Here's the question you clicked on:
UnkleRhaukus
\[\int_0^\infty e^{-3t}\int_0^te^{t-u}\sin(u)\text du\text dt\]
\[\begin{align*} \\\int_0^\infty e^{-3t}\int_0^te^{t-u}\sin(u)\text du\text dt&=\mathcal L\left\{\int_0^te^{t-u}\sin(u)\text du\right\}_{p\rightarrow3}\\ \\&=\mathcal L\left.\left\{\mathcal L\{e^t*\sin(u)\}\right\}\right|_{p\rightarrow3}\\ \\&=\mathcal L\left.\left\{\frac{1}{p-1}\times\frac{1}{p^2+1^2}\right\}\right|_{p\rightarrow3}\\ \\&= \end{align*}\]
I could literally mash the keyboard for 12 straight minutes, and I would understand it better than I would that.
\[u = \frac{1}{2} i (t-\log(e^{t}))\] right?
i have this , but it dosent seam right to me for some reason \[\begin{align*} \\&=\mathcal L\left\{\frac{1}{3-1}\times\frac{1}{3^2+1}\right\} \\&=\mathcal L\left\{\frac{1}{2}\times\frac{1}{10}\right\} \\&=\frac 1{20} \end{align*}\]
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. ;)
ah, this makes more sense , i see where i was getting confused now , \[\begin{align*} \int_0^\infty e^{-3t}\int_0^te^{t-u}\sin(u)\text du\text dt \\&=\mathcal L\left\{\int_0^te^{t-u}\sin(u)\text du\right\}_{p\rightarrow3} \\&=\mathcal L\left.\left\{e^t*\sin(u)\right\}\right|_{p\rightarrow3} \\&=\left(\left.\mathcal L\left\{e^t\right\}\times\mathcal L\left\{\sin(u)\right\}\right)\right|_{p\rightarrow3} \\&=\left(\left.\frac{1}{p-1}\times\frac{1}{p^2+1^2}\right)\right|_{p\rightarrow3} \\&=\frac{1}{3-1}\times\frac{1}{3^2+1}\\&=\frac{1}{2}\times\frac{1}{10}\\ \\&=\frac 1{20} \end{align*}\]