Why is a miracle happening on this double integral? I am trying to solve a double integral of the form
$$\int_0^\infty du\int_0^\infty dx \frac{1}{x}\left(\frac{\partial}{\partial x}\right)^{2n+1}f\left(\sqrt{x^2+u}\right),$$
where $f$ is a well-behaved even function that decays rapidly at infinity. The content of this question (and miracle) does not depend on the specific $f$, as long as it is even and well behaved, but $f(t)=e^{-t^2}$ is a fine example. Note that since $1/x$ multiplies an odd function, the integrand is nonsingular as $x\to0$.
I would like to have a solution for a generic integer $n$, in terms of the known quantities $f^{(m)}(0)$ (i.e., arbitrary $m$th derivatives of $f$ evaluated at zero).
Amazingly, such a solution seems to always exist (I have tested up to $n = 5$), but the method I am using is extremely tedious, relies on Mathematica, and results in infinite terms that miraculously cancel. So the question is, what's a more efficient and clear method to get the same result?

Step 1: Evaluate all $2n+1$ derivatives.
Step 2: Transform the first integral with the change of variables $u=y^2,du=2ydy$.
Step 3: Convert to polar coordinates.
Step 4: Evaluate the $\theta$-integral.
Step 5: What remains is an integral of the form $\int_0^\infty dr\left(c_1\frac{1}{r^{p_1}}f^{(q_1)}(r) + \cdots + c_r r^{p_N}f^{(q_N)}(r)\right)$. Some of the terms with powers of $x$ in the denominator are infinite, but integrating by parts on some terms always makes these cancel. The terms with positive powers of $x$ can be integrated by parts repeatedly until they become $\int_0^\infty dr f^{(m)}(r) = - f^{(m-1)}(0)$.

Steps 1–4 can be performed by Mathematica using the single line (here $n=2$):
Integrate[
 Assuming[{r, \[Theta]} \[Element] PositiveReals, 
  2r Tan[\[Theta]] D[f[Sqrt[x^2 + u]], {x, 2*2 + 1}] /. {x -> 
      r Cos[\[Theta]], u -> r^2 Sin[\[Theta]]^2} // 
   Simplify], {\[Theta], 0, \[Pi]/2}]

I invite you to copy/paste this line in, examine the output and verify what I have claimed in step 5. It may seem pointless for me to ask this question, given that I already have a method for getting to a solution; however, Mathematica starts taking a very long time to compute this even for $n=5$, and even still the above line does not consider the final integrations by parts that make several more terms vanish and combine, so I think a better method is needed for both efficiency and understanding's sake.
 A: Let $I$ denote the integral. Then by the Fubini's theorem and Leibniz's integral rule,
\begin{align*}
I 
&= \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \frac{\mathrm{d}^{2n+1}}{\mathrm{d}x^{2n+1}} \int_{0}^{\infty} \mathrm{d}u \, f(\sqrt{x^2+u}) \\
&= \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \frac{\mathrm{d}^{2n+1}}{\mathrm{d}x^{2n+1}} \int_{x}^{\infty} \mathrm{d}s \, 2s f(s) \tag{$s=\sqrt{x^2+u}$} \\
&= -2 \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \frac{\mathrm{d}^{2n}}{\mathrm{d}x^{2n}} x f(x) \\
&= -2 \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \bigl( x f^{(2n)}(x) + 2n f^{(2n-1)}(x) \bigr) \\
&= - 4n \int_{0}^{\infty} \mathrm{d}x \, \frac{f^{(2n-1)}(x)}{x}
\end{align*}
In the fourth step, we utilized the general Leibniz's rule, and in the last step $f^{(2n-1)}(0) = 0$ is used. Given this formula, I see no reason for this to reduce to a linear combination of $f^{(k)}(0)$'s. For instance,

*

*If $f(x) = e^{-x^2}$ and $n = 2$, then $I = -32\sqrt{\pi}$.


*If $f(x) = \frac{1}{1+x^2}$ and $n = 2$, then $I = -24\pi$.


*If $f(x) = \operatorname{sech} x$ and $n = 2$, then $I = 8\pi - \frac{1}{\pi^3}\psi^{(3)}\bigl(\tfrac{1}{4}\bigr) $, where $\psi^{(m)}$ is the polygamma function of order $m$.
