I need to evaluate the following real convergent improper integral using residue theory (vital that i use residue theory so other methods are not needed here) I also need to use the following contour (specifically a keyhole contour to exclude the branch cut):

$$\int_0^\infty \frac{\sqrt{x}}{x^3+1}\ \mathrm dx$$

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    $\begingroup$ If it is homework just say so :-) $\endgroup$ – Aryabhata Mar 7 '12 at 18:42
  • $\begingroup$ Yep i wasnt trying to hide that fact, im new to this forum so im not sure what your policies are, but yes this is homework :) $\endgroup$ – Bany Mar 7 '12 at 18:55
  • $\begingroup$ @Bany If you go to the FAQ in the menu, there's a link to the homework FAQ: meta.math.stackexchange.com/questions/1803/… $\endgroup$ – dls Mar 7 '12 at 19:13
  • $\begingroup$ @Bany Although it doesn't use contour integration, this answer might help to check the value, which is $\pi/3$ $\endgroup$ – Pedro Tamaroff Mar 7 '12 at 19:31
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    $\begingroup$ Nice picture :) $\endgroup$ – AD. Mar 7 '12 at 19:33

Consider the integral of $\sqrt{z}/(z^3+1)$ around the given contour, using a branch of $\sqrt{z}$ with branch cut on the positive real axis. This can be evaluated using residues. Note that (in the appropriate limit) the integrals over $L_1$ and $L_2$ both approach $\int_0^\infty \frac{\sqrt{x}}{x^3+1}\ dx$ (for $L_2$ you're going from right to left, but also $\sqrt{z}$ approaches $-\sqrt{x}$ as $z$ approaches $x$ from below the branch cut). The integrals over the circular arcs should both go to $0$.


Since a solution involving contour integration has been given, I am providing an alternative method without contour integration. Let $u:=\sqrt{x}$. Then, the integral $I:=\displaystyle\int_0^\infty\,\frac{\sqrt{x}}{x^3+1}\,\text{d}x$ equals $$I=2\,\int_0^\infty\,\frac{u^2}{u^6+1}\,\text{d}u=\int_{-\infty}^{+\infty}\,\frac{u^2}{u^6+1}\,\text{d}u\,.$$ Note that $$\frac{u^2}{u^6+1}=\frac{1}{3}\,\left(\frac{u^2+1}{u^4-u^2+1}\right)-\frac13\,\left(\frac{1}{u^2+1}\right)\,.$$ Now, let $v:=u-\frac{1}{u}$. Then, $$\frac{u^2+1}{u^4-u^2+1}=\frac{1+\frac{1}{u^2}}{\left(u-\frac{1}{u}\right)^2+1}=\left(\frac{1}{v^2+1}\right)\,\frac{\text{d}v}{\text{d}u}\,.$$ Thus, $$\begin{align}\int\,\frac{u^2+1}{u^4-u^2+1}\,\text{d}u&=\int\,\frac{1}{v^2+1}\,\text{d}v\\&=\text{arctan}(v)+C\\&=\text{arctan}\left(u-\frac{1}{u}\right)+C\,,\end{align}$$ where $C$ is a constant of integration. Thus, $$\int_0^{+\infty}\,\frac{u^2+1}{u^4-u^2+1}\,\text{d}u=\pi=\int_{-\infty}^0\,\frac{u^2+1}{u^4-u^2+1}\,\text{d}u\,.$$ On the other hand, $$\int_{-\infty}^{+\infty}\,\frac{1}{u^2+1}\,\text{d}u=\Big.\big(\text{arctan}(u)\big)\Big|_{u=-\infty}^{u=+\infty}=\pi\,,$$ making $$I=\frac{1}{3}\,(2\pi)-\frac{1}{3}\,\pi=\frac{\pi}{3}\,.$$ This result agrees with the computation made by Amir Alizadeh approximately six years ago.


Close format for this type of integrals: $$ \int_0^{\infty} x^{\alpha-1}Q(x)dx =\frac{\pi}{\sin(\alpha \pi)} \sum_{i=1}^{n} \,\text{Res}_i\big((-z)^{\alpha-1}Q(z)\big) $$ $$ I=\int_0^\infty \frac{\sqrt{x}}{x^3+1} dx \rightarrow \alpha-1=\frac{1}{2} \rightarrow \alpha=\frac{3}{2}$$ $$ g(z) =(-z)^{\alpha-1}Q(z) =\frac{(-z)^{\frac{1}{2}}}{z^3+1} =\frac{i \sqrt{z}}{z^3+1}$$ $$ z^3+1=0 \rightarrow \hspace{8mm }z^3=-1=e^{i \pi} \rightarrow \hspace{8mm }z_k=e^{\frac{\pi+2k \pi}{3}} $$ $$z_k= \begin{cases} k=0 & z_1=e^{i \frac{\pi}{3}}=\frac{1}{2}+i\frac{\sqrt{3}}{2} \\ k=1 & z_2=e^{i \pi}=-1 \\k=2 & z_3=e^{i \frac{5 \pi}{3}}=\frac{1}{2}-i\frac{\sqrt{3}}{2} \end{cases}$$ $$R_1=\text{Residue}\big(g(z),z_1\big)=\frac{i \sqrt{z_1}}{(z_1-z_2)(z_1-z_3)}$$ $$R_2=\text{Residue}\big(g(z),z_2\big)=\frac{i \sqrt{z_2}}{(z_2-z_1)(z_2-z_3)}$$ $$R_3=\text{Residue}\big(g(z),z_3\big)=\frac{i \sqrt{z_3}}{(z_3-z_2)(z_3-z_1)}$$ $$ I=\frac{\pi}{\sin\left( \frac{3}{2} \pi\right)} (R_1+R_2+R_3) = \frac{\pi}{-1} \left(\frac{-1}{3}\right)=\frac{\pi}{3}$$

Matlab Program

     syms x
     ans =


Compute R1,R2,R3 with Malab

  • $\begingroup$ when computing the residue of $R_2$, do you take $i$ or $-i$ for $\sqrt{-1}$ and why? $\endgroup$ – fysh Sep 23 '18 at 15:40

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