Why does the Laplace transform shift theorem not work with a positive shift?

Question

I am trying to compute the integral

$$ I = \int\limits_1^\infty e^{-t x} \sqrt{t} J_1\!\left(2 \sqrt{t}\right) \mathrm{d} t $$

which I encountered in my research. Now, I noticed that it has the same form as the Laplace transform, if only that lower integration limit were 0 instead of 1. I thought I could apply the shift theorem:

$$ f\!\left(t\right) = \sqrt{t} J_1\!\left(2 \sqrt{t}\right) $$

$$ I = \int\limits_1^\infty e^{-t x} f\!\left(t\right) \Theta\!\left(t\right) \mathrm{d} t = e^{-x} \int\limits_0^\infty e^{-t x} f\!\left(t + 1\right) \Theta\!\left(t + 1\right) \mathrm{d} t $$

My logic is that, in the first integral, since $t > 0$ always, it doesn't matter if we insert the Heaviside theta function; the second step is just a linear change of variables.

As far as I can tell, this is the right form for the Laplace transform of a shifted function except that the shift $a$ is supposed to satisfy $a > 0$ where $f\!\left(t - a\right)$ is the shifted function. Numerically, if I choose a value for $x$, I can do this integral, and the formula above holds:

(* the target function *)
f[t_] := Sqrt[t] BesselJ[1, 2 Sqrt[t]]

(* the original integral *)
With[{x = 1}, 
 NIntegrate[Exp[-t x] f[t], {t, 1, Infinity}, AccuracyGoal -> 5]]

(* the shifted integral *)
With[{x = 1}, 
 Exp[-x] NIntegrate[
   Exp[-t x] f[t + 1] HeavisideTheta[t + 1], {t, 0, Infinity}, 
   AccuracyGoal -> 5]]

>> 0.175032
>> 0.175032

However, the Laplace transform of $f\!\left(t\right)$ is given by

$$ F\!\left(x\right) = \frac{e^{-1/x}}{x^2} $$

and the Wikipedia table says that

$$ \mathcal{L}\!\left[f\!\left(t - a\right) \Theta\!\left(t - a\right)\right] = e^{-a x} F\!\left(x\right) $$

The factors of $e^{-x}$ and $e^x$ cancel, and I'm left with

With[{x = 1}, Exp[-1/x] / x^2 // N]

>> 0.135335

I get a different answer. I suspect this all has something to do with my shift being positive, but I don't understand conceptually why that should be a problem, since I never assumed $a > 0$ explicitly. I appreciate any help solving this problem.

Answer 1

The Laplace transform of $t \mapsto u(t-c) f(t)$ for $c>0$ has no simple relationship with that of $f$, intuitively because some of the information about $f$ on $[0,\infty)$ has been destroyed. The shift theorem applies to $u(t-c) f(t-c)$.

Answer 2

Just for the integral of your interest.

$$I = \int\limits_1^\infty e^{-t x} \sqrt{t}\, J_1\!\left(2 \sqrt{t}\right) \,dt=\int\limits_0^\infty e^{-t x} \sqrt{t}\, J_1\!\left(2 \sqrt{t}\right) \,dt-\int\limits_0^1 e^{-t x} \sqrt{t}\, J_1\!\left(2 \sqrt{t}\right) \,dt$$ As you wrote, assuming $x >0$, $$\int\limits_0^\infty e^{-t x} \sqrt{t}\, J_1\!\left(2 \sqrt{t}\right) \,dt=\frac{e^{-1/x}}{x^2}$$ Now, for the second integral, use $$\sqrt{t}\, J_1\!\left(2 \sqrt{t}\right)=\sum_{n=1}^\infty (-1)^{n+1}\frac {t^n}{n! \,(n-1)! }$$ which makes $$\int\limits_0^1 e^{-t x} \sqrt{t}\, J_1\!\left(2 \sqrt{t}\right) \,dt=\sum_{n=1}^\infty (-1)^n \frac{\Gamma (n+1,x)-\Gamma (n+1,0)}{n!\,(n-1)! }\, x^{-(n+1)}$$ For $x=1$, the partial sums (from $n=1$ to $n=p$) converge extremely fast $$\left( \begin{array}{cc} p & \sum_{n=1}^p \\ 1 & 0.2642411177 \\ 2 & 0.1839397206 \\ 3 & 0.1934337990 \\ 4 & 0.1928238246 \\ 5 & 0.1928485823 \\ 6 & 0.1928478886 \\ 7 & 0.1928479028 \\ 8 & 0.1928479026 \end{array} \right)$$