How to calculate this integral

[mark.laidlaw19]

Homework Statement

As part of a larger problem involving integrating over a circular wedge contour in the complex plane starting at the origin, I have parametrised the integral, as is asked to do so in the question.

Homework Equations

I end up with one integral that has proved, for me, to be extremely difficult to find $$\lim_{R\rightarrow +\infty} {\int_0^R 1/(x^3+1)^3\,dx}$$

The Attempt at a Solution

I have tried a variety of methods to compute this. I know, from working which uses the residue theorem, that this integral should equal $${\frac{10\pi}{27\sqrt{3}}}$$, and have been able to compute all other parts of the question.
I tried partial fractions, but because the denominator is of degree 9, this would take an incredibly long time, longer than I believe is necessary, because other people have completed this in fewer steps.
I have tried integration by parts, but this does not seem to work, as I do not get the answer that I know to be correct from using the residue theorem.

Basically, what I would appreciate is a suggestion of another method, and perhaps a overview of the initial steps, to do an integral like this.

Any help is greatly appreciated.

Many thanks,

Mark

 

[Saitama]

Homework Statement

As part of a larger problem involving integrating over a circular wedge contour in the complex plane starting at the origin, I have parametrised the integral, as is asked to do so in the question.

Homework Equations

I end up with one integral that has proved, for me, to be extremely difficult to find ∫dx/(x^3+1), where the terminals go from 0 to R, and we are integrating along the real axis.

The Attempt at a Solution

I have tried a variety of methods to compute this. I know, from working which uses the residue theorem, that this integral should equal 10∏/(27√3), and have been able to compute all other parts of the question.
I tried partial fractions, but because the denominator is of degree 9, this would take an incredibly long time, longer than I believe is necessary.
I have tried integration by parts, but this does not seem to work, as I do not get the answer that I know to be correct from using the residue theorem.

Basically, what I would appreciate is a suggestion of another method, and perhaps a overview of the initial steps, to do an integral like this.

Any help is greatly appreciated.

Many thanks,

Mark

Hi Mark! Welcome to PF!

Can you please post the integral you are working with? It is difficult to see what you are trying to find.

 

[George Jones]

Have you tried partial fractions? I haven't pursued this, but the denominator does factor into a product of two lower degree polynomials.

 

[Ray Vickson]

Homework Statement

As part of a larger problem involving integrating over a circular wedge contour in the complex plane starting at the origin, I have parametrised the integral, as is asked to do so in the question.

Homework Equations

I end up with one integral that has proved, for me, to be extremely difficult to find ∫dx/(x^3+1), where the terminals go from 0 to R, and we are integrating along the real axis.

The Attempt at a Solution

I have tried a variety of methods to compute this. I know, from working which uses the residue theorem, that this integral should equal 10∏/(27√3), and have been able to compute all other parts of the question.
I tried partial fractions, but because the denominator is of degree 9, this would take an incredibly long time, longer than I believe is necessary.
I have tried integration by parts, but this does not seem to work, as I do not get the answer that I know to be correct from using the residue theorem.

Basically, what I would appreciate is a suggestion of another method, and perhaps a overview of the initial steps, to do an integral like this.

Any help is greatly appreciated.

Many thanks,

Mark

Questions:
(1) How can an integral from 0 to R be a number like $10 \pi /(27\sqrt{3})$ that does not depend on R?
(2) Why do you say that $x^3+1$ is of degree 9? It looks like degree 3 to me.

If you mean that the denominator is $x^9+1$, you get an integral that is "doable", but nasty and lengthy. Maple's answer for $\int_0^R dx/(x^9+1)$ takes 10 pages to write out.

 

[mark.laidlaw19]

Hi Mark! Welcome to PF!

Can you please post the integral you are working with? It is difficult to see what you are trying to find.

I'm not sure how to get it to look proper, but I end up with one integral that has proved, for me, to be extremely difficult to find ∫dx/(x^3+1), where the terminals go from 0 to R, and we are integrating along the real axis.

 

[Saitama]

I'm not sure how to get it to look proper, but I end up with one integral that has proved, for me, to be extremely difficult to find ∫dx/(x^3+1), where the terminals go from 0 to R, and we are integrating along the real axis.

I don't think I understand "integrating along the real axis", this is probably something related to countour integration.

What I asked was to post the exact definite integral you are dealing with, if possible, I can help you solve the problem using real methods as I don't know anything about complex analysis or countour integration.

 

[mark.laidlaw19]

We need to calculate this:
[tex]\lim_{R\rightarrow +\infinity} {\int_0^R 1/(x^3+1)^3\,dx}[\tex]
(sorry i am still learning how to type these properly). I have used complex analysis and contour integration to get to this stage, but I think this can be treated a a real integral.
I know the answer is meant to look similar to
[tex]{\frac{10\pi}{27\sqrt{3}}}[\tex]
as i have already calculated what it should be via the Residue theorem, but to calculate it by parametrisation, i somehow need to evaluate this integral, and then take the limit as R goes to infinity.

I have tried partial fractions, but i know that, as it is degree 9, the arithmetic will take ages, and I am certain it can be done faster, as I know other people who have completed this exercise.

 

[LCKurtz]

We need to calculate this:
$$\lim_{R\rightarrow +\infty} {\int_0^R 1/(x^3+1)^3\,dx}$$
(sorry i am still learning how to type these properly). I have used complex analysis and contour integration to get to this stage, but I think this can be treated a a real integral.
I know the answer is meant to look similar to
$${\frac{10\pi}{27\sqrt{3}}}$$
as i have already calculated what it should be via the Residue theorem, but to calculate it by parametrisation, i somehow need to evaluate this integral, and then take the limit as R goes to infinity.

I have tried partial fractions, but i know that, as it is degree 9, the arithmetic will take ages, and I am certain it can be done faster, as I know other people who have completed this exercise.

Fixed your tex. Use [/tex] not [\tex] to close the tex statements.

 

[Saitama]

We need to calculate this:
[tex]\lim_{R\rightarrow +\infinity} {\int_0^R 1/(x^3+1)^3\,dx}[\tex]
(sorry i am still learning how to type these properly). I have used complex analysis and contour integration to get to this stage, but I think this can be treated a a real integral.
I know the answer is meant to look similar to
[tex]{\frac{10\pi}{27\sqrt{3}}}[\tex]
as i have already calculated what it should be via the Residue theorem, but to calculate it by parametrisation, i somehow need to evaluate this integral, and then take the limit as R goes to infinity.

I have tried partial fractions, but i know that, as it is degree 9, the arithmetic will take ages, and I am certain it can be done faster, as I know other people who have completed this exercise.

That's certainly better. I have come up with a method but I am sure that faster ones exist.

Consider an easier problem first. Can you evaluate the following:
$$\int_0^{\infty} \frac{dx}{a^3+x^3}$$

 

[mark.laidlaw19]

That's certainly better. I have come up with a method but I am sure that faster ones exist.

Consider an easier problem first. Can you evaluate the following:
$$\int_0^{\infty} \frac{dx}{a^3+x^3}$$

Well, i would factorise it into three linear factors and then use partial fractions, integrating each fraction separately.

 

[Saitama]

Well, i would factorise it into three linear factors and then use partial fractions, integrating each fraction separately.

Yep, so what do you get? You should get a nice expression. :)

EDIT: Erm...wait, three? $x^3+a^3=(x+a)(x^2-ax+a^2)$.

 

[mark.laidlaw19]

Yep, so what do you get? You should get a nice expression. :)

EDIT: Erm...wait, three? $x^3+a^3=(x+a)(x^2-ax+a^2)$.

Well, if we assume a > 0, then to factorise the quadratic factor we would need to introduce complex numbers. But this should be okay, because the integrand can be complex, even if the variable can only be real.

But if I were to keep the integrand real to go along with what you're saying (which I really appreciate), I would end up with A/(x+a) + (Bx+C)/(x^2-ax+a^2), and I would evaluate A, B and C by equating coefficients.

 

[Saitama]

But if I were to keep the integrand real to go along with what you're saying (which I really appreciate), I would end up with A/(x+a) + (Bx+C)/(x^2-ax+a^2), and I would evaluate A, B and C by equating coefficients.

Yes, now what result do you get? Its quite straightforward once you evaluate the definite integral I ask you.

 

[mark.laidlaw19]

Well integrating the first fraction would give me Alog|R|-Alog(0)
But I'm having a bit of trouble with the second. It doesn't seem to require a trig substitution, and I don't seem to be able to get it to work by substitution. For example, if I take u=x^2-ax+a^2, then du=(2x-a)dx, but I can't then substitute this in easily because I will still have the B and the C, and I can't add coefficients to satisfy B=2 without making c not equal to -a and vice versa.

 

[Saitama]

Well integrating the first fraction would give me Alog|R|-Alog(0)
But I'm having a bit of trouble with the second. It doesn't seem to require a trig substitution, and I don't seem to be able to get it to work by substitution. For example, if I take u=x^2-ax+a^2, then du=(2x-a)dx, but I can't then substitute this in easily because I will still have the B and the C, and I can't add coefficients to satisfy B=2 without making c not equal to -a and vice versa.

I am unable to follow you here.
$$\frac{1}{x^3+a^3}=\frac{A}{x+a}+\frac{Bx+C}{x^2-ax+a^2}$$
$$\Rightarrow \frac{1}{x^3+a^3}=\frac{x^2(A+B)+x(aB-aA+C)+a^2A+aC}{x^3+a^3}$$
So you have the following three equations:
$$A+B=0$$
$$aB-aA+C=0$$
$$a^2A+aC=1$$
What do you get for A,B and C?

 

[mark.laidlaw19]

Did you mean that $$a^2A=aC=1$$? I assumed that you did, and If I solved these I get: $$A=1/3a^2 and B=-1/3a^2 and C=2/(3a)$$

 

[mark.laidlaw19]

The substitution $u = 1/(x^3 + 1)$ yields $$\int_0^\infty \frac{1}{(x^3 + 1)^3}\,dx = \frac{1}{3} \int_0^1 u^{-1/3}(1 - u)^{-2/3}\,du = \frac13 B(\tfrac23, \tfrac 13)$$ where $B(p,q)$ is the beta function. You can now use the standard results $$B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p + q)}$$ and $$\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin \pi z}$$ together with $\Gamma(1) = 1$ (see Gamma function) to complete the calculation.

Thank you very much for your response. In our course, however, we have not yet covered the gamma, beta and zeta functions, and this leads me to believe that the integral can be done using parametrisation.

 

[Saitama]

Did you mean that $$a^2A=aC=1$$? I assumed that you did, and If I solved these I get: $$A=1/3a^2 and B=-/3a^2 and C=2/(3a)$$

I meant what I wrote but you got the correct coefficients.

So you have:
$$\int_0^{R} \frac{dx}{a^3+x^3}=\int_0^{R} \frac{dx}{3a^2(a+x)}+\int_0^{R} \frac{2a-x}{3a^2(x^2-ax+a^2)}dx$$

I suppose the first integral is easy to evaluate.

To evaluate the second one, try splitting it in the following way:
$$\int_0^R \frac{2a-x}{3a^2(x^2-ax+a^2)}\,dx=\frac{1}{6a^2}\left(\int_0^R \frac{3a}{x^2-ax+a^2}\,dx-\int_0^R \frac{2x-a}{x^2-ax+a}\,dx\right)$$

The first one can be handled by completing the square and the second one by the substitution $u=x^2-ax+a^2$.

 

[pasmith]

Thank you very much for your response. In our course, however, we have not yet covered the gamma, beta and zeta functions, and this leads me to believe that the integral can be done using parametrisation.

On checking, I realize that I made an error in the initial substitution, so I deleted my post.

However you do end up with a beta function, and I think that you have $$\int_0^\infty \frac{1}{(1 + x^3)^3}\,dx = \frac13 \int_0^1 u^{5/3} (1 - u)^{-2/3}\,du = \tfrac13 B(\tfrac 83, \tfrac 13).$$
You will need the additional result $$\Gamma(z + 1) = z \Gamma(z).$$

 

[BruceW]

looks pretty good. But to continue from here, he needs to know quite a few more formulas related to the beta and gamma functions. Or simply use beta(8/3,1/3) in an advanced calculator :)

 

[Ray Vickson]

looks pretty good. But to continue from here, he needs to know quite a few more formulas related to the beta and gamma functions. Or simply use beta(8/3,1/3) in an advanced calculator :)

The integral
$$F(R) =\int_0^R \frac{dx}{(1+x^3)^3}$$
is doable in elementary terms (no Gamma, Beta functions, etc) and then the limit can be taken. The answer happens to coincide with $\frac{1}{3} B(\frac{8}{3},\frac{1}{3}).$

The indefinite integral is a bit nasty and long, but nowhere near as nasty as the one in my original response; the Maple printout is only about 1/2 page now, rather than 10 pages from before. Doing it by hand would be challenging but possible. It blows away Wolfram Alpha inasmuch as the default cpu time is exceeded.

 

[Saitama]

The indefinite integral is a bit nasty and long, but nowhere near as nasty as the one in my original response; the Maple printout is only about 1/2 page now, rather than 10 pages from before. Doing it by hand would be challenging but possible. It blows away Wolfram Alpha inasmuch as the default cpu time is exceeded.

Yes, the indefinite one is a bit nasty but still doable by hand. Instead of directly working with $1/(1+x^3)^3$, one can evaluate the indefinite integral of $1/(a^3+x^3)$ (not cubed!) and take out the necessary info. I hope you see what I am trying to do here. :)

@mark.laidlaw19: Did you evaluate the definite integral? You are quite close to the final answer now.

 

[mark.laidlaw19]

Yes, the indefinite one is a bit nasty but still doable by hand. Instead of directly working with $1/(1+x^3)^3$, one can evaluate the indefinite integral of $1/(a^3+x^3)$ (not cubed!) and take out the necessary info. I hope you see what I am trying to do here. :)

@mark.laidlaw19: Did you evaluate the definite integral? You are quite close to the final answer now.

Thank you for your help everyone, just letting you know I am in class right now but will reply as soon as I can. I really appreciate all your assistance

 

[Ray Vickson]

Yes, the indefinite one is a bit nasty but still doable by hand. Instead of directly working with $1/(1+x^3)^3$, one can evaluate the indefinite integral of $1/(a^3+x^3)$ (not cubed!) and take out the necessary info. I hope you see what I am trying to do here. :)

@mark.laidlaw19: Did you evaluate the definite integral? You are quite close to the final answer now.

The "trick" you are aiming at would go even easier if you started with
$$\int \frac{1}{x^3+a} \, dx$$

 

[Saitama]

The "trick" you are aiming at would go even easier if you started with
$$\int \frac{1}{x^3+a} \, dx$$

I am sorry but I will have to disagree here (maybe its a matter of choice). Your suggestion would give nasty cuberoots and the subsequent steps would be much harder to deal with than they already are. Using $a^3$ gives a nicer expression.

 

[mark.laidlaw19]

Yes, the indefinite one is a bit nasty but still doable by hand. Instead of directly working with $1/(1+x^3)^3$, one can evaluate the indefinite integral of $1/(a^3+x^3)$ (not cubed!) and take out the necessary info. I hope you see what I am trying to do here. :)

@mark.laidlaw19: Did you evaluate the definite integral? You are quite close to the final answer now.

Well I try to, I've worked through it several times but end up with this (messy) answer which I'm pretty sure does not give the correct answer.

$$1/3a^2log(1+R/a)+√3/6a^2*(log(\frac{2R-(1+√3)a}{2R-(1-√3)a})-log|-2-√3|)-\frac{1}{6a^2}\log(\frac{R^2-aR+a^2}{R})$$

I am really interested in how you will use this to generate the integral for $$\int_0^R \frac{1}{(a^3+x^3)^3},dx$$

But it might be useful for this stage if I knew what integral I was supposed to generate?

Many thanks

 

[Saitama]

Well I try to, I've worked through it several times but end up with this (messy) answer which I'm pretty sure does not give the correct answer.

$$1/3a^2log(1+R/a)+√3/6a^2*(log(\frac{2R-(1+√3)a}{2R-(1-√3)a})-log|-2-√3|)-\frac{1}{6a^2}\log(\frac{R^2-aR+a^2}{R})$$

I think what you have got is incorrect. There must be some terms involving $\pi$ and $\arctan$.

The first integral is:
$$\int_0^R \frac{dx}{3a^2(a+x)}=\frac{1}{3a^2}\ln(a+x)$$
and I hope you got this one correct. The other ones are:
$$\frac{1}{6a^2}\int_0^R \frac{3a}{x^2-ax+a^2}\,dx$$
$$\frac{1}{6a^2}\int_0^R \frac{2x-a}{x^2-ax+a}\,dx$$
Can you show what do you get for the above two?

I am really interested in how you will use this to generate the integral for $$\int_0^R \frac{1}{(a^3+x^3)^3},dx$$

But it might be useful for this stage if I knew what integral I was supposed to generate?

Many thanks

Once the definite integral is evaluated, you will get something like
$$\int_0^{\infty} \frac{dx}{a^3+x^3}=f(a)$$
Since you need $(a^3+x^3)^3$ in the denominator, you simply differentiate both the sides with respect to $a$ twice and then substitute $a=1$.

And $f(a)$ comes out to be a very nice expression. :)

 

[mark.laidlaw19]

I think what you have got is incorrect. There must be some terms involving $\pi$ and $\arctan$.

The first integral is:
$$\int_0^R \frac{dx}{3a^2(a+x)}=\frac{1}{3a^2}\ln(a+x)$$
and I hope you got this one correct. The other ones are:
$$\frac{1}{6a^2}\int_0^R \frac{3a}{x^2-ax+a^2}\,dx$$
$$\frac{1}{6a^2}\int_0^R \frac{2x-a}{x^2-ax+a}\,dx$$
Can you show what do you get for the above two?


Once the definite integral is evaluated, you will get something like
$$\int_0^{\infty} \frac{dx}{a^3+x^3}=f(a)$$
Since you need $(a^3+x^3)^3$ in the denominator, you simply differentiate both the sides with respect to $a$ twice and then substitute $a=1$.

And $f(a)$ comes out to be a very nice expression. :)

Haha yes I got the first integral fine. And yes, I can see now how to get the final integral and that's a very clever technique!

For the second integral, I completed the square for the denominator, which gave two factors, so I used partial fractions and integrated both. So this integral gave: $$√3/6a^2*(log(\frac{2R-(1+√3)a}{2R-(1-√3)a})-log|-2-√3|)$$

I used substitution for the third integral, and ended up with $$\frac{1}{6a^2}\log(\frac{R^2-aR+a^2}{R})$$

I'm interested to see how arctan will come about, as we don't have $\frac{1}{x^2+1}$, in the integrands, right?

 

[Saitama]

For the second integral, I completed the square for the denominator, which gave two factors, so I used partial fractions and integrated both. So this integral gave: $$√3/6a^2*(log(\frac{2R-(1+√3)a}{2R-(1-√3)a})-log|-2-√3|)$$

What you are saying is the correct method but you are getting the wrong result. :/

Completing the square you should get:
$$\frac{1}{6a^2}\int_0^R \frac{3a}{x^2-ax+a^2}\,dx=\frac{1}{2a}\int_0^R \frac{dx}{\left(x-a/2\right)^2+(\sqrt{3}a/2)^2}$$
It is of the form $dx/(x^2+b^2)$, do you see how the $\arctan$ comes in the picture? :)

I used substitution for the third integral, and ended up with $$\frac{1}{6a^2}\log(\frac{R^2-aR+a^2}{R})$$

Quite close, that should be $a^2$ instead of $R$ in the denominator.

 

[statdad]

I meant what I wrote but you got the correct coefficients.

So you have:
$$\int_0^{R} \frac{dx}{a^3+x^3}=\int_0^{R} \frac{dx}{3a^2(a+x)}+\int_0^{R} \frac{2a-x}{3a^2(x^2-ax+a^2)}dx$$

I suppose the first integral is easy to evaluate.

To evaluate the second one, try splitting it in the following way:
$$\int_0^R \frac{2a-x}{3a^2(x^2-ax+a^2)}\,dx=\frac{1}{6a^2}\left(\int_0^R \frac{3a}{x^2-ax+a^2}\,dx-\int_0^R \frac{2x-a}{x^2-ax+a}\,dx\right)$$

The first one can be handled by completing the square and the second one by the substitution $u=x^2-ax+a^2$.

The integral he needs is
$$\int_0^R \frac 1 {\left(1 + x^3\right)^3} \, dx$$

and not


$$\int_0^R \frac 1 {\left(1 + x^3\right)} \, dx$$

 

[Saitama]

The integral he needs is
$$\int_0^R \frac 1 {\left(1 + x^3\right)^3} \, dx$$

and not


$$\int_0^R \frac 1 {\left(1 + x^3\right)} \, dx$$

Yes, I am very well aware of that. And anyways, we are not evaluating $\int_0^R \frac{dx}{1+x^3}$ but rather, we are evaluating $\int_0^R \frac{dx}{a^3+x^3}$. :)

You should check my post #27, it states why we are doing so.

 

[LCKurtz]

Once the definite integral is evaluated, you will get something like
$$\int_0^{\infty} \frac{dx}{a^3+x^3}=f(a)$$
Since you need $(a^3+x^3)^3$ in the denominator, you simply differentiate both the sides with respect to $a$ twice and then substitute $a=1$.

And $f(a)$ comes out to be a very nice expression. :)

That's very nice Pranav-Arora. I've been watching this thread and I didn't see that coming!

 

[Saitama]

That's very nice Pranav-Arora. I've been watching this thread and I didn't see that coming!

Thank you LCKurtz for the appreciation.

 

[ehild]

Once the definite integral is evaluated, you will get something like
$$\int_0^{\infty} \frac{dx}{a^3+x^3}=f(a)$$
Since you need $(a^3+x^3)^3$ in the denominator, you simply differentiate both the sides with respect to $a$ twice and then substitute $a=1$.

And $f(a)$ comes out to be a very nice expression. :)

Very clever, Pranav!

ehild

 

[Saitama]

Very clever, Pranav!

ehild

Thank you very much ehild!