Question

\[\int_0^\infty e^{-3t}\int_0^te^{t-u}\sin(u)\text du\text dt\]

Answer 1

\[\begin{align*}
\\\int_0^\infty e^{-3t}\int_0^te^{t-u}\sin(u)\text du\text dt&=\mathcal L\left\{\int_0^te^{t-u}\sin(u)\text du\right\}_{p\rightarrow3}\\
\\&=\mathcal L\left.\left\{\mathcal L\{e^t*\sin(u)\}\right\}\right|_{p\rightarrow3}\\
\\&=\mathcal L\left.\left\{\frac{1}{p-1}\times\frac{1}{p^2+1^2}\right\}\right|_{p\rightarrow3}\\
\\&=
\end{align*}\]

Answer 2

I could literally mash the keyboard for 12 straight minutes, and I would understand it better than I would that.

Answer 3

\[u = \frac{1}{2} i (t-\log(e^{t}))\] right?

Answer 4

i have this , but it dosent seam right to me for some reason \[\begin{align*}
\\&=\mathcal L\left\{\frac{1}{3-1}\times\frac{1}{3^2+1}\right\}
\\&=\mathcal L\left\{\frac{1}{2}\times\frac{1}{10}\right\}
\\&=\frac 1{20}
\end{align*}\]

Answer 5

Oh, I honestly have no clue. I put it in Wolfram Alpha and it told me that was a root. It makes no sense to me, but 1/20 is a nice answer. ;)

Answer 6

ah, this makes more sense , i see where i was getting confused now ,
\[\begin{align*}
\int_0^\infty e^{-3t}\int_0^te^{t-u}\sin(u)\text du\text dt
\\&=\mathcal L\left\{\int_0^te^{t-u}\sin(u)\text du\right\}_{p\rightarrow3}
\\&=\mathcal L\left.\left\{e^t*\sin(u)\right\}\right|_{p\rightarrow3}
\\&=\left(\left.\mathcal L\left\{e^t\right\}\times\mathcal L\left\{\sin(u)\right\}\right)\right|_{p\rightarrow3}
\\&=\left(\left.\frac{1}{p-1}\times\frac{1}{p^2+1^2}\right)\right|_{p\rightarrow3}
\\&=\frac{1}{3-1}\times\frac{1}{3^2+1}\\&=\frac{1}{2}\times\frac{1}{10}\\
\\&=\frac 1{20}
\end{align*}\]