Integral representation for the solution of a difference equation 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.
$_{\text{Source: Advanced Mathematical Methods for Scientists and Engineers I, Asymptotic Methods and Perturbation Theory}}$
$_{\text{by Carl Bender and Steven Orzag}}$

 A: 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.
