Complexifying the Integral (Arthur Mattuck, MIT)

What people said

Cites elleciel · 2013-03 · Mathematics Stack Exchange · on “How to verify method used to solve integral was actually the fastest?”

I noticed that you updated your question. No, there is no preferred method to solve an integral analytically, provided an elementary solution exists. What you're most comfortable with is the fastest. I could, however, advise you on certain tricks that are useful. Since lookup is always theoretically fastest, i.e. $\mathcal{O}(1)$, you should: Memorize as many identities in Gradshteyn et al [LINK] as possible. Try to reduce the problem to a known one, e.g. using differentiation under the integral sign [LINK] , complexifying the integral [this video] , series expansion etc. Practice. I don't know of a good problem book purely on integration practice, but you could look at some problem sets in quant

Cites Ahmed S. Attaalla · 2015-12 · Mathematics Stack Exchange · on “When can we apply the method of complexifying the integral.”

Recently I was scrolling through YouTube, and saw the method of complexifying the integral [this video] . I tried some integrals out and it worked just fine. However, I tried to take it up a notch, and tried finding. $$\int \frac{e^x}{\mathrm{cos}x} dx$$ Which didn't work out. My guess was that it didn't work out because the function we are integrating is discontinuous at some points. So my question is under what circumstances can we apply the method of complexifying the integral? My work: $$=Re{\int \frac{e^x}{e^{ix}} dx}$$ $$=Re{\int e^{(1-i)x} dx}$$ $$=Re{\frac{e^{(1-i)x}}{1-i}}+c$$ With a little more algebra (and verified through wolphy I get): $$\frac{1}{2}e^x(\mathrm{sin}x+\mathrm{cos}x)+c$

Cites user2550228 · 2021-08 · Mathematics Stack Exchange · on “Self-Study: Laplace Transform $\sin(t)$ Different Results for Different Approaches”

I calculated the Laplace Transform $\mathcal{L}\{sin(t)\}$ using the complexification of the integral and using the linearity property in combination with $\mathcal{L}\{e^{iat}\} = \frac{1}{s-ia}$ and get different results. The second approach is takeing from Wiki Proofs [LINK] and is rather short: Because: $\mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\}$ and $\mathcal{L}\{e^{iat}\} = \frac{1}{s-ia}$ : $$ \begin{align} \mathcal{L}\{e^{iat}\} &= \mathcal{L}\{\cos(t) + i\sin(t)\} = \mathcal{L}\{\cos(t)\} + i\mathcal{L}\{\sin(t)\} \\ &\Rightarrow \mathcal{L}\{\sin(t)\} = IM(\mathcal{L}\{e^{iat}\}) = IM(\frac{s+ia}{s^2 + a^2}) = \frac{a}{s^2 + a^2} \end{align} $$ Thus