\mathscr{L}^{-1}(\frac{F(s)}{s}).$$

Here $$F(s)=\int_s^\infty \frac{1}{1+u^2}du= 1-tan^{-1}(s) (\because \mathscr{L}(\frac{f(t)}{t})=\int_s^\infty \mathscr{L}(f(t))(u) du.$$

Hence $$I=\int_0^t\frac{\sin(\tau)}{\tau}d\tau=\mathscr{L}^{-1}(\frac{1-tan^{-1}(s)}{s})=\mathscr{L}^{-1}(\frac{1}{s}-\frac{tan^{-1}(s)}{s})=1-\mathscr{L}^{-1}(\frac{tan^{1}(s)}{s}).$$

$$\mathscr{L}^{-1}(\frac{tan^{1}(s)}{s})=\int_0^ t \mathscr{L}^{-1}(tan^{1}(s))(u)du.$$ I don't know how to find $$\mathscr{L}^{-1}(tan^{1}(s)).$$ Is there any shorter way to evaluate the above iterated integral using laplace transform?

• Is there a typo? is not $\mathcal{L}^{-1}\left( \frac{\tan^{-1}(s)}{s}\right)$?
– user798113
Commented Dec 23, 2020 at 6:53
• No@АлександрПальма Commented Dec 23, 2020 at 16:23

Well, we know that the Laplace transform is defined as:

$$\text{Y}\left(\text{s}\right)=\mathscr{L}_x\left[\text{y}\left(x\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{y}\left(x\right)\exp\left(-\text{s}x\right)\space\text{d}x\tag1$$

In your case we are trying to find:

$$\mathcal{I}:=\left.\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}\right\vert_{\text{s}\space=\space1}\tag2$$

Using the time-domain integration property of the Laplace transform, we know that:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\mathscr{L}_x\left[\frac{\sin\left(x\right)}{x}\right]_{\left(\text{s}\right)}\tag3$$

Using the frequency-domain integration property of the Laplace transform, we know that:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\int_\text{s}^\infty\mathscr{L}_x\left[\sin\left(x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag4$$

Using the table of selected Laplace transforms, we know that:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\int_\text{s}^\infty\frac{1}{1+\sigma^2}\space\text{d}\sigma\tag5$$

We know that this integral represent the inverse tangent, so:

$$\mathscr{L}_x\left[\int_0^x\frac{\sin\left(\tau\right)}{\tau}\space\text{d}\tau\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\cdot\lim_{\text{n}\to\infty}\left[\arctan\left(\sigma\right)\right]_\text{s}^\infty=$$ $$\frac{1}{\text{s}}\cdot\lim_{\text{n}\to\infty}\left(\arctan\left(\text{n}\right)-\arctan\left(\text{s}\right)\right)=\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\text{s}\right)\right)\tag6$$

So, for your problem we get:

$$\mathcal{I}=\left.\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\text{s}\right)\right)\right\vert_{\text{s}\space=\space1}=\frac{1}{1}\cdot\left(\frac{\pi}{2}-\arctan\left(1\right)\right)=$$ $$\frac{1}{1}\cdot\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{4}\tag7$$

Using the same method as above, we can show that:

$$\mathscr{S}_\omega\left(\text{s}\right):=\int_0^\infty\int_0^t\exp\left(-\text{s}t\right)\cdot\frac{\sin\left(\omega\tau\right)}{\tau}\space\text{d}\tau\space\text{d}t=\frac{\omega}{\text{s}}\cdot\int_\text{s}^\infty\frac{1}{\omega^2+\sigma^2}\space\text{d}\sigma\tag8$$

Subsitute $$\text{u}:=\frac{\text{s}}{\omega}$$, so we get (assuming that $$\omega>0$$):

$$\mathscr{S}_\omega\left(\text{s}\right)=\frac{1}{\text{s}}\cdot\int_\frac{\text{s}}{\omega}^\infty\frac{1}{1+\text{u}^2}\space\text{du}=\frac{1}{\text{s}}\cdot\lim_{\text{p}\to\infty}\left[\arctan\left(\text{u}\right)\right]_\frac{\text{s}}{\omega}^\text{p}=$$ $$\frac{1}{\text{s}}\cdot\lim_{\text{p}\to\infty}\left(\arctan\left(\text{p}\right)-\arctan\left(\frac{\text{s}}{\omega}\right)\right)=\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\frac{\text{s}}{\omega}\right)\right)\tag9$$

So:

$$\mathscr{S}_\omega\left(\text{s}\right):=\int_0^\infty\int_0^t\exp\left(-\text{s}t\right)\cdot\frac{\sin\left(\omega\tau\right)}{\tau}\space\text{d}\tau\space\text{d}t=\frac{1}{\text{s}}\cdot\left(\frac{\pi}{2}-\arctan\left(\frac{\text{s}}{\omega}\right)\right)\tag{10}$$