Presumably Simple Laplace Transform Question I have a (presumably simple) Laplace Transform problem which I'm having trouble with:
$$\mathcal L\big\{t \sinh(4t)\big\} = ?$$
How would I go about solving this? Would you please show working if possible, or alternatively point me in the right direction regarding how to go about solving this?
Also, I'm studying Electrical Engineering at University, and in my ElectEng lectures, my lecturers are referring to Laplace transforms and the Laplace domain (with regard to the frequency response of circuits) in a way which I haven't been exposed to in my mathematics courses (as we've only briefly covered Laplace). Do you know any good resources which I could look at (either on the web or in the library) to get a better understanding of Laplace transforms (and, in particular, their application to circuit analysis)?
Sorry if the last part of this question is out of the scope of this website.
Thanks in advance.
 A: These notes here look good.
The definition of the Laplace transform is $\mathcal{L}(f)(s) := \int_0^\infty e^{-st} f(t) dt$ and for $\sinh$ the following holds: $\sinh x = \frac12 (e^x -e^{-x})$. Now you put these two things together and compute 
$$\frac12 \int_0^\infty xe^{4x} e^{-xs} dx - \frac12 \int_0^\infty xe^{-4x} e^{-xs} dx$$ 
Hope this helps.
A: You got $t\operatorname{sinh}{4t}=t\frac{e^{4t}-e^{-4t}}{2}$ , so:
$$\mathcal{L}\big\{t\operatorname{sinh}4t\big\}=\tfrac{1}{2}\left(\mathcal{L}\left\{{te^{4t}}\right\}+\mathcal{L}\left\{te^{-4t}\right\}\right)$$
We know that $\mathcal{L}\big\{e^{at}f(t)\big\}=F(s-a)$. Furthermore $\mathcal{L}\left\{t\right\}=\frac{1}{s^2}$. So in our case $\mathcal{L}\left\{te^{4t}\right\}=F(s-4)=\frac{1}{(s-4)^2}$. 
Finally:$$\mathcal{L}\big\{t \text{sinh} 4t\big\}=\frac{0.5}{(s-4)^2}+\frac{0.5}{(s+4)^2}$$
A: *

*There is a generic formula to write $G(s)=\mathcal{L}\{tf(t)\}(s)$ in terms of $F(s)=\mathcal{L}\{f(t)\}(s)$; seeing it involves integration by parts. Have you covered this rule?

*Hyperbolic sine is a difference of exponentials; can you find the Laplace transform of these?

A: $\mathcal{L}\{f(t)\}' = F'(s)= \int_0^\infty\frac{d}{ds}[e^{-st} f(t)] dt = \int_0^\infty e^{-st}[-t f(t)]dt$, 
so $\mathcal{L}\{tf(t)\} = -F'(s).$
Since $\mathcal{L}\{\sinh(4t)\} = \frac{4}{s^2 - 8}$, (I "cheated" and checked a table!)
Then $\mathcal{L}\{t*\sinh(4t)\} = -\mathcal{L}\{\sinh(4t)\}' = -\frac{d}{ds}(\frac{4}{s^2 - 16})$.
In general, $\mathcal{L}\{t^nf(t)\} = (-1)^n\frac{d}{ds}^nF(s)$.
