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The following is an example from my textbook book on asymptotic studies. How is this true? I've never worked with the Bessel functions directly and the author isn't clear on which definition of $J_0$ he uses (I've googled and found different ones).

Example 5 The solution of the initial value problem $$(n + 1)a_{n+1} =2na_n- na_{n-1}, ~~~~~ a_0=1, a_1 =0,$$ may be expressed as an integral which is equivalent to the difference equation together with the initial conditions: $$\displaystyle a_n = \frac{1}{n!} \int_0^\infty e^{1-t}t^nJ_0(2\sqrt{t})dt$$ where $J_0$ is the Bessel function of order zero.
$_{\text{Source: Advanced Mathematical Methods for Scientists and Engineers I, Asymptotic Methods and Perturbation Theory}}$ $_{\text{by Carl Bender and Steven Orzag}}$

p3ngu1n
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  • While there's a lot of definitions, they all describe the same function... As for whether or not this integral satisfies the recurrence relation, we can check this, usually integration by parts is involved. It's quite a different matter how to obtain it in the first place. Could you give a reference to the textbook? – Yuriy S Nov 12 '18 at 07:15
  • I did have a typo. Edited with source included. – p3ngu1n Nov 12 '18 at 17:35
  • I see. As I said, you should try integrating by parts and use the recurrence relations for Bessel functions – Yuriy S Nov 12 '18 at 18:02
  • See the answer for the full proof – Yuriy S Nov 13 '18 at 10:22

1 Answers1

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Since the solution is given (and it seems correct), let's just show the integral satisfies the recurrence. The factor $e$ doesn't change anything, so we will use:

$$a_n=\frac{1}{n!} \int_0^\infty e^{-t}t^nJ_0(2\sqrt{t})dt$$

There are two possible ways I see here. First is to use integration by parts and the known recurrence relations for the Bessel functions.

Another would be to use the series for $J_0$, which is simple enough:

$$J_0(2\sqrt{t})=\sum_{k=0}^\infty (-1)^k \frac{t^k}{k!^2}$$

We need to prove that:

$$\frac{1}{n!} \int_0^\infty e^{-t}t^{n+1} J_0(2\sqrt{t})dt= \\ =\frac{2}{(n-1)!} \int_0^\infty e^{-t}t^{n} J_0(2\sqrt{t})dt-\frac{n}{(n-1)!} \int_0^\infty e^{-t}t^{n-1} J_0(2\sqrt{t})dt$$

Or:

$$ \int_0^\infty e^{-t}t^{n+1} J_0(2\sqrt{t})dt=2n \int_0^\infty e^{-t}t^{n} J_0(2\sqrt{t})dt-n^2 \int_0^\infty e^{-t}t^{n-1} J_0(2\sqrt{t})dt$$

$$I_{n+1}=2n I_n-n^2 I_{n-1}$$

Let's see how we could use integration by parts. We have:

$$\int J_0(2\sqrt{t}) dt=\sqrt{t} J_1(2\sqrt{t}) \\ J_0(2\sqrt{t})'=-\frac{J_1 (2\sqrt{t})}{\sqrt{t}}$$

This is quite fortunate, as we obtain a similar expression in both cases. Now, clearly, we could try integration by parts with different choice of functions.

I'll just use the method of exhaustion and consider every possible choice of $u,dv$ for integration by parts. This was very exhausting, believe me. We start with an integral:

$$I_a=\int_0^\infty e^{-t} t^a J_0(2 \sqrt{t}) dt$$

$1) \quad u=e^{-t} J_0 (2 \sqrt{t}), \quad dv=t^a dt$

$$I_a=\frac{1}{a+1} I_{a+1}-\frac{1}{a+1} \int_0^\infty e^{-t} t^{a+1/2} J_1 dt$$

$2) \quad u=t^a J_0 (2 \sqrt{t}), \quad dv=e^{-t} dt$

$$I_a=a I_{a-1}- \int_0^\infty e^{-t} t^{a-1/2} J_1 dt$$

$3) \quad u=e^{-t}t^a, \quad dv=J_0 (2 \sqrt{t}) dt$

$$I_a=\int_0^\infty e^{-t} t^{a+1/2} J_1 dt- a\int_0^\infty e^{-t} t^{a-1/2} J_1 dt$$

$4) \quad u=e^{-t}t^a J_0 (2 \sqrt{t}), \quad dv= dt$

$$I_a=\int_0^\infty e^{-t} t^{a+1/2} J_1 dt- aI_a+I_{a+1}$$

$$(a+1)I_a=\int_0^\infty e^{-t} t^{a+1/2} J_1 dt+I_{a+1}$$

The last relation allows us to represent the other bothersome integral as:

$$J_a=\int_0^\infty e^{-t} t^{a+1/2} J_1 dt=(a+1)I_a-I_{a+1}$$

We can now see that $1)$ and $2)$ represent the same relation. From $3)$ we get:

$$I_a=J_a-aJ_{a-1}=(a+1)I_a-I_{a+1}-a(a I_{a-1}-I_a)$$

From which we obtain:

$$I_{a+1}=2aI_a-a^2 I_{a-1}$$

Which is just the relation we needed to prove.

Yuriy S
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