We can show that [tex] \int_{0}^\infty e^{-kx}dx=\frac{1}{k} [/tex] for real $$k>0.$$ Does this result hold for $$\Re k>0$$ belonging to complex numbers? The reason I have this question is because $$i\times\infty$$ is not $$\infty$$ and so u substitution would not work.
Let's try this. We'll extend k to a+ib, and see what happens to our integral: [tex]e^{-(a+ib)x} = e^{-a x} e^{-i b x} = e^{-a x} cos(b x) - i e^{-a x} sin(b x)[/tex] Integrating this, we get the following two integrals: [tex]\int_0^\infty e^{-a x} cos(b x) dx = \frac{a}{a^2 + b^2}[/tex] and [tex]\int_0^\infty e^{-a x} sin(b x) dx = \frac{b}{a^2 + b^2}[/tex] Summing these two, we get [itex]\frac{a - i b}{a^2 + b^2}[/itex], or 1/(a + i b). Note that to get this, we DID assume that Re(k)>0, and we got as our answer 1/k. So we can say that e^(-k x), integrated from 0 to infinity, will give 1/k, where k is any complex number with real part greater than zero. In other words, yes, that's correct.
Thanks for responding I talked to my professor and he said that if you look at the Riemann sphere, you could just assume that i*infinity is equal to infinity. The reason I got confused was because there is a similar notation which appears in the formula for the inverse Laplace transform.