# 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 $(4)$.

• How do we know that (4) is wrong? Oct 5, 2017 at 14:06
• @Chappers I checked it numerically. Oct 5, 2017 at 15:13
• @ Nick : Thanks a lot for your good critic, the criteria which I've used doesn't work. (I've deleted.) Hope someone find the correct theorems. Oct 5, 2017 at 18:27
• Please, where do you’ve read about $(2)$ ? --- Be $n\in\mathbb{N}_0$ . It seems to be $\displaystyle \int\limits_0^\infty x^{2n}\frac{\sin(\frac{\pi x}{2})\sinh(\frac{\pi x}{2})}{\cos(\pi x)+\cosh(\pi x)}dx = \frac{(-1)^n E_{2n}}{2^{n+2}}\sqrt{2}\sin\left(\frac{\pi}{4}(2n+1)\right)$ where $E_n$ are the Euler numbers coming from the Taylor series (development around $0$) of $\displaystyle \frac{2}{e^x+e^{-x}}$ and I'd like to know the formula for $\displaystyle \int\limits_0^\infty x^{4n+1}\frac{\sin(\frac{\pi x}{2})\sinh(\frac{\pi x}{2})}{\cos(\pi x)+\cosh(\pi x)}dx$ . Thanks in advance! :-) Oct 6, 2017 at 13:44
• @user90369 Note that the fraction is the real part of $\frac{1}{2i}\operatorname{sech}{\left( \frac{1+i}{2}\pi z \right)}$. You can then calculate the integral by moving the contour to where the hyperbolic secant is real using Cauchy's Theorem on a wedge, then using the Hurwitz zeta integral. ($\int_0^{\infty} t^{s-1} \operatorname{sech}{at} \, dt = 2^{1-2s}a^{-s} \Gamma(s) (\zeta(s,1/4)-\zeta(s,3/4))$ for $\Re s,\Re a>0$, for example.) Oct 6, 2017 at 13:54

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.