# Thread: stuck during computation of an integral

1. ## stuck during computation of an integral

Hi everybody,

I have a function:
$\displaystyle h(x,y,z)=\frac{z(x^2+2y^2)}{8x^4\sqrt{x^2+y^2}}ata n\left(\frac{z}{\sqrt{x^2+y^2}}\right)$,
and I would like to integrate it twice against x to obtain $\displaystyle h_2(x,y,z)=\iint h(x,y,z)dx dx$.
More precisely, I need to compute
$\displaystyle \int_{x_1=0}^{a_1}dx_1\int_{x=d+a1-x_1}^{d+a_1+a_2-x_1}dx\,h(x,y,z)$,
which can also be written
$\displaystyle \left[h_2(x,y,z)\right]_{x=d+a_1}^{x=d+a_1+a_2}+\left[h_2(x,y,z)\right]_{x=d+a_2}^{x=d}$.
However there are no symmetries that can be exploited. $\displaystyle d, a_1, a_2$ can have any positive real value.

Until now I have obtained a quite complicated expression for $\displaystyle h_2$ with explicit terms and two simple integrals which I can't figure out how to compute, namely :
$\displaystyle \int \frac{\sqrt{x^2+y^2}}{x}atan\left(\frac{z}{\sqrt{x ^2+y^2}}\right)dx$,
and
$\displaystyle \int \frac{\sqrt{x^2+y^2}}{x^3}atan\left(\frac{z}{\sqrt {x^2+y^2}}\right)dx$.

Any idea to integrate this function ? I have not been lucky with variable changes...

Thanks,
Kind regards,
Arixo

PS: for those interested, it is an integration involved in the computation of the van der Waals interaction energy between two solid cubes of size a1 and a2 and separated by a distance d in the x direction...

2. ## Re: stuck during computation of an integral

Before trying to do the integrals, let's make sure that you've set the problem up correctly. What's the reference for this problem?

3. ## Re: stuck during computation of an integral

of course, I understand.
I would like to extend the calculations presented in this paper :
http://www.ncbi.nlm.nih.gov/pmc/arti...00207-0051.pdf

I'd like to obtain relation (25) for a general case similar to that of figure 6, but with a=c=a1 for the first cube, a=c=a2 for the second cube, and the distances between them being hx, hy and hz in the x, y, and z directions respectively (instead of being hx=hy=d and hz=0 in the paper).

Thus it is necessary to compute
$\displaystyle K6=\int_0^{a_1}dx_1\int_{hx+a_1-x_1}^{hx+a_1+a_2-x_1}dx \int_0^{a_1}dy_1\int_{hy+a_1-y_1}^{hy+a_1+a_2-y_1}dy \int_0^{a_1}dz_1\int_{hz+a_1-z_1}^{hz+a_1+a_2-z_1}dz (x^2+y^2+z^2)^{-3}$

If I set $\displaystyle f(x,y,z)=(x^2+y^2+z^2)^{-3}$, I have computed (actually my computer did it) $\displaystyle f_2(x,y,z)=\frac{3z}{8(x^2+y^2)^{5/2}}atan\left[\frac{z}{\sqrt{x^2+y^2}}\right]-\frac{1}{8(x^2+y^2)(x^2+y^2+z^2)}$ so that $\displaystyle \partial^2_ {zz}f_2=f$.

Then evaluating f2 at the bounds, the first two integrals produce 4 terms (following the remark in my first post) : $\displaystyle f_2(x,y,hz+a_1+a_1)+f_2(x,y,hz)-f_2(x,y,hz+a_1)-f_2(x,y,hz+a_2)$
Since it seems that the sum of these 4 terms produce no simplification, I just continue with a generic function $\displaystyle f_2(x,y,z)$, keeping in mind that z is one of the 4 aforementioned values.

Then I compute the 3rd and 4th integrals to obtain f4 so that $\displaystyle \partial^2_{yy}f_4=f_2$ and it is possible to find :
$\displaystyle f_4(x,y,z) = -\frac{3y}{16x^3}atan(y/x) + \frac{z(2y^2+x^2)}{8x^4\sqrt{x^2+y^2}}atan\left[\frac{z}{\sqrt{x^2+y^2}}\right] + \text{last term with y and z interchanged}$

The middle term in f4 is the h(x,y,z) of my first post. I am stuck when attempting to compute f6 such that $\displaystyle \partial^2_{xx}f_6=f_4$.

By the way, is $\displaystyle [_{a,b}^{c,d}f]$ standard notation in English to mean $\displaystyle [_{a}^{c}f]+[_{b}^{d}f]=f(c)-f(a)+f(d)-f(b)$ ?

Thank you for the attention you showed to this problem.
Kind regards,
Arixo

4. ## Re: stuck during computation of an integral

OK, give me a little time to look into this.

5. ## Re: stuck during computation of an integral

Just to clarify what you're asking: in your original post you asked for help evaluating an integral. Integrals of this type are evaluated in the paper, although the steps are skipped. Are you asking for the steps?

I'll need to check your calculation of $\displaystyle K_6$ for the general case but its evaluation should be similar to cases covered in the paper.

6. ## Re: stuck during computation of an integral

Sorry for the delay, I must have missed the e-mail notifying your answer. I am still interested of course.

You got exactly the problem : these integral were computed in the paper, but I have difficulties to recover the steps. And it is likely that the integration in the general case will be similar. If I could understand how to go from (24) to (25) in the paper, I think I could finish by myself for the general case.
Actually my first post was just about going from (24) to (25).

7. ## Re: stuck during computation of an integral

The notation in this paper is a little strange. What do the commas in the limits mean?

8. ## Re: stuck during computation of an integral

So you confirm it is non-standard notation... From the computations I did myself for the beginning of the paper I guessed it is $[_{a,b}^{c,d}f]$=$[_{a}^{c}f]+[_{b}^{d}f]=f(c)-f(a)+f(d)-f(b)$.
I worked a little on one of the integrals (the one with 1/x, not the one with 1/x^3) and I found prettier forms by replacing x/y by Z and than 1+Z^2 by U^2. But I still have to integrate. I'll have a look later this evening.

9. ## Re: stuck during computation of an integral

The notation is definitely not standard but then again it is an old paper.

I'm referring to integrals like in equation (20):

$\displaystyle \int_{d+c, d+c}^{d+2c, d}$

I'll continue to look online and see if I can find something.

10. ## Re: stuck during computation of an integral

ok. I found that $\displaystyle \int_{d+c,d+c}^{d+2c,d}=\int_{d+c}^{d+2c}+\int_{d+ c}^{d}$.
The expression inside (19) and inside (20) is the same: say f(z).
(19) is $\displaystyle \int_{0}^{c}(f(z=d+2c-z1)-f(z=d+c-z1))dz1$
$\displaystyle =\int_{0}^{c}f(z=d+2c-z1)dz1 -\int_{0}^{c}f(z=d+c-z1))dz1$$\displaystyle =-\int_{d+2c}^{d+c}f(Z1)dZ1 +\int_{d+c}^{d}f(Z1)dZ1$
with $\displaystyle Z1=d+2c-z1$ in the left integral and $\displaystyle Z1=d+c-z1$ in the right one.
so that (19) $\displaystyle =\int_{d+c}^{d+2c}f(Z1)dZ1 +\int_{d+c}^{d}f(Z1))dZ1$

11. ## Re: stuck during computation of an integral

This looks plausible. I've written to someone to try and confirm that this is the meaning of the notation.

12. ## Re: stuck during computation of an integral

OK, I agree with your interpretation. So what exactly is the problem going from (24) to (25)? It's similar to going from (20) to (21).

13. ## Re: stuck during computation of an integral

you are right. The problem is also in integrating the last two terms of (20).
But now I think I could try to guess their primitive, because I can compute the primitives of the other terms in (20), and by difference I may be able to obtain what I want. But I think the coefficients in the brackets of the last terms in (20) are different of those I must consider, and it can complicate things.
I need some time to try this.

14. ## Re: stuck during computation of an integral

I found something interesting using the step (20)->(21) in the paper:
$\displaystyle \int a\left[\frac{1}{6b^2z}-\frac{1}{3z^3}\right]\sqrt{z^2+b^2}atan\left[\frac{a}{\sqrt{z^2+b^2}}\right]dz + \text{same term with a and b interchanged} =a\left[\frac{1}{6z^2}+\frac{1}{6b^2}\right]\sqrt{z^2+b^2}atan\left[\frac{a}{\sqrt{z^2+b^2}}\right]+ \text{same term with a and b interchanged}+\frac{1}{6}ln(z)+ln(z^2+a^2+b^2) \left[-\frac{1}{4}+\frac{(a^2+b^2)^2}{12a^2b^2}\right]$
Now I have to split the two contributions in the integral because my coefficients are not the same...

Edit : Actually the coefficients are the same -> I may be able to finish. Let's spend a few unhappy hours with a pen and a paper...

15. ## Re: stuck during computation of an integral

I'll take a look at this in a few days; bit busy at the moment.

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