How can I prove this?



I will use complex analysis in the form of Cauchy's theorem, just for those who want to see it here. (Maybe that is just me.) Consider the contour integral

$$\oint_C dz \frac{\log{(1+z^4)}}{z^4}$$

where $C$ is the following contour:

enter image description here

Note that this is a semicircle in the upper half plane, deformed to avoid the branch points at $z=e^{i \pi/4}$ and $z=e^{i 3 \pi/4}$. By Cauchy's theorem, this contour integral is equal to zero. (The singularity at $z=0$ is removable.) Thus, taking a few liberties with certain limits, the contour integral is equal to

$$\int_{-R}^R dx \frac{\log{(1+x^4)}}{x^4} + i R \int_0^{\pi} d\theta \, e^{i \theta}\, \frac{\log{(1+R^4 e^{i 4 \theta})}}{R^4 e^{i 4 \theta}} \\ +e^{i \pi/4} \int_1^R dt \frac{\log{(1-t^4)}}{t^4} + i \epsilon \int_{2 \pi}^{0} d\theta \, e^{i \theta}\, \frac{\log{(1+\epsilon^4 e^{i 4 \theta})}}{\epsilon^4 e^{i 4 \theta}} \\ -e^{i \pi/4} \int_1^R dt \frac{\log{(1-t^4)}-i 2 \pi}{t^4} + e^{i 3\pi/4} \int_1^R dt \frac{\log{(1-t^4)}}{t^4}\\ + i \epsilon \int_{2 \pi}^{0} d\theta \, e^{i \theta}\, \frac{\log{(1+\epsilon^4 e^{i 4 \theta})}}{\epsilon^4 e^{i 4 \theta}} -e^{i 3 \pi/4} \int_1^R dt \frac{\log{(1-t^4)}-i 2 \pi}{t^4}$$

Note that the characteristic $i 2 \pi$ jump from the logs is negative because we are traversing the branch point in a clockwise (i.e., negative) sense.

As $R \to \infty$ and $\epsilon \to 0$, the second, fourth, and seventh integrals vanish. Thus by Cauchy's theorem, we are left with

$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^4)}}{x^4} = -i 2 \pi \left (e^{i \pi/4} + e^{i 3 \pi/4} \right ) \int_1^{\infty} \frac{dt}{t^4} = \frac{2 \sqrt{2}\, \pi}{3}$$


With real analysis techniques only:

Let $I$ denote the integral to be computed. An integration by parts based on $u(x)=\log(1+x^4)$ and $v(x)=-1/(3x^3)$ yields $$ I=\frac43\int_{-\infty}^\infty\frac{\mathrm dx}{1+x^4}. $$ To compute the integral on the RHS, note that $$ \frac{2\sqrt2}{x^4+1}=\frac{x+\sqrt2}{x^2+\sqrt2x+1}-\frac{x-\sqrt2}{x^2-\sqrt2x+1}. $$ Using the change of variable $t=\sqrt2x+1$ in the first fraction and $t=\sqrt2x-1$ in the second fraction yields $\mathrm t=\sqrt2\mathrm dx$ in both cases and $$ \frac{2\sqrt2}{x^4+1}=\frac{\frac1{\sqrt2}(t+1)}{\frac12(t^2+1)}-\frac{\frac1{\sqrt2}(t-1)}{\frac12(t^2+1)}. $$ Thus, $$ I=\frac43\frac1{2\sqrt2}\int_{-\infty}^\infty 2\sqrt2\frac1{\sqrt2}\frac{\mathrm dt}{1+t^2}=\frac{2\sqrt2}3\int_{-\infty}^\infty\frac{\mathrm dt}{1+t^2}. $$ The integral on the RHS is classic since a primitive of $t\mapsto1/(1+t^2)$ is $\arctan$, hence its value is $\left.\arctan\right|_{-\infty}^{+\infty}=\pi$, and the value of $I$ follows.

  • 2
    $\begingroup$ Many thanks, Sir. You really made this accessible to people(or pupil?) like me who don't know any complex analysis. $\endgroup$ – Silent Nov 4 '13 at 13:56
  • $\begingroup$ @Sush You are welcome. $\endgroup$ – Did Nov 4 '13 at 18:05

You can use another way to do calculation. It is much easier. Let $$ I(\alpha)=\int_{-\infty}^\infty\frac{\log(1+\alpha x^4)}{x^4}dx. $$ Then $$ I'(\alpha)=\int_{-\infty}^\infty\frac{1}{1+\alpha x^4}dx=\frac{\pi}{\sqrt{2}\sqrt[4]{\alpha}}. $$ So \begin{eqnarray*} I(1)&=&\int_0^1\frac{\pi}{\sqrt{2}\sqrt[4]{\alpha}}d\alpha\\ &=&\frac{2\sqrt{2}\pi}{3}. \end{eqnarray*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy