Computing a definite integral not using symmetry I have calculated an integral:
$$\int^\infty _{-\infty}\frac{x^3}{1 + x^8} \, dx = 0 $$
By the fact the the integral is of an odd function there for as it is from $-\infty \rightarrow +\infty$, the answer is zero.
Now what I want to know, is it possible to do this using Cauchy's Integral formula or Cauchy's Residue Theorem.
Thanks
 A: This should be a comment. But sorry I don't have enough reputation to comment on this. 
I think you should be careful using your symmetry argument. For example, the function $\frac{x}{1+x^2}$ is also an odd function, but we cannot say that $\int_{-\infty}^{\infty}\frac{x}{1+x^2}dx=0$ except when the limit is understood as P.V, i.e., if we define $\int_{-\infty}^{\infty}\frac{x}{1+x^2}dx $ as the limit $\lim_{n\rightarrow \infty}\int_{-n}^n\frac{x}{1+x^2}dx$.
For your example, one can also compute the antiderivative of the function $\frac{x^3}{1+x^8}$ directly.
A: As always consider the semicircle:
$$B_R:=\{Re^{i\varphi}\mid\varphi\in[0,\pi]\}$$
Then 
$$\left|\int_{B_R}\frac{z^3}{1+z^8}dz\right|\leq \pi R^{-4}\to0$$
as $R\to\infty$. Thus
$$\int_{-\infty}^\infty\frac{x^3}{1+x^8}=\lim_{R\to\infty}\int_{[-R,R]\cup B_R}\frac{z^3}{1+z^8}dz.$$
The poles in the upper half plane are at $z_k=e^{\pi/8i+k2\pi/8i},k\in0,1,2,3.$
And the residue there is given by
$$Res[\frac{z^3}{1+z^8},z_k]=\frac{z_k^3}{8z_{k}^7}=\frac{1}{8}z_k^{-4}=\frac18 e^{-\pi/2i}e^{-k\pi i}$$
Thus 
$$\int_{-\infty}^\infty\frac{x^3}{1+x^8}=2\pi i(e^{-\pi/2i}\sum_{k=0}^3e^{-k\pi i})=2\pi i(\frac18 e^{-\pi/2i}(1-1+1-1))=0$$
