Trouble in proving $\int_0^\infty x^3\frac{\sin\frac{\pi x}{2}\sinh\frac{\pi x}{2}}{\cos\pi x+\cosh\pi x}(J_0(x)+I_0(x))dx=0$ I want to prove that
$$
\int_0^\infty x^3\frac{\sin\frac{\pi  x}{2}\sinh\frac{\pi  x}{2}}{\cos\pi x+\cosh\pi x}(J_0(x)+I_0(x))dx=0,\tag{1}
$$
where $J_0(x)$ and $I_0(x)$ are Bessel function and modified Bessel function of the first kind.
What I know? This
$$
\int_0^\infty x^{4n-1}\frac{\sin\frac{\pi  x}{2}\sinh\frac{\pi  x}{2}}{\cos\pi x+\cosh\pi x}dx=0,\,\,\,\,\, n\in\mathbb{N},\tag{2}
$$
and this
$$
J_0(x)+I_0(x)=\sum_{n=0}^\infty a_nx^{4n}.\tag{3}
$$
However the trouble is I don't know how to combine $(2)$ and $(3)$ rigorously to get $(1)$.
For example this is not true:
$$
\int_0^\infty x^3\cos x^2\frac{\sin\frac{\pi  x}{2}\sinh\frac{\pi  x}{2}}{\cos\pi x+\cosh\pi x}dx=0.\,\,\,\,\,(\text{wrong})\tag{4}
$$

Q: Explain rigorously why $(1)$ is true and $(4)$ is wrong.

Note. I believe closed forms of $(1)$ and $(4)$ can be derived by residue theorem. However in answering this question I want to avoid explicit calculations and understand how to get from $(2)$ and $(3)$ to $(1)$ and also to understand why the same reasoning fails in case of $(4)$.
 A: The short answer is that the Dominated Convergence Theorem may be applied to one, but not the other. We know that
$$ f(x) = \sin{(\pi x/2)}g(x) = \frac{\sin{(\pi x/2)}\sinh{(\pi x/2)}}{\cos{\pi x}+\cosh{\pi x}} \sim \sin{(\pi x/2)} e^{-\pi x/2} $$
as $x \to \infty$. Therefore even if we replace the sine by $1$, most of the integrals we are interested in will still converge; in particular $ \int_0^{\infty} x^{4k+3} g(x) \, dx $ and $\int_0^{\infty} x^3( J_0+I_0) g(x)$ exist. We from now on take the measure to be $g(x) \, dx$ to save writing.
Also, for (1), since in (3) the $a_n$ are all positive (easy to verify),
$$0 \leq \sum_{k=0}^n a_k x^{4k+3} \leq x^3(J_0(x)+I_0(x)). $$
Putting all this together, $\sin{(\pi x/2)}\sum_{k=0}^n a_k x^{4k+3}$ are a sequence of integrable functions that converge pointwise (by Taylor's theorem) to $x^3(J_0(x)+I_0(x))\sin{(\pi x/2)}$ and are dominated by the integrable function $x^3(J_0(x)+I_0(x))$, and hence the Dominated Convergence Theorem applies and the limit and the sum may be interchanged.

This does not work for $x^3\cos{(x^2)}$. There are two problems: firstly the coefficients in the Taylor series are not nonnegative, so the limit function is no use for domination. Secondly, the obvious guess for a dominating function is $\cosh{(x^2)}$, which obviously doesn't work since it is not integrable. One can surely prove that there is no dominating function (probably using that the sum of the absolute values looks like $\cosh{(x^2)}$ somewhere), but since you have numerical evidence that the limit and the integral don't commute, there can't be one.
