Complex Integrals: Evaluating \int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx

[Tangent87]

In our Complex Methods lecture today, our lecturer went through the example of evaluating the integral $$\int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx$$ and then proceeded to do the whole contour calculation using the complex function $$\frac{(z^2-1)^{1/2}}{z^2+1}$$. I'm concerned that the answer will be a factor of i out because $$(1-z^2)^{1/2}=i(z^2-1)^{1/2}$$ but he said in the lecture it didn't matter, can anyone explain why?

 

[dingo_d]

Maybe because in the end you are only interested in real solution (part)?

Just guessing here...

 

[Tangent87]

Maybe because in the end you are only interested in real solution (part)?

Just guessing here...

You're probably on the right lines but the real part of $$(z^2-1)^{1/2}$$ is def not $$(1-x^2)^{1/2}$$ because at the very least there'll be loads of y's involved as well since z=x+iy?

 

[dingo_d]

Oh, I was thinking that at the end of the integration you get something real and sth imaginary and then you take the real part...

Don't take that as a concrete answer, I too don't see how that wouldn't matter at first...

Because it really is not the same integral (by that factor).

Altho the residues do seem to cancel each other, so maybe there's the catch :\

 

[paranormal]

As a scientist, it is important to understand the reasoning behind mathematical calculations and how they relate to real-world applications. In this case, the complex integral \int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx is being evaluated using the complex function \frac{(z^2-1)^{1/2}}{z^2+1}. This is a common technique in complex analysis, where it is often easier to work with complex functions rather than real-valued functions.

One concern raised in the lecture was that the solution may include a factor of i because (1-z^2)^{1/2}=i(z^2-1)^{1/2}. However, in this case, it does not matter because the integral is being evaluated along a real interval, from -1 to 1. This means that z is restricted to real numbers, and therefore the imaginary component of (z^2-1)^{1/2} will be zero. This results in the same answer as if we had used the real-valued function (1-x^2)^{1/2}.

In complex analysis, it is common to use complex functions to simplify calculations and solve problems. As long as the integral is being evaluated along a real interval, the use of complex functions will not affect the final result. It is important to note, however, that when evaluating integrals along complex contours or in other complex scenarios, the use of complex functions may result in different solutions. Therefore, it is important to carefully consider the context and restrictions of the problem when choosing whether to use real or complex functions.