Given the integral $ I:= \int_0^\infty \frac{1}{x^3+1} dx$.

There are various ways of solving it such as Cauchy's residues theorem, partial fraction. I am interested in solving it using the residues theorem but choosing a different contour than the ones i was able to find here (Integrating $\int_0^{\infty} \frac{dx}{1+x^3}$ using residues.). In addition, I would focus only in the part in which i can spot i made the mistake. Now lets jump into the problem:

The contour i want to use is half circle centered at z=R and let R approach infinity as shown in the picture. img The integral over the contour is equal to the sum of the integrals over $c_1$ and $c_2$ as shown in the picture. The integral over $c_1$ is simply I as defined above. For the integral over $c_2$ lets use z=R+R$e^{i\theta}$, $\theta$ goes from 0 to $\pi$ ,dz=$Rie^{i\theta}d\theta$ hence the integral over $c_2$ (lets name it $I_2$) becomes $I_2=\int_0^\pi \frac{Rie^{i\theta}}{R^3(1+e^{i\theta})^3+1} d\theta$ , as R approaches infinity $I_2$ approaches zero leaving us with the integral over the contour equals I. Now lets use the residues theorem to obtain I=$2{\pi}iRes(\frac{1}{z^3+1})$.

I have calculated the residues and used it in a different calculation using a different contour and got the correct answer which means the mistake is somewhere in my contour choosing but i dont know where. I would like to know what i did wrong.

Thanks in advance.


2 Answers 2


Your error is estimating the semi-circular integral. The integrand is $\frac{Rie^{i\theta}}{R^3(1+e^{i\theta})^3+1}$. As $\theta \to \pi$, we have $1+e^{i\theta} \to 0$ and the denominator $\to 1$. In fact, $\lim_{R\to\infty}\int_0^\pi \frac{Rie^{i\theta}}{R^3(1+e^{i\theta})^3+1}\,d\theta$ is something like $0.6-1.0 i$, not $0$.

In fact, according to Maple, $\int \frac{Rie^{i\theta}}{R^3(1+e^{i\theta})^3+1}\,d\theta$ has a symbolic antiderivative in terms of logarithms and arctangents. From that we get $$ \lim_{R\to\infty}\int_0^\pi \frac{Rie^{i\theta}}{R^3(1+e^{i\theta})^3+1}\,d\theta = \frac{\pi}{3\sqrt3} - i\,\frac{\pi}{3} . $$ Conclusion: it is not zero.

  • $\begingroup$ Thanks for the answer. However, i think you are wrong. Let u=R(1+exp(iTHETA)) and you will see that the integral does not equal 0.6-1i, in fact we get that this integral equals -I where I is the integral i defined above. $\endgroup$
    – Nitay
    Dec 7, 2023 at 19:08
  • $\begingroup$ See here ... https://www.wolframalpha.com/input?i=integrate+100*I*exp%28I*theta%29%2F%281000000*%281%2Bexp%28I*theta%29%29%5E3%2B1%29+from+theta%3D0+to+theta%3DPI $\endgroup$
    – GEdgar
    Dec 7, 2023 at 21:05
  • $\begingroup$ It is not a proof, go ahead and do the subtitution i suggest and you will see or put R as a larger number in your example and see.... for R=10^20 wolfram said its 0: wolframalpha.com/input?i=integrate+10%5E20*Iexp%28Itheta%29%2F%28%2810%5E60%29*%281%2Bexp%28I*theta%29%29%5E3%2B1%29+from+theta%3D0+to+theta%3DPI $\endgroup$
    – Nitay
    Dec 8, 2023 at 6:42
  • $\begingroup$ @Nitay Precision errors. Clicking "More digits" gives wildly different (non-zero!) values each time. $\endgroup$
    – user170231
    Dec 8, 2023 at 7:53
  • $\begingroup$ Still, this is not correct. This strategy is risky but can be rewarded yet in this case it is simply not correct because you can easily show that the integral (With the R involved..) is equal to -I as defined above $\endgroup$
    – Nitay
    Dec 8, 2023 at 18:59

To reiterate GEdgar's post, the upper bound on the integral in question does not converge to $0$ for all $\theta\in[0,\pi]$:

$$\left\lvert \int_0^\pi \frac{i R e^{i\theta}}{R^3 \left(1+e^{i\theta}\right)^3 + 1} \, d\theta \right\rvert \le \frac{\pi R}{\left\lvert R^3 \left(1+e^{i\theta}\right)^3 + 1\right\rvert} \le \frac{\pi R}{\left\lvert 2^{3/2} R^3 (1 + \cos\theta)^{3/2} - 1 \right\rvert}$$

As $\theta\to\pi^-$, $1+\cos\theta\to0$ and the upper bound on the integral would be $\pi R \not \to 0$.


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