December 13th
Today I learned about Laplace integration. Recall that the Laplace transform $\mathcal L\{f(t)\}$ is a function in $s$ defined by\[\mathcal L\{f(t)\}=\int_0^\infty f(t)e^{-st}\,dt.\]Now the idea is that we can integrate $F(s)$ to be\[\int_0^\infty\mathcal L\{f(t)\}\,ds=\int_0^\infty\left(\int_0^\infty f(t)e^{-st}\,dt\right)ds=\int_0^\infty\left(\int_0^\infty e^{-st}\,ds\right)f(t)\,dt\]by waving our hands. The integral evaluates to $\frac1t,$ so we are left with\[\int_0^\infty\mathcal L\{f(t)\}\,ds=\int_0^\infty\frac{f(t)}t\,dt.\]Of course, this is just a manifestation of the fact that $\mathcal L\{t\cdot f(t)\}=-\frac d{ds}\mathcal L\{f(t)\}.$
As an example of doing this something interest, we can recall that $\mathcal L\{\sin(t)\}=\frac1{1+t^2}$ (say, write $\sin(t)=\frac1{2i}\left(e^{it}+e^{-it}\right)$), we find that\[\int_0^\infty\frac{\sin(t)}t\,dt=\int_0^\infty\frac1{1+t^2}\,dt=\arctan(t)\bigg|_0^\infty=\frac\pi2.\]Normally this integral is computed with the Residue theorem or some other technique, but it's quite painless with this theory, which is nice.