I've been asked to evaluate the following integral using complex analysis.
$$I=\int_{0}^{\infty}\frac{x^a}{(x^2+1)^2}dx$$
Where I have been given $-1<a<3$.
What my teacher does is, consider this to be a general multivalued function, and take a branch cut from $0$ to $\infty$ on the positive real axis, and uses the fact that the function has an additional phase on the bottom part, and gets something like
$$I=\frac{1}{1-e^{2\pi ia}}\int_C \frac{z^a}{(z^2+1)^2}dz$$
Here C is the hairpin contour with the branch cut along the positive real axis. The poles are at $\pm i$ and we can use the residue theorem.
However, my issue is that this technique should hold for only non-integer values of $a$. For integer values of $a$, this relation shouldn't work.
So, what I did was, I did this exact thing for all the non-integer values of $a$, and then solved for the integer values of a separately, using the properties of the natural log.
Thus, for integer values of $a$, I found that $$I=-2\pi i\int _C\frac{z^a \ln(z)}{(z^2+1)^2}dx$$
Here I've chosen the exact same contour and the branch cut. However, unlike before, my branch cut is now due to the properties of the natural log. Thus I've solved the two problems separately for integer and non-integer values of $a$.
However, my friends claim that the first method is general and should hold for all values of $a$.
Can someone tell me if I'm making a mistake here, or if my friends are correct? This problem is from the Churchill book, and even there, a single answer is given, while I'm getting two different expressions of the solution in two separate cases i.e. for integer and non-integer values of $a$.
Is it really possible to solve this integral using the first method for all general values of $a$, or do we need to solve it separately for integer and non-integer $a$ as I've done.