contour integral, limiting contour theorem with residue

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May 20th 2011, 06:44 PM
tsang

contour integral, limiting contour theorem with residue

Hi, can anyone please help me with this question, thanks a lot.

$\displaystyle \int_{0}^{\infty} \frac{x^{-1/3}}{x^2+1} dx$

I did try to take the contour, and take notice the three "bad points" are $0$ , $i$ , and $-i$ .

I used residue theorem that $\displaystyle\oint_{\Gamma_{R,\epsilon}} \frac{dz}{\sqrt[3]{z}(z^2+1)}=2\pi i\displaystyle \sum_{poles\ in\ the\ plane}Res(f(z), a_j)$ .

I can use limiting contour theorem to get one integral is $0$ .

However, I'm really having trouble solve this question, I thought my methods are right, but I can't get the right answer which is $\frac{\sqrt{3}{\pi}}{3}$ . One friend told me I need to worry about choosing branch because of that $\sqrt[3]{z}$ , but I don't quite understand it and what I supposed to do.

Can anyone please show me some precise steps on solving this question? Thanks a lot.
May 20th 2011, 07:13 PM
chisigma

Quote:

Originally Posted by tsang

Hi, can anyone please help me with this question, thanks a lot.

$\displaystyle \int_{0}^{\infty} \frac{x^{-1/3}}{x^2+1} dx$

I did try to take the contour, and take notice the three "bad points" are $0$ , $i$ , and $-i$ .

I used residue theorem that $\displaystyle\oint_{\Gamma_{R,\epsilon}} \frac{dz}{\sqrt[3]{z}(z^2+1)}=2\pi i\displaystyle \sum_{poles\ in\ the\ plane}Res(f(z), a_j)$ .

I can use limiting contour theorem to get one integral is $0$ .

However, I'm really having trouble solve this question, I thought my methods are right, but I can't get the right answer which is $\frac{\sqrt{\pi}}{3}$ . One friend told me I need to worry about choosing branch because of that $\sqrt[3]{z}$ , but I don't quite understand it and what I supposed to do.

Can anyone please show me some precise steps on solving this question? Thanks a lot.

The function to be integrated has two 'simple poles' in z=i and z=-i and one 'brantch point' in z=0. The last type of singularity has to be excluded from the integration path so that may be that the best integration path is illustrated in the figure...

http://digilander.libero.it/luposabatini/MHF116.bmp

Kind regards

$\chi$ $\sigma$
May 20th 2011, 11:21 PM
tsang

Quote:

Originally Posted by chisigma

The function to be integrated has two 'simple poles' in z=i and z=-i and one 'brantch point' in z=0. The last type of singularity has to be excluded from the integration path so that may be that the best integration path is illustrated in the figure...

http://digilander.libero.it/luposabatini/MHF116.bmp

Kind regards

$\chi$ $\sigma$

Hi, thank you for your help. Yes, I used the same contour as your graph.
But, could you please give me a bit more details? I'm still confused on what I should do. I can get the whole contour is made by four parts, but I still can't get the right answer. Please help me a bit more. Thanks a lot.
May 21st 2011, 12:17 AM
chisigma

Quote:

Originally Posted by tsang

Hi, thank you for your help. Yes, I used the same contour as your graph.
But, could you please give me a bit more details? I'm still confused on what I should do. I can get the whole contour is made by four parts, but I still can't get the right answer. Please help me a bit more. Thanks a lot.

All right!... may be it is useful to examine the more general case...

$I= \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx\ ,\ -1<p<0$ (1)

As said above the solution is in the computation of the integral...

$\int_{\gamma} f(z)\ dz = \int_{\gamma} \frac{z^{p}}{1+z^{2}}\ dz$ (2)

... along the 'red path' $\gamma$ of the figure...

http://digilander.libero.it/luposabatini/MHF116.bmp

The procedure is a lottle long and it is better to devide it into two steps. The first step is the computation of the integral (2) that is done by the Cauchy integral formula...

$\int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i}$ (3)

In our case f(z) inside $\gamma$ has two simple poles: $z=i$ and $\z=-i$ and their residues are...

$r_{1}= \lim_{z \rightarrow i} (z-i)\ \frac{z^{p}}{1+z^{2}}= \frac{e^{i\ \frac{\pi\ p}{2}}}{2i}$ (4)

$r_{2}= \lim_{z \rightarrow -i} (z+i)\ \frac{z^{p}}{1+z^{2}}= -\frac{e^{-i\ \frac{\pi\ p}{2}}}{2i}$ (5)

... so that the integral (3) is...

$\int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i}= 2 \pi\ i\ \sin \frac{\pi\ p}{2}$ (6)

... and the first step is done. Are You able to proceed?...

Kind regards

$\chi$ $\sigma$
May 21st 2011, 08:46 AM
chisigma

Quote:

Originally Posted by chisigma

All right!... may be it is useful to examine the more general case...

$I= \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx\ ,\ -1<p<0$ (1)

As said above the solution is in the computation of the integral...

$\int_{\gamma} f(z)\ dz = \int_{\gamma} \frac{z^{p}}{1+z^{2}}\ dz$ (2)

... along the 'red path' $\gamma$ of the figure...

http://digilander.libero.it/luposabatini/MHF116.bmp

The procedure is a lottle long and it is better to devide it into two steps. The first step is the computation of the integral (2) that is done by the Cauchy integral formula...

$\int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i}$ (3)

In our case f(z) inside $\gamma$ has two simple poles: $z=i$ and $\z=-i$ and their residues are...

$r_{1}= \lim_{z \rightarrow i} (z-i)\ \frac{z^{p}}{1+z^{2}}= \frac{e^{i\ \frac{\pi\ p}{2}}}{2i}$ (4)

$r_{2}= \lim_{z \rightarrow -i} (z+i)\ \frac{z^{p}}{1+z^{2}}= -\frac{e^{-i\ \frac{\pi\ p}{2}}}{2i}$ (5)

... so that the integral (3) is...

$\int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i}= 2 \pi\ i\ \sin \frac{\pi\ p}{2}$ (6)

... and the first step is done...

The second step is the division of integral (6) into four distinct integrals...

$\int_{\gamma} f(z)\ dz = \int_{r}^{R} \frac{x^{p}}{1+x^{2}}\ dx + i\ \int_{0}^{2 \pi} \frac{R^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+R^{2}\ e^{2 i \theta}}\ d \theta +$

$+ \int_{R}^{r} \frac{x^{p}\ e^{2\ \pi\ i\ p}}{1+x^{2}\ e^{4\ \pi\ i}}\ dx + i\ \int_{2\ \pi}^{0} \frac{r^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+r^{2}\ e^{2 i \theta}}\ d \theta$ (7)

Now if $-1<p<0$ the second integral in (7) vanishes if R tends to infinity and the fourth integral in (7) also vanishes if r tends to 0, so that, taking into account (6), is...

$(1- e^{2\ \pi\ i\ p})\ \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = 2\ \pi\ i\ \sin \frac{\pi\ p}{2}$ (8)

... and from (8) with simple steps we arrive at the result...

$\int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = (-1)^{1-p}\ \pi\ \frac{\sin \frac{\pi\ p}{2}}{\sin \pi\ p}$ (9)

The result (9) is in some way 'a little enbarassing'... for $p=-\frac{1}{3}$ we have the correct result...

$\int_{0}^{\infty} \frac{x^{-\frac{1}{3}}}{1+x^{2}}\ dx = \frac{\pi}{\sqrt{3}}$ (10)

... as in...

int x^(-1/3)/(1+x^2) dx, x=0..infinity - Wolfram|Alpha

... but for other values of p (9) [for istance $p=-\frac{1}{2}$ ...] that is not true... an interesting problem for the 'experts' [unless some mistake of me]...

Kind regards

$\chi$ $\sigma$
May 23rd 2011, 05:29 PM
tsang

Quote:

Originally Posted by chisigma

The second step is the division of integral (6) into four distinct integrals...

$\int_{\gamma} f(z)\ dz = \int_{r}^{R} \frac{x^{p}}{1+x^{2}}\ dx + i\ \int_{0}^{2 \pi} \frac{R^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+R^{2}\ e^{2 i \theta}}\ d \theta +$

$+ \int_{R}^{r} \frac{x^{p}\ e^{2\ \pi\ i\ p}}{1+x^{2}\ e^{4\ \pi\ i}}\ dx + i\ \int_{2\ \pi}^{0} \frac{r^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+r^{2}\ e^{2 i \theta}}\ d \theta$ (7)

Now if $-1<p<0$ the second integral in (7) vanishes if R tends to infinity and the fourth integral in (7) also vanishes if r tends to 0, so that, taking into account (6), is...

$(1- e^{2\ \pi\ i\ p})\ \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = 2\ \pi\ i\ \sin \frac{\pi\ p}{2}$ (8)

... and from (8) with simple steps we arrive at the result...

$\int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = (-1)^{1-p}\ \pi\ \frac{\sin \frac{\pi\ p}{2}}{\sin \pi\ p}$ (9)

The result (9) is in some way 'a little enbarassing'... for $p=-\frac{1}{3}$ we have the correct result...

$\int_{0}^{\infty} \frac{x^{-\frac{1}{3}}}{1+x^{2}}\ dx = \frac{\pi}{\sqrt{3}}$ (10)

... as in...

int x^(-1/3)/(1+x^2) dx, x=0..infinity - Wolfram|Alpha

... but for other values of p (9) [for istance $p=-\frac{1}{2}$ ...] that is not true... an interesting problem for the 'experts' [unless some mistake of me]...

Kind regards

$\chi$ $\sigma$

Hi, sorry I just realised there's something confusing from your (8) to (10). This is because I do understand everything you did up to part (8), but I can't see how you get from part (8) to part (9), then, I used Matlab to double check, if I subsititute $p=-\frac{1}{3}$ in your equation (9), I don't actually get $\frac{\pi}{\sqrt{3}}$ , it is different answer.
May 23rd 2011, 10:00 PM
chisigma

Quote:

Originally Posted by tsang

Hi, sorry I just realised there's something confusing from your (8) to (10). This is because I do understand everything you did up to part (8), but I can't see how you get from part (8) to part (9), then, I used Matlab to double check, if I subsititute $p=-\frac{1}{3}$ in your equation (9), I don't actually get $\frac{\pi}{\sqrt{3}}$ , it is different answer.

In my opinion for $-1<p < 1$ is 'with great probability'...

$\int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = \pi\ \frac{\sin \frac{\pi\ p}{2}}{\sin \pi\ p} = \frac{\frac{\pi}{2}}{\cos \frac{\pi\ p}{2}}$ (1)

For example $p=\pm \frac{1}{2}$ gives...

$\int_{0}^{\infty} \frac{x^{\pm \frac{1}{2}}}{1+x^{2}} = \frac{\pi}{\sqrt{2}}$ (2)

... as in...

int x^(1/2)/(1+x^2&#41 ; dx, x=0..infinity - Wolfram|Alpha

... and in...

int x^(-1/2)/(1+x^2) dx, x=0..infinity - Wolfram|Alpha

My only problem is... I am unable to demonstrate that (Headbang)... if the result I arrived to is correct...

$(1-e^{2 \pi i p})\ \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = 2 \pi i \sin \frac{\pi p}{2}$ (3)

... then I'm in a trouble because is...

$1-e^{2 \pi i p} = - e^{\pi i p}\ (e^{\pi i p}-e^{-\pi i p}) = 2 i e^{\pi i (p-1)}\ \sin \pi p$ (4)

... and the fastidious term $e^{\pi i (p-1)}= (-1)^{p-1}$ is not eliminable... that's why I'm expecting some 'help' from the 'experts' of the forum(Itwasntme)...

Kind regards

$\chi$ $\sigma$