Integration of functions of Complex Variablesby Charles49 Tags: complex, functions, integration, variables 

#1
Feb2212, 12:11 PM

P: 87

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. 



#2
Feb2212, 01:40 PM

PF Gold
P: 1,930

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. 



#3
Feb2412, 05:02 PM

P: 87

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. 


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