Calculate $\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx$

Question

I wanted to practice using the Residue theorem to calculate integrals. I chose an integral of the form $$\int_{-\infty}^\infty\frac{x^n}{1+x^m}dx$$and then choose random numbers for $n$ and $m$, which happened to be $$\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx$$I know there is a general formula for integrals of this form, which is $$\int_{-\infty}^\infty\frac{x^n}{1+x^m}dx=\frac{\pi}{m}\csc\left(\frac{\pi(n+1)}{m}\right)$$But let's forget about this right now. I used the Residue theorem and the semicircular contour to get that $$\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx=\Re\left(\frac{\pi i}{4}\left(\frac{1}{e^{\frac{5\pi i}{8}}}+\frac{1}{e^{\frac{15\pi i}{8}}}+\frac{1}{e^{\frac{25\pi i}{8}}}+\frac{1}{e^{\frac{35\pi i}{8}}}\right)\right)$$

Is there any other ways to calculate this integral?

Answer 1

Let $$ I=\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx=2\int_{0}^\infty\frac{x^2}{1+x^8}dx.$$ Under $x\to\frac1x$, one has $$ I=\int_{-\infty}^\infty\frac{x^2}{1+x^8}dx=2\int_{0}^\infty\frac{x^4}{1+x^8}dx.\tag1$$ Clearly $$ I=-2\int_{0}^\infty\frac{x^{4}}{1+x^{8}}d\bigg(\frac1x\bigg). \tag2$$ Adding (1) to (2) gives \begin{eqnarray} 2I&=&2\int_{0}^\infty\frac{x^{4}}{1+x^{8}}d\bigg(x-\frac1x\bigg)\\ &=&2\int_{0}^\infty\frac{1}{x^4+x^{-4}}d\bigg(x-\frac1x\bigg)\\ &=&2\int_{0}^\infty\frac{1}{(x^2+x^{-2})^2-2}d\bigg(x-\frac1x\bigg)\\ &=&2\int_{0}^\infty\frac{1}{((x-x^{-1})^2+2)^2-2}d\bigg(x-\frac1x\bigg)\\ &=&2\int_{-\infty}^\infty\frac{1}{(x^2+2)^2-2}dx\\ &=&2\cdot\frac1{2\sqrt2}\int_{-\infty}^\infty\bigg(\frac{1}{x^2+2-\sqrt2}-\frac{1}{x^2+2+\sqrt2}\bigg)dx\\ &=&\frac1{\sqrt2}\bigg(\frac1{\sqrt{2-\sqrt2}}\arctan\bigg(\frac{x}{\sqrt{2-\sqrt2}}\bigg)-\frac1{\sqrt{2+\sqrt2}}\arctan\bigg(\frac{x}{\sqrt{2+\sqrt2}}\bigg)\bigg)\bigg|_{-\infty}^\infty\\ &=&\frac\pi{\sqrt2}\bigg(\frac1{\sqrt{2-\sqrt2}}-\frac1{\sqrt{2+\sqrt2}}\bigg)\\ &=&\frac\pi{2}(\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2}). \end{eqnarray} So $$ I= \frac\pi{4}(\sqrt{2+\sqrt2}-\sqrt{2-\sqrt2})=\frac\pi{2}\sqrt{1-\frac1{\sqrt2}}. $$

Answer 2

Alternative complex solution:

$$\begin{align*} I &= \int_{-\infty}^\infty \frac{x^2}{1+x^8} \, dx \\[1ex] &= \int_0^\infty \frac{2x^2}{1+x^8} \, dx \tag{1} \\[1ex] &= \int_0^\infty \frac{\sqrt x}{1+x^4} \, dx \tag{2} \\[1ex] &= i \pi \sum_{k\in\{1,3,5,7\}} \operatorname{Res}\left(\frac{\sqrt z}{1+z^4}, z=e^{i\frac{k\pi}4}\right) \tag{3} \\[1ex] &= \frac\pi{\sqrt2}\sin\left(\frac\pi8\right) \end{align*}$$

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• $(1)$ : symmetry
• $(2)$ : substitute $x\mapsto \sqrt x$
• $(3)$ : integrate in the complex plane along a deformed circular contour that avoids a branch cut taken along the positive real axis

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We can also reduce the powers further to halve the number of residues one has to compute. For instance, with $x\mapsto \sqrt[4]{x}$ we get

$$I = \frac12 \int_0^\infty \frac{dx}{\sqrt[4]{x}(1+x^2)}$$

By the residue theorem (using the same contour and branch),

$$i2\pi \sum_{\zeta=\pm i} \operatorname{Res}\left(\frac1{\sqrt[4]{z}(1+z^2)}, z=\zeta\right) = (1+i) \int_0^\infty \frac{dx}{\sqrt[4]{x}(1+x^2)} \, dx \\ \implies I = \frac\pi2\left(\underbrace{\cos\left(\frac\pi8\right)-\sin\left(\frac\pi8\right)}_{=\sqrt2\sin\left(\frac\pi8\right)}\right)$$

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Once more, under $x\mapsto \sqrt[8]{x}$,

$$I = \int_0^\infty \frac{2x^2}{1+x^8} \, dx = \frac14 \int_0^\infty \frac{dx}{x^{5/8}(1+x)}$$

and we can use the residue theorem again, or simply recognize the beta integral, so that $I=\frac14\operatorname{B}\left(\frac38,\frac58\right)$.