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 $(4)$.

  • $\begingroup$ How do we know that (4) is wrong? $\endgroup$
    – Chappers
    Oct 5, 2017 at 14:06
  • $\begingroup$ @Chappers I checked it numerically. $\endgroup$
    – Tyrell
    Oct 5, 2017 at 15:13
  • $\begingroup$ @ 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. $\endgroup$
    – user90369
    Oct 5, 2017 at 18:27
  • 1
    $\begingroup$ 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! :-) $\endgroup$
    – user90369
    Oct 6, 2017 at 13:44
  • $\begingroup$ @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.) $\endgroup$
    – Chappers
    Oct 6, 2017 at 13:54

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .