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I'm a student who just got done with their calculus sequence and will be taking differential equations next semester. I already know of quite a few methods of solving these including using the Laplace transform. My problem is that I prefer to be able to derive a result versus going by a table of Laplace transforms. I ended up reading a bit on the Laplace inversion formula, but I couldn't find a derivation, and the formula itself looks like it will require some level of understanding of complex analysis, a class I have yet to have taken. I was wondering if anyone could provide a derivation and maybe an explanation for the parts of which may be a bit beyond the scope of the typical calculus sequence.

Mittens
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Jacob Hulse
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    My only advice to you is that this attitude will not be helpful. You'll have practice to recognizing patterns in mathematics in order to arrive at answers to problems, some of which is outside-the-box thinking where the head on approach is unenlightening and provides no theoretical understanding. I'm not saying it wouldn't be useful to see the derivation of the Laplace inversion formula (but a solid grasp of complex analysis would be useful, I suspect you would not be happy with someone yelling "Residue Theorem"), but you need to change your perspective on what it means to "derive a result". – Ninad Munshi Dec 28 '19 at 23:53
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    I understand that this kind of thinking may not always be helpful, but seeing a derivation would more beneficial than using the formula or tables without really being sure of what you're actually doing. I do fully intend to take a class in complex analysis in the future and try to derive this myself, but at the moment, I'm only a senior in high school and I'm taking these courses at a university that is 50 miles away from where I live so I don't have time for a whole lot of reading between the university courses, school, and work. – Jacob Hulse Dec 29 '19 at 00:19
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    That's exactly the point I was making, though. Using a table $\neq$ not being sure of what you are doing. When you are taking the class, you'll see that the table is not as unintuitive or as rote as you think it is right now. Trust me, I was in almost the exact same boat when I was a senior in high school, and I had the same attitude you did towards rote looking things. But there is beauty in the "tables" because it involves theoretical linear algebra. – Ninad Munshi Dec 29 '19 at 00:54
  • anything unclear, cc @NinadMunshi – reuns Dec 30 '19 at 19:34

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The inverse Laplace transform has another name : the inverse Fourier transform. $$F(s)= \int_0^\infty f(x)e^{-sx}dx$$ Assume $f$ is bounded and continuously differentiable with $f'$ bounded. For $\sigma> 0$

$$\int_{-A}^A F(\sigma+it) e^{i t x}dt = \int_0^\infty f(y)e^{-\sigma y} \int_{-A}^A e^{it(x-y)}dt dy = \int_0^\infty f(y)e^{-\sigma y} 2A\frac{\sin( A (y-x))}{ A(y-x)}dy$$

$$ = f(0) g(-Ax) - \int_0^\infty (f(y)e^{-\sigma y})' g(A(y-x))dy$$ where $g(x) = \int_{-\infty}^x 2\frac{\sin(y)}{y}dy$ and the last step was an integration by parts.

Since $g$ is bounded, continuous and $g(\infty)$ exists and $g(-\infty)=0$ we get that for $x > 0$ $$\lim_{A\to \infty} \int_{-A}^A F(\sigma+it) e^{i t x}dt =\lim_{A\to \infty} f(0) g(-Ax) - \int_0^\infty (f(y)e^{-\sigma y})' g(A(y-x))dy$$ $$= f(0)g(-\infty) - \int_x^\infty (f(y)e^{-\sigma y})' g(\infty)dy=f(x)e^{-\sigma x}g(\infty)$$ And the Fourier/Laplace inversion theorem follows from $g(\infty)=2\pi$.

reuns
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