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I have $$\int_0^{\infty}\frac{dz}{z^6+1}$$

I am a bit confused about residues result in my calculus, but what I've done:

$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{dz}{z^6+1} = 2\pi i \left(\sum_{j=1}^n res_j\right)$$

therefore $z = \sqrt[6]{-1}$

I figured out 3 roots that are in the countier

$\varphi_1 = \frac{\sqrt{3}}{2} + \frac{i}{2} $

$ \varphi_2 =-\frac{1}{2} + i$

$ \varphi_3 = -\frac{\sqrt{3}}{2} + \frac{i}{2}$

to find the residues I used the formula for the fraction:

$$res_{z = a} = \frac{\xi(a)}{\psi'(a)}$$

so therefore $$F(z) =\frac{1}{6z^5} = \Bigg\lvert_{z= \varphi_1..\varphi_3}$$

and after that:

$$res_1 = \frac{1}{6\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5}$$

$$res_2 = \frac{1}{-6\left(\frac{1}{2}-i\right)^5}$$

$$res_3 = \frac{1}{-6\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5}$$

but after all $i$ just did not dissaper in the final answer, so maybe I am doing something wrong? because I know that when we do real integration with residues, we should get rid of $i$ in the final answer.

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    $\begingroup$ Showing the whole elaboration helps much, because now nobody can see where it went wrong except maybe if we would elaborate it ourself, but I guess nobody will do that. ... $\endgroup$ – Shashi Dec 10 '17 at 13:38
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    $\begingroup$ "in the countier" should be "inside the contour"? But you haven't really committed to a specific contour (which has to be a closed curve or a limiting family of closed curves). $\endgroup$ – hardmath Dec 10 '17 at 13:47
  • $\begingroup$ your $\varphi_2$ isn't a pole. $\endgroup$ – Lord Shark the Unknown Dec 10 '17 at 13:50
  • $\begingroup$ You need to specify a contour that you're integrating over, otherwise your solution is missing important stuff. See my explanation down below. $\endgroup$ – Rebellos Dec 10 '17 at 14:00
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Hint:

The contour should consist of a semicircle of radius $R$ and the segment of the real axis from $-R$ to $R$.

The three roots under consideration are: $$z_1=e^{\frac{\pi i}{6}} \,, z_2=e^{\frac{3\pi i}{6}} \,, z_3=e^{\frac{5\pi i}{6}}$$

Then as you have calculated correctly, $$\text{ Res } f(z)_{z=z_i} = \frac{1}{6z^5} \Bigg\lvert_{z=z_i}$$

Thus, by residue theorem, we then have, $$I=2\pi i \sum_{z=z_1, z_2, z_3} \text{ Res } f(z)$$

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  • $\begingroup$ There must be a contour specified in order to consider certain roots (poles). $\endgroup$ – Rebellos Dec 10 '17 at 13:56
  • $\begingroup$ yes, I got the same roots, I just had written it in the real form, I was told I have to do it because I have to check $i$ sign for each root, and take the roots that have positive $i$ $\endgroup$ – M.Mass Dec 10 '17 at 13:58
  • $\begingroup$ what is the shortest methot to calculate say $$6(e^{\frac{5\pi i}{6}})^5$$, I never use exponents before $\endgroup$ – M.Mass Dec 10 '17 at 14:15
  • $\begingroup$ @M.Mass Note that $e^{2\pi i}=1=(e^{2\pi i})^2=e^{4\pi i}=e^{\frac{24\pi i}{6}}$. Thus, $$6e^{\frac{25\pi i}{6}}=6e^{\frac{\pi i}{6}}=6[\frac{\sqrt3}{2}+i\frac{1}{2}]$$ $\endgroup$ – Rohan Dec 10 '17 at 14:27
  • $\begingroup$ I was more confused if there's the shortest way of raising it to the 5th power (wgich I have in denumenator), but De Moivre's formula $\endgroup$ – M.Mass Dec 10 '17 at 15:06
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In order to perform such an integration, you need to specify a specific contour over which you're integrating, since the singularities of the integrating function vary.

A standard way of solving such integrals is :

The singularities of the integrating function :

$$f(x) = \frac{1}{x^6+1}$$

come as : $x^6 + 1 = 0$.

But we're integrating over the bounds $0$ and $\infty$, which means we come upon only the roots that are in the upper plane (there are 3 there).

We will integrate the function $f(x)$ with respect to $x$, over the partially smooth curve that is consisted of the half-moon of the upper level of $γ_R$, with $z(θ) = Re^{iθ}$, $0 \geq θ \geq π$ and the line segment $[-R,R]$, where we take $R$ to be big enough, so that the singularity points in the upper plane are inside $γ_R$.

From the Residue-Theorem we have :

$$\int_{-R}^R \frac{1}{x^6+1}dx + \int_{γ_R}\frac{1}{z^6+1}dz = 2\pi i \sum_{x=x_1, x_2, x_3} \text{ Res } f(x)$$

The function also satisfies Jordan's Lemma, so it is :

$$\lim_{R\to \infty}\int_{γ_R}\frac{1}{z^6+1}=0$$

thus :

$$ \int_0^\infty\frac{1}{x^6+1}dx=\frac{1}{2} \lim_{R\to \infty} \int_{-R}^R \frac{1}{x^6+1}dx= \pi i \sum_{x=x_1, x_2, x_3} \text{ Res } f(x)$$

where $x_1,x_2,x_3$ are the roots of $x^6+1=0$ in the upper-plane.

I'll leave the residue calculation to you!

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  • $\begingroup$ This is a great answer and I hate to be that guy but OP needs to divide final answer by two $\endgroup$ – Lanier Freeman Dec 10 '17 at 18:38
  • $\begingroup$ True and thanks for noticing ! I'll edit it shortly just by adding a fraction. $\endgroup$ – Rebellos Dec 10 '17 at 22:28
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Let us tackle the more general problem of finding, for some $m\in\mathbb{N}^+$, $$ I(m)=\int_{-\infty}^{+\infty}\frac{dx}{x^{2m}+1}. $$ By exploiting parity, the substitution $\frac{1}{x^{2m}+1}=u$, Euler's Beta function and the reflection formula for the $\Gamma$ function we get $I(m)=\frac{\pi}{m\sin\frac{\pi}{2m}}$. Let us prove this identity through the residue theorem, too. $f(z)=\frac{1}{z^{2m}+1}$ has simple poles at $\zeta_k=\exp\left(\frac{2\pi i}{4m}(2k-1)\right)$ for $k=1,2,\ldots,2m$.

$$\operatorname*{Res}_{z=\zeta_k}\frac{1}{z^{2m}+1}=\lim_{z\to \zeta_k}\frac{z-z_k}{z^{2m}+1}\stackrel{d.H.}{=}\lim_{z\to z_k}\frac{z}{2m z^{2m}}=-\frac{\zeta_k}{2m} $$ hence $$ I(m)=-\frac{\pi i}{m}\sum_{k=1}^{m}\zeta_k=-\frac{\pi i}{m}\sum_{k=1}^{m}\zeta_1^{2k-1}=-\frac{\pi i}{m}\cdot\frac{\zeta_1(\zeta_1^{2m}-1)}{\zeta_1^2-1}=\frac{2\pi i}{m\left(\zeta_1-\zeta_1^{-1}\right)}$$ and by De Moivre's formula we get $$\boxed{ I(m)=\int_{-\infty}^{+\infty}\frac{dx}{x^{2m}+1}=\frac{\pi}{m\sin\frac{\pi}{2m}}.}$$

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