How to calculate this integral

[Saitama]

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

and not


<br /> \int_0^R \frac 1 {\left(1 + x^3\right)} \, dx<br />

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!

 

[BruceW]

haha, yeah me too. I was really unsure about how your method would get back to the original problem. That is one of the lesser-used methods of calculating difficult integrals, for sure! I've probably only used a method like that once or twice.

 

[Saitama]

haha, yeah me too. I was really unsure about how your method would get back to the original problem. That is one of the lesser-used methods of calculating difficult integrals, for sure! I've probably only used a method like that once or twice.

Thanks Bruce! Yes, you are right, this is one of the lesser-used methods but I recently came to know how powerful it is in evaluating some of the most awkward and difficult integrals. :)

Oh and BTW, I came up with a faster method to evaluate the definite integral.

In the definite integral, $\int_0^{\infty} \frac{dx}{x^3+a^3}$, use the substitution $x^3=a^3t$ to obtain:
$$\frac{1}{3a^2}\int_0^{\infty} \frac{t^{-2/3}}{1+t}\,dt$$
Then I use the result:
$$\int_0^{\infty} \frac{t^{-b}}{1+t}\,dt=\Gamma(b)\Gamma(1-b)=\frac{\pi}{\sin(\pi b)}$$
Hence,
$$\frac{1}{3a^2}\int_0^{\infty} \frac{t^{-2/3}}{1+t}\,dt=\frac{2\pi}{3\sqrt{3}a^2}$$

@Mark: The result I used is easy enough to prove once you know the gamma function, I suppose you are going to start with it soon. :)

 

[Ray Vickson]

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.

For the indefinite integral, you are absolutely correct. However, for the definite integral it does not make much difference; in fact, IMHO the $x^3+a$ form really is a bit easier than the $x^3+a^3$ form, inasmuch as there are fewer opportunities for errors, etc.

 

[mark.laidlaw19]

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. :)

Thank you very much. I see how this will get the correct denominator, but when you twice differentiate \int_0^R{\frac{1}{a^3+x^3}}wouldn't you end up with \int_0^R \frac{12a^4-6ax^3}{(a^3+x^3)^3}\,dx which will still have an x in the numerator even after setting a=1?

 

[Ray Vickson]

Thank you very much. I see how this will get the correct denominator, but when you twice differentiate \int_0^R{\frac{1}{a^3+x^3}}wouldn't you end up with \int_0^R \frac{12a^4-6ax^3}{(a^3+x^3)^3}\,dx which will still have an x in the numerator even after setting a=1?

Your second $a$-derivative is
\frac{\partial^2}{\partial a^2} \frac{1}{x^3+a^3} <br /> = \frac{18 a^4}{(x^3+a^3)^3} - \frac{6a}{(x^3+a^3)^2}
and you get the integral of the second term from
\frac{\partial}{\partial a} \frac{1}{x^3+a^3} = -\frac{3a^2}{(x^3+a^3)^2}

However, as I said before, if you look at the definite integral you can just use
\frac{\partial^2}{\partial b^2} \frac{1}{x^3+b} = \frac{2}{(x^3+b)^3}
in the integral
\int_0^{\infty} \frac{dx}{x^3+b} = \frac{2 \pi \sqrt{3}}{9\, b^{2/3}}

 

[vela]

Is there some reason you didn't evaluate this integral using contour integration?

 

[mark.laidlaw19]

Is there some reason you didn't evaluate this integral using contour integration?

questions asked explicitly for parametrization. Otherwise it would be much easier i agree

 

[Saitama]

Thank you very much. I see how this will get the correct denominator, but when you twice differentiate \int_0^R{\frac{1}{a^3+x^3}}wouldn't you end up with \int_0^R \frac{12a^4-6ax^3}{(a^3+x^3)^3}\,dx which will still have an x in the numerator even after setting a=1?

No. You can simplify a bit after the first derivative. :)

You have:
$$\int_0^{\infty} \frac{dx}{a^3+x^3}=\frac{2\pi}{3\sqrt{3}a^2}$$
Differentiating once both the sides give:
$$\int_0^{\infty} \frac{3a^2}{(a^3+x^3)^2}\,dx=\frac{4\pi}{3\sqrt{3}a^3} \Rightarrow \int_0^{\infty}\frac{dx}{(a^3+x^3)^2}= \frac{4\pi}{9\sqrt{3}a^5}$$
Differentiate again:
$$\int_0^{\infty} \frac{6a^2}{(a^3+x^3)^3}\,dx=\frac{20\pi}{9\sqrt{3}a^6} \Rightarrow \int_0^{\infty} \frac{dx}{(a^3+x^3)^2}=\frac{10\pi}{27\sqrt{3}a^8}$$

Now just substitute $a=1$. ;)