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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 :(

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    $\begingroup$ What integral did you compare it to? $\endgroup$ – saulspatz Mar 15 '20 at 18:21
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    $\begingroup$ Note that the integrand is an odd function. $\endgroup$ – Gary Mar 15 '20 at 18:30
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    $\begingroup$ (Apologies if there are too many people making comments): You just want to prove convergence, so, what if you just prove $\int_1^{\infty} \frac{x}{x^4+a^4}dx$ converges using some comparison? $\endgroup$ – Michael Mar 15 '20 at 18:31
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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 $$ \int_{-\infty}^\infty \frac x{x^4+a^4}\ \mathsf dx = \lim_{x\to\infty} \left[\frac{\tan ^{-1}\left(\frac{x^2}{a^2}\right)}{2 a^2}\right] - \lim_{x\to-\infty} \left[\frac{\tan ^{-1}\left(\frac{x^2}{a^2}\right)}{2 a^2}\right] = 0. $$

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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.

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    $\begingroup$ But ... merely showing that the integral converges is MUCH easier than this. $\endgroup$ – GEdgar Mar 15 '20 at 22:02
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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 $\int_{-\infty}^{-1}$.

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