Double Integration of Bessel Functions

[csprh]

Hi
I have proved (through educated guess-work and checking analytically) the following identity

$\int\limits_0^\infty\int\limits_0^\infty s_1 \exp\left(-\gamma s_1\right) s_2 \exp\left(-\gamma s_2\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

$J_0$ is a Bessel function of the first kind.

However I can't find this identity in any book and therefore stuck in how to give a rigorous reference (rather than say I guessed it, and I was right).

However I have found a related single integral identity (that IS in a book) and wonder if it could be used to prove the above.

$\int_0^\infty \! s \exp\left(-\gamma s\right)J_0\left(sr\right)ds = \frac{\gamma}{\left(r^2+\gamma^2\right)^{3/2}}$


I have tried Mathematica, Maple, Matlab,... you name it but to no avail. I've trawled all the usual book suspects (of integral tables) but also... zilch.

Any ideas of where to start or any leads here?

thanks in advance.

Paul

 

[Mandelbroth]

Hi
I have proved (through educated guess-work and checking analytically) the following identity

$\int\limits_0^\infty\int\limits_0^\infty s_1 \exp\left(-\gamma s_1\right) s_2 \exp\left(-\gamma s_2\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

$J_0$ is a Bessel function of the first kind.

However I can't find this identity in any book and therefore stuck in how to give a rigorous reference (rather than say I guessed it, and I was right).

However I have found a related single integral identity (that IS in a book) and wonder if it could be used to prove the above.

$\int_0^\infty \! s \exp\left(-\gamma s\right)J_0\left(sr\right)ds = \frac{\gamma}{\left(r^2+\gamma^2\right)^{3/2}}$I have tried Mathematica, Maple, Matlab,... you name it but to no avail. I've trawled all the usual book suspects (of integral tables) but also... zilch.

Any ideas of where to start or any leads here?

thanks in advance.

Paul

Let's try. $$I=\int\limits_{[0,+\infty)}\int\limits_{[0,+\infty)} s_1 \exp\left(-\gamma s_1\right) s_2 \exp\left(-\gamma s_2\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right)~ds_1~ds_2\\ I=\int\limits_{[0,+\infty)}s_2 \exp\left(-\gamma s_2\right)J_0\left(s_2r_2\right)\int\limits_{[0,+\infty)} s_1 \exp\left(-\gamma s_1\right) J_0\left(s_1r_1\right)~ds_1~ds_2.$$ Can you take it from there?

 

[csprh]

Thanks Mandlebroth.

I was looking at this and wondering how I could be so stupid not to see something so obvious. However noticing that

$\frac{\gamma}{\left(r_1^2+\gamma^2\right)^{3/2}} \times \frac{\gamma}{\left(r_2^2+\gamma^2\right)^{3/2}} \neq \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

I rechecked and found an error in the first equation.

$\int\limits_0^\infty\int\limits_0^\infty s_1 \exp\left(-\gamma s_1\right) s_2 \exp\left(-\gamma s_2\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

should actually be
$\int\limits_0^\infty\int\limits_0^\infty s_1 s_2 \exp\left(-\gamma \sqrt{s_1^2+s_2^2}\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

As this is not separable (in the same way) it makes it a bit more tricky. Any ideas?

 

[jackmell]

$\int\limits_0^\infty\int\limits_0^\infty s_1 s_2 \exp\left(-\gamma \sqrt{s_1^2+s_2^2}\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

As this is not separable (in the same way) it makes it a bit more tricky. Any ideas?

So you walk into class one day and the professor, who's very eccentric, writes it on the board and says, "ok, come back next class with something on your own and I'll be checking PF to see if you're cheating" and then dismisses class. He's teaching you a valuable lesson. Do you know?

An integral part of success in math is trying things and your ability to tolerate failure in that endeavor. So next class, I'm going to give our poor dear professor exactly what he's asking for: something (meaningful) I tried even if it's wrong because that is the exercise he's really giving to us and not so much the solution. :)

 

[csprh]

Jackmell

I do understand that this forum is used by Maths undergrads to skip doing their homework. However, I'm a engineering postdoc who is completely hopeless at (this type of) maths and uses it as black box solutions. There's nobody in this department who works with this stuff so I thought I'd come here.

I did get an expression for the single inner integral but the resulting integral was so hideous that I don't know where to start to do the second. The paper I have written does not really depend on this as the identity has been proved (analytically).

any help would be appreciated.

Paul

 

[Mandelbroth]

Thanks Mandlebroth.

I was looking at this and wondering how I could be so stupid not to see something so obvious. However noticing that

$\frac{\gamma}{\left(r_1^2+\gamma^2\right)^{3/2}} \times \frac{\gamma}{\left(r_2^2+\gamma^2\right)^{3/2}} \neq \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

I rechecked and found an error in the first equation.

$\int\limits_0^\infty\int\limits_0^\infty s_1 \exp\left(-\gamma s_1\right) s_2 \exp\left(-\gamma s_2\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

should actually be
$\int\limits_0^\infty\int\limits_0^\infty s_1 s_2 \exp\left(-\gamma \sqrt{s_1^2+s_2^2}\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2 = \frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

As this is not separable (in the same way) it makes it a bit more tricky. Any ideas?

Try the "separation" in a similar way. Though I disagree with jackmell's somewhat frequent untactful use of anecdotes, he has a point. You should show a little more work, according to the rules of the forum. I appreciate your situation, and I'd be happy to help you.

 

[jackmell]

Jackmell

I do understand that this forum is used by Maths undergrads to skip doing their homework. However, I'm a engineering postdoc who is completely hopeless at (this type of) maths and uses it as black box solutions. There's nobody in this department who works with this stuff so I thought I'd come here.

I did get an expression for the single inner integral but the resulting integral was so hideous that I don't know where to start to do the second. The paper I have written does not really depend on this as the identity has been proved (analytically).

any help would be appreciated.

Paul

Hi Paul,

I really meant no offense; mostly play but I do make a point afterall. I tried integration by parts and failed miserably. However after reading Mandelbroth's post, perhaps I missed something and will work with it some more.

 

[Bill Simpson]

Maybe I'm missing something here, but in a few seconds I get the following

In[1]:= $Assumptions = {r1 > 0 && r2 > 0 && gamma > 0};
Integrate[Integrate[s1 Exp[-gamma s1] s2 Exp[-gamma s2] BesselJ[0, s1 r1] BesselJ[0, s2 r2], {s1, 0, Infinity}], {s2, 0, Infinity}]

Out[2]= gamma^2/((gamma^2 + r1^2)*(gamma^2 + r2^2))^(3/2)

Substituting some (positive) random values for r1, r2 and gamma into this result gives very different values for my result and for the result that you found.

I assume this means I have broken or misunderstood something.

If you can point out my errors or post your Mathematica code then perhaps we can track this down.

 

[csprh]

Hi Bill

Sorry, this is my fault. I got the initial question wrong (see my second post).

The mathematica code would be (from the corrected integral).

in[1] := $Assumptions = {r1 > 0 && r2 > 0 && gamma > 0};
Integrate[ Integrate[ s1 s2 Exp[-gamma Sqrt[s1*s1 + s2*s2]] BesselJ[0, s1 r1] BesselJ[0,
s2 r2], {s1, 0, Infinity}], {s2, 0, Infinity}]

Mathematica gives up and gives the question back as the answer.

 

[csprh]

So my stab at this.

The original double integral is
$\int\limits_0^\infty\int\limits_0^\infty s_1 s_2 \exp\left(-\gamma \sqrt{s_1^2+s_2^2}\right) J_0\left(s_1r_1\right) J_0\left(s_2r_2\right) ds_1ds_2$

Taking this one integral at a time (and extracting terms that need not be included in the second integral).

$\int\limits_0^\infty s_2 J_0\left(s_2r_2\right) \left(\int \limits_0^\infty s_1 \exp\left(-\gamma \sqrt{s_1^2+s_2^2}\right) J_0\left(s_1r_1\right) ds_1 \right) ds_2$

Using an identity from

http://pi.physik.uni-bonn.de/~dieckman/IntegralsDefinite/DefInt.html

Setting $m = 0, x = s_1, c = r_1, b = 1, a = s_2, p = \gamma$ The middle integral

$\int \limits_0^\infty s_1 \exp\left(-\gamma \sqrt{s_1^2+s_2^2}\right) J_0\left(s_1r_1\right) ds_1$

can be expressed as

$\sqrt{\frac{2}{r_1 \pi}}\frac{s_2 \gamma}{\left(\gamma^2+r_1^2\right)^{3/4}}\sqrt{s_2 r_1}K_{1.5}\left(s_2 \sqrt{\gamma^2+r_1^2}\right)$

The final double integral should therefore be able to be expressed as the single integral

$\int\limits_0^\infty s_2 J_0\left(s_2r_2\right) \left(\sqrt{\frac{2}{r_1 \pi}}\frac{s_2 \gamma}{\left(\gamma^2+r_1^2\right)^{3/4}}\sqrt{s_2 r_1}K_{1.5}\left(s_2 \sqrt{\gamma^2+r_1^2}\right)\right) ds_2$

I then plugged this into mathematica (using the following code):

$Assumptions = {g >= 0 && r1 >= 0 && r2 > 0}; Integrate[
s2*BesselJ[0, s2*r2]*
Sqrt[2/(Pi*r1)]*((s2*g)/((g^2 + r1^2)^(0.75)))*(Sqrt[s2*r1])*
BesselK[1.5, s2*Sqrt[g^2 + r1^2]], {s2, 0, Infinity}]

Giving the result of (after simplifying and running through wolfram alpha)

$= 3\frac{\gamma}{\left(r_1^2+r_2^2+\gamma^2\right)^{5/2}}$

which is what I had originally found analytically (minus the multiplication factor of 3). The mathematica code seemed very sensitive to the assumptions and gave all sorts of rubbish before I used the specific combinations above.

Thanks for the replies (Jackmell, Mandlebroth etc). Actually, you were quite right, I needed to be prodded into looking into this a bit more thoroughly. The result is actually really cute and is probably entirely useless to anyone but me, but is a new identity for anyone else.

Thanks for putting up with me here, I feel I have restored faith in my meagre mathematical ability.

 

[jackmell]

Outstanding crpsh! I couldn't do it and thanks for posting your results.