Question

Y = ∫e^(−t) . sin^2(t) . 1/ t with laplace

Answer 1

I'm guessing your supposed to compute the value of the integral

`I=\displaystyle\int_0^\infty\frac{e^(-t)\sin^2t}t\,\mathrm dt`

by way of Laplace transform.

The integral is equivalent to the transform of
`\frac{\sin^2t}t` when
`s=1`, since

`\mathscr L_s\left\{\frac{\sin^2t}t\right\}=\displaystyle\int_0^\infty\frac{\sin^2t}te^(-st)\,\mathrm dt`

Recall the frequency-domain integration property of the transform:

`\mathscr L_s\left\{\frac{f(t)}t\right\}=\displaystyle\int_s^\infty F(\sigma)\,\mathrm d\sigma`

where
`F(s)` is the Laplace transform of
`f(t)`.

Let
`f(t)=\sin^2t=\frac{1-\cos(2t)}2`; then

`F(s)=\frac12\left(\frac1s-\frac s{s^2+4}\right)`

Next, it follows that

`\mathscr L_s\left\{\frac{\sin^2t}t\right\}=\displaystyle\int_s^\infty F(\sigma)\,\mathrm d\sigma`

`\displaystyle=\frac{\ln\sigma^2-\ln(\sigma^2+4)}4\bigg|_(\sigma=s)^(\sigma\to\infty)`

`\displaystyle=\frac14\ln\left(1+\frac4{s^2}\right)`

Get the value of
`I` by substituting
`s=1`:

`\displaystyle\int_0^\infty\frac{e^(-t)\sin^2t}t\,\mathrm dt=\boxed{\frac{\ln5}4}`