I want to compute $\displaystyle \int^{\infty}_{0}\frac{x}{x^4+1}dx$ using the residue theorem.

The poles in the upper half plane are:

location: $\large e^{\frac{\pi i}4}$, order: 1, residue: $\large\frac{1}{4}e^{\frac{3\pi i}2}$
location: $\large e^{\frac{3\pi i}4}$, order: 1, residue: $\large \frac{1}{4}e^{\frac{\pi i}2}$

The problem is that the integral from $-\infty$ to $\infty$ vanishes for symmetry reasons, so I cannot apply the standard approach of putting the half of a 1-sphere on top of the real axis and letting its radius go to infinity. If x was replaced with $x^2$ for instance, I could just divide the result by two. Is there another way of contour integration to evaluate the upper expression?

  • $\begingroup$ Try the contour $\Gamma_R = \gamma_1+\gamma_2+\gamma_3$ where $\gamma_1(t) = t, 0\leq t\leq R$, $\gamma_2(\theta) = Re^{i\theta}, 0\leq\theta\leq \frac{\pi}3$ and $\gamma_3(t) = (R-t)e^{i\frac{\pi}3}, 0\leq t\leq R$. $\endgroup$ – Davide Giraudo Jul 9 '11 at 18:52
  • $\begingroup$ Can't you substitute $u=x^2$ first? $\endgroup$ – Emre Jul 9 '11 at 18:52
  • 1
    $\begingroup$ I think you mean "improper integral," not "indefinite integral." $\endgroup$ – Corey Jul 9 '11 at 19:06
  • $\begingroup$ @Corey: Yes, of course; My undergraduate courses are not yet in english, so I still have to learn the correct terms.. @girdav: Thanks, I might look into that after I went through math.stackexchange.com/questions/44928/… $\endgroup$ – Eli Jul 9 '11 at 19:10
  • $\begingroup$ See here math.stackexchange.com/questions/60443/… $\endgroup$ – Pablo Sep 12 '11 at 21:32

Those integrals were discussed by us in detail already.

To be more explicit you are asking for the special case of Interesting integral formula for $m=2$, $n=4$ and $a=1$. Just directly plugging in those values in the proof you get what you want (they are not really used anyways).

  • $\begingroup$ Thanks, I wasn't aware of this discussion. All the other problems I had to solve regarding contour integration were considerably easier so far, so I guess I expected a short solution - but a more abstract formula is much nicer, of course. $\endgroup$ – Eli Jul 9 '11 at 19:10
  • $\begingroup$ This is why I wrote this into the community wiki, there might be a shortcut to get your result but I doubt it. Basically the exponents $m$ and $n$ are not essential to the calculation as you can see in the problems i linked in the question, they are solved in a similar way. $\endgroup$ – Listing Jul 9 '11 at 19:17

Suppose we are interested in $$I = \int_0^\infty \frac{x}{1+x^4} dx$$

and evaluate it by integrating $$f(z) = \frac{z}{1+z^4}$$

around a pizze slice contour with the horizontal side $\Gamma_1$ of the slice on the positive real axis and the slanted side $\Gamma_3$ parameterized by $z= \exp(2\pi i/4) t = \exp(\pi i/2) t = it$ with the two connected by a circular arc $\Gamma_2$ parameterized by $z= R \exp(it)$ with $0\le t\le \pi i/2.$ We let $R$ go to infinity.

The integral along $\Gamma_1$ is $I$ in the limit. Furthermore we have in the limit $$\int_{\Gamma_3} f(z) dz = - \int_0^\infty \frac{\exp(\pi i/2) t}{1+t^4 \exp(2\pi i)} \exp(\pi i/2) dt \\ = - \exp(\pi i) \int_0^\infty \frac{t}{1+t^4} dt = I.$$

Furthermore $$\int_{\Gamma_2} f(z) dz \rightarrow 0$$ by the ML bound which yields $$\lim_{R\rightarrow \infty} \pi i/2 R \frac{R}{R^4-1} = 0.$$

The four poles are at $$\rho_k = \exp(\pi i/4 + 2\pi i k/4).$$

Considering the one pole $\rho_0$ inside the slice ($\rho_1 = \exp(\pi i/4 + \pi i/2) = \exp(3\pi i/4)$ so it is not inside the contour) we thus obtain

$$2 I = 2\pi i \mathrm{Res}_{z=\rho_0} f(z)$$


$$I = \pi i \frac{\rho_0}{4\rho_0^3} = \pi i \frac{\rho_0^2}{4\rho_0^4} = -\frac{1}{4} \pi i \exp(\pi i/2) = \frac{\pi}{4}.$$

Remark. We see that the pizza slice is in fact a quarter slice and the parameterization may use the rotation $it$ by $\pi/2$ throughout. The four poles are centered on the diagonals of the four quadrants.


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