Evaluate $\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$

Evaluate :$$\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$

using residue theorem. The integral is even and it can be written as :

$$\frac{1}{2} \int_{-\infty}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$

The only pole with a positive imaginary part is $$\frac{i}{\sqrt2}$$ and its degree is $$4$$. How can I evaluate this without differentiating 3 times when calculating the residue?

I also tried calculating the infinite residue but that didn't help much.

• Have you considered expanding the integrand as$$\frac{x^4}{(1+2x^2)^4}=\frac14\left(\frac1{(1+2x^2)^4}-\frac2{(1+2x^2)^3}+\frac1{(1+2x^2)^2}\right)\quad?$$ Commented Nov 11, 2019 at 20:15

You can instead find the laurent series, and pick up the coefficient of the $$z^{-1}$$ term. i.e. $$(1+2z^2)^{-4} = (1+2(z-\frac{i}{\sqrt2})^2-2(\frac{-1}{2})+4\frac{iz}{\sqrt2})^{-4}= (1+2(z-\frac{i}{\sqrt2})^2+1+2\sqrt2iz)^{-4} = (2\sqrt{2}i(z-\frac{i}{\sqrt2})+2\sqrt2i\frac{i}{\sqrt2}+2+2(z-\frac{i}{\sqrt2})^2)^{-4}=(2\sqrt2i(z-\frac{i}{\sqrt2})+2(z-\frac{i}{\sqrt2})^2)^{-4}$$

Put $$y=z-\frac{i}{\sqrt2}$$, so you need to find the laurent series of $$(2\sqrt{2}iy+2y^2)^{-4}=(2\sqrt2iy)^{-4}(1+\frac{-i}{\sqrt2}y)^{-4}$$.

Writing $$\frac{z^4}{(1+2z^2)^{4}}=\frac{(y+\frac{i}{\sqrt2})^4}{(2\sqrt2iy)^{4}(1+\frac{-i}{\sqrt2}y)^{4}}=\frac{(y+\frac{i}{\sqrt2})^4}{64y^4(1+\frac{-i}{\sqrt2}y)^{4}}$$

Therefore you are only interested in the $$y^{-1}$$ coefficient which corresponds to $$y^3$$ coefficient of $$\frac{(y+\frac{i}{\sqrt2})^4}{(1+\frac{-i}{\sqrt2}y)^{4}}$$ scaled by $$\frac{1}{64}$$.

Looking at $$(y+\frac{i}{\sqrt2})^4(1+\frac{-i}{\sqrt2}y)^{-4}=(1/4 - i \sqrt2 y - 3 y^2 + 2 i \sqrt2 y^3 + y^4)(1+2\sqrt2iy-5y^2-5\sqrt2iy^3+O(y^4))=....-\frac{i}{2\sqrt2}y^3+....$$

Scaling that by $$\frac{1}{64}$$, we find the residue to be $$\frac{-i}{128\sqrt2}$$

• Wolfram agrees with my answer. I suggest you revisit your calculation to work out where the $5$ came from.
– J.G.
Commented Nov 11, 2019 at 20:21
• I did not see the $z^4$ term in the numerator. Commented Nov 11, 2019 at 20:24

Note first that$$\frac{x}{1+2x^2}=\frac14\left(\frac{1}{x-i/\sqrt{2}}+\frac{1}{x+i/\sqrt{2}}\right).$$Hence$$\left(\frac{x}{1+2x^2}\right)^2=\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+\frac{2}{x^2+1/2}\right)\\\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+i\sqrt{2}\left(\frac{1}{x-i/\sqrt{2}}-\frac{1}{x+i/\sqrt{2}}\right)\right).$$Squaring this again, none of the diagonal terms matter, and many of the cross terms don't either. Let $$f\sim g$$ denote the condition that $$f-g$$ has zero residue at $$\frac{i}{\sqrt{2}}$$, so$$\left(\frac{x}{1+2x^2}\right)^4\sim\frac{i\sqrt{2}}{128}\frac{1}{(x+i/\sqrt{2})^2(x-i/\sqrt{2})}\\\implies\operatorname{Res}_{-i/\sqrt{2}}\left(\frac{x}{1+2x^2}\right)^4=\frac{-i\sqrt{2}}{256}.$$Now we just need to multiply by $$\frac122\pi i$$ to give $$\frac{\pi\sqrt{2}}{256}$$ as the value of the integral.

Just to double-check, let's solve the problem with $$x=\frac{1}{\sqrt{2}}\tan t$$ so the original integral is$$\frac{\sqrt{2}}{8}\int_0^{\pi/2}\sin^4t\cos^2tdt=\frac{\sqrt{2}}{16}\operatorname{B}\left(\frac52,\,\frac32\right)=\frac{\sqrt{2}}{16}\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)}{\Gamma(4)}=\frac{\pi\sqrt{2}}{256}.$$

• So after we square again we look up for the coefficient of $(x - \frac{i}{\sqrt2})$ ? Commented Nov 11, 2019 at 20:57
• @SADBOYS Yes. See if you can derive general residues for $\oint\frac{f(z)dz}{(z-a)^n}$ for $f$ analytic with $f(a)$ finite but nonzero.
– J.G.
Commented Nov 11, 2019 at 20:59
• Well it's $2\pi i$ if $n = -1$ Commented Nov 11, 2019 at 20:59
• @SADBOYS Sorry, I've fixed my comment.
– J.G.
Commented Nov 11, 2019 at 21:03
• I don't get it. Isn't that $2 \pi i f(a)$ if we write Cauchy integral theorem? Commented Nov 11, 2019 at 21:13