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this is the last from me

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): http://imgur.com/LP7Cu

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

Thanks :)

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If it is homework just say so :-) –  Aryabhata Mar 7 '12 at 18:42
    
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 :) –  Bany Mar 7 '12 at 18:55
    
@Bany If you go to the FAQ in the menu, there's a link to the homework FAQ: meta.math.stackexchange.com/questions/1803/… –  dls Mar 7 '12 at 19:13
    
@Bany Although it doesn't use contour integration, this answer might help to check the value, which is $\pi/3$ –  Pedro Tamaroff Mar 7 '12 at 19:31
    
Nice picture :) –  AD. Mar 7 '12 at 19:33
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2 Answers

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 $0$.

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

Matlab Program

     syms x
     f=sqrt(x)/(x^3+1);
     int(f,0,inf) 
     ans =

     pi/3

Compute R1,R2,R3 with Malab

     z1=exp(i*pi/3);
     z2=exp(i*pi);
     z3=exp(5*i*pi/3);
     R1=i*sqrt(z1)/((z1-z2)*(z1-z3));
     R2=i*sqrt(z2)/((z2-z1)*(z2-z3));
     R3=i*sqrt(z3)/((z3-z2)*(z3-z1));
     I=(-pi)*(R1+R2+R3);
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