question on integral using Bessel function of zero order $\int_0^1 x J_0^2(\lambda x) \,dx$

Question

according to this PDF at Georgia Tech University math dept web site http://people.math.gatech.edu/~herod/conted/M33.pdf at page 11 of the above PDF file, it says that

But I am not able to verify the above. Using Maple and Mathematica, I get that

$$ \int_0^1 x J_0^2(\lambda x) \,dx = \frac{1}{2} \left( J_0^{2}(\lambda) + J_1^2(\lambda) \right) $$

And not as the above PDF says (using $a=1$)

$$ \int_0^1 x J_0^2(\lambda x) \,dx = \frac{1}{2} (J_0^{'}(\lambda))^2 $$

I tried using the following recurrence relations from https://dlmf.nist.gov/10.51 to see if I can convert one result to the other, but not able to

Here is Maple and Mathematica result

ClearAll[lam, x, a]
Integrate[ x (BesselJ[0, lam x])^2, {x, 0, 1}]

restart;
int(x*(BesselJ(0,lambda*x))^2,x=0..1);

So the question is, how to show that

$$ \frac{1}{2} (J_0^{'}(\lambda))^2 = \frac{1}{2} \left( J_0^{2}(\lambda) + J_1^2(\lambda) \right) $$

For example, replacing $J_0^{'}(\lambda)$ by $-J_1(\lambda)$ using (10.51.2) the above becomes

$$ \frac{1}{2} J_1^2(\lambda) = \frac{1}{2} \left( J_0^{2}(\lambda) + J_1^2(\lambda) \right) $$

Which is true only if $J_0(\lambda)$ is zero.

Is it possible the PDF file above has an error in it?

Answer 1

• Obviously the PDF has a typo: in the first relation $i$ should be $k$.

• In your case with $J=J_0$ the $\lambda_k$ are the zeroes of $J_0$. So if you choose a special $\lambda = \lambda_p$ you have in deed $J_0(\lambda)=0$.

• For the relation, see also https://en.wikipedia.org/wiki/Bessel_function#Properties.

• If you use a Bessel zero in the integral int(x*BesselJ(0,BesselJZeros(0,n)*x))^2,x=0..1); Maple returns 1/2 BesselJ(1, BesselJZeros(0, n))