Can Contour Integrals Simplify Real Antiderivative Calculations?

[ice109]

so suppose i wanted to calculate the antiderivatives of e^x\sin{x} and just for the hell of it also e^x\cos{x}. well i could perform integration by parts twice recognize that the original integral when it reappears, subtract from one side to the other blah blah blah.

or i could pervert a contour integral: i could integrate e^z along the line in the (1,i) direction:

e^{(1+i)x}=e^{x+ix}=e^x\cos{x}+ie^x\sin{x}

and leave off the limits

\int e^{(1+i)x}dx=\frac{1}{1+i}e^{(1+i)x}=\frac{1}{2}\left(e^x(\cos{x}+\sin{x})+ie^x(\sin{x}-\cos{x})\right)

and compare the real and imaginary parts of the integrand and the "antiderivative" and conclude:

\int e^x \cos{x} = \frac{1}{2}e^x(\cos{x}+\sin{x})
\int e^x \sin{x} = \frac{1}{2}e^x(\sin{x}-\cos{x})

which are of course the correct answers.

what i don't know is what an indefinite integral in the complex plane even is. I've only so far learned that i can use this "trick" to perform contour integrals <=> with end points, and then make conclusions about definite integrals of the real and imaginary parts of the integrand.

so how legit is this?

 

[lurflurf]

That is convoluted.
let s be the antiderivative opperator
s*exp(x)cos(x)=Re[s*exp(x(1+i))]
s*exp(x)cos(x)=Im[s*exp(x(1+i))]
s*exp(x(1+i))=exp(x(1+i))/(1+i)

 

[ice109]

That is convoluted.
let s be the antiderivative opperator
s*exp(x)cos(x)=Re[s*exp(x(1+i))]
s*exp(x)cos(x)=Im[s*exp(x(1+i))]
s*exp(x(1+i))=exp(x(1+i))/(1+i)

and how do i know s commutes with Re and Im?

 

[HallsofIvy]

Surly "perverted" is not the right word here!

 

[ice109]

can one of you more knowledgeable people just tell me if this a legitimate technique?

 

[Ben Niehoff]

It is legitimate for entire functions. There may be subtleties if a function has poles.

 

[dvs]

A function f:R->C can be decomposed into its real and imaginary parts, say f=u+iv, where u=Re(f) and v=Im(f). The definition of the Riemann/Lebesgue integral of such a function is

\int f = \int u + i \int v.

The integrals \int u and \int v are just the usual real integrals.

Does this clear anything up for you?