Integrating from $-\infty$ to $\infty$ improper integral

Question

I am trying to do a practice problem identifying if the following integral converges or diverges: $$\int_{-\infty}^\infty \frac{x}{x^4 + a^4}\, dx$$ for some $a > 0$. I have tried applying the limit comparison test and basic comparison test on some intervals, but this seems to fail. Is there a way to do this without integrating the function, I have tried integrating it but even online calculators seem to fail :(

Answer 1

First note that the map $f(x) = \frac x{x^4+a^4}$ is odd, that is, $$ f(-x) = \frac{-x}{(-x)^4+a^4} = \frac{-x}{x^4+a^4} = - f(x). $$ So it follows immediately that $$ \lim_{M\to\infty}\int_{-M}^M \frac x{x^4+a^4}\ \mathsf dx= 0, $$ that is, the Cauchy principal value of this integral is zero.

However, this is not enough to prove that the integral itself is zero. Observe that $$ \frac{\mathsf d}{\mathsf dx} \left[\frac{\tan ^{-1}\left(\frac{x^2}{a^2}\right)}{2 a^2}\right] = \frac x{x^4 + a^4}. $$ It follows then from the fundamental theorem of calculus that $$ \int_{-\infty}^\infty \frac x{x^4+a^4}\ \mathsf dx = \lim_{x\to\infty} \left[\frac{\tan ^{-1}\left(\frac{x^2}{a^2}\right)}{2 a^2}\right] - \lim_{x\to-\infty} \left[\frac{\tan ^{-1}\left(\frac{x^2}{a^2}\right)}{2 a^2}\right] = 0. $$

Answer 2

You need to do two things.$$*$$ (i)Find the antiderivative Note that $$x^4+a^4=(x^2+a^2)^2-(\sqrt 2 ax)^2$$ $$=(x^2-\sqrt 2 ax+a^2)(x^2+\sqrt 2 ax+a^2)$$ Thus $$\frac{x}{x^4+a^4}=\frac{Bx+C}{(x^2-\sqrt 2 ax+a^2)}+\frac{Dx+E}{(x^2+\sqrt 2 ax+a^2)}.$$ You can find $B,C,D \text { and } E .$ Then find the ani-derivative. It will involve ln and arctan functions. $$*$$ (ii) Find the limit $$\int_R^S\frac{x}{x^4+a^4}dx$$ as $R \rightarrow -\infty, S \rightarrow \infty $ by inserting the values of $R$ and $S$ into your antiderivative and see whether or not the limit exists.

Answer 3

The integral $\int_{-1}^1$ exits because the integrand is continuous. For the integral $\int_1^\infty$ compare to $1/x^3$. Similar for the integral $\int_{-\infty}^{-1}$.