Question

Pls help:)

Answer 1

sorry but i not can opening it - do you not can update it in a different format ?

Answer 2

(1) is good

(2) I'm being a bit picky here, but the problem statement should be
\[\large\int_s^\infty \left(\frac{u}{u^2+a^2}-\frac{u}{u^2+b^2}\right)~d\color{red}u\]
Your work is correct though, so don't worry about it.

(3) There's an easier way about this one without using the inverse transform of \(\dfrac{1}{s}\), but one thing that comes to mind is the convolution theorem
\[\mathcal{L}\left\{\int_0^tf(t-\tau)g(\tau)~d\tau\right\}=F(s)G(s)\]
From this, you can set \(F(s)=G(s)=\dfrac{1}{s}\), so that
\[\mathcal{L}\left\{\int_0^tf(t-\tau)g(\tau)~d\tau\right\}=\frac{1}{s^2}\]
with \(f(t)=g(t)=1\), so
\[\mathcal{L}\left\{\int_0^t1\times1~d\tau\right\}=\mathcal{L}\left\{\left[\tau\right]_0^t\right\}=\mathcal{L}\left\{t\right\}=\frac{1}{s^2}\]
Next, set \(F(s)=\dfrac{1}{s}\) and \(G(s)=\dfrac{1}{s^2}\), then
\[\mathcal{L}\left\{\int_0^tf(t-\tau)g(\tau)~d\tau\right\}=\frac{1}{s^3}\]
with \(g(t)=1\) and \(f(t)=t\).
\[\mathcal{L}\left\{\int_0^tt~d\tau\right\}=\frac{1}{2}\mathcal{L}\left\{\left[\tau^2\right]_0^t\right\}\iff\mathcal{L}\left\{t^2\right\}=\frac{2}{s^3}\]
You can use induction to show that
\[\frac{1}{n!}\mathcal{L}\left\{t^n\right\}=\frac{1}{s^{n+1}}~~\iff~~\frac{1}{(n-1)!}\mathcal{L}\left\{t^{n-1}\right\}=\frac{1}{s^n}\]
and so
\[\frac{t^{n-1}}{(n-1)!}=\mathcal{L}^{-1}\left\{\frac{1}{s^n}\right\}\]
(This is usually listed in a table of transforms, which would be the easier method to establish this.)

Then apply the shifting theorem and you're done.

Answer 3

thnx a lot. but how do i do it without using convolution theorem.

Answer 4

@myininaya @ganeshie8