Fascinating property of the Dirichlet Beta function

Question

I was originally investigating the definite integral

$$\ I = \int_0^{\frac{\pi}{4}} \cos(\ln(\tan x))dx$$

After substituting $\ln(\tan x) = u$ and changing the upper and lower bounds, the integral simplified into

$$\ I = \int_0^{\infty} \frac{\cos u \cdot e^u}{1+e^{2u}}du$$

Which is nothing else but

$$\ I = \frac{1}{2} \cdot \int_0^{\infty} \frac{\cos u}{\cosh u} du$$

Now, I considered the Maclaurin series of $\cos x$:

$$ \cos x = \sum_{n=0}^{\infty}(-1)^n \cdot \frac{x^{2n}}{(2n)!}$$

Then, I substituted this summation inside the integral:

\begin{align} I &= \int_0^{\infty} \frac{e^u}{1+e^{2u}} \cdot \sum_{n=0}^{\infty}(-1)^n \cdot \frac{u^{2n}}{(2n)!} du \\ &= \int_0^{\infty} \frac{e^u}{1+e^{2u}}du - \frac{1}{2!}\cdot \int_0^{\infty} \frac{u^2e^u}{1+e^{2u}}du + \frac{1}{4!} \cdot \int_0^{\infty} \frac{u^4e^u}{1+e^{2u}}du +\ldots \end{align}

Rearranging the terms, we get:

$$ I = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!} \cdot \int_0^{\infty} \frac{u^{2n}e^u}{1+e^{2u}} du $$

As seen here and substituting $\ 2n$ instead of $\ n$, we get:

$$\int_0^{\infty} \frac{u^{2n}e^u}{1+e^{2u}} du = \Gamma(2n+1)\beta(2n+1) = (2n)!\beta(2n+1)$$

Substituting this equality into our original integral, note that the $(2n)!$ terms cancel and we obtain

$$\ I = \sum_{n=0}^{\infty}(-1)^n\beta(2n+1)$$

Meanwhile, from the comments in the answer of the same question, setting $\ t =1$, we have: $$\ I = \frac{1}{2}\int_0^{\infty} \frac{\cos u}{\cosh u} du = \frac{\pi}{4}\cdot \operatorname{sech}(\frac{\pi}{2})$$

Then, is it valid to assert the below equality?

$$\sum_{n=0}^{\infty}(-1)^n\beta(2n+1) = \frac{\pi}{4} \cdot \operatorname{sech}(\frac{\pi}{2})$$

Which seems to be true, as per WolframAlpha at least.

Can we derive it via any other method? The sum on the LHS feels like a daunting one, but it turns out to be equal to this rather simplified expression on the right.

Also - as pointed out in the comments, the limit as k approaches ${\infty}$ of $\beta(2k+1)$ should be $\ 1$. Given that this is true, how do we explain the fact that this sum converges?

Thank you for reading. Any suggestions are appreciated.

Answer 1

It seems the answer is pretty straightforward if one has knowledge about the Taylor expansion of $\operatorname{sech(z)}$ and its relation to the Euler numbers.

We have:

$$\operatorname{sech(z)} = \sum_{n=0}^{\infty}\frac{E_{2n}}{(2n)!}\cdot z^{2n}$$

Also, as pointed out in the comments, it is known that

$$\beta(2k+1) = \frac{(-1)^k\cdot E_{2k}}{2(2k)!}(\frac{\pi}{2})^{2k+1}$$

Multiplying $(-1)^k$ on both sides and taking the sum from $\ n = 0$ to $\infty$, we get the final result:

$$S = \sum_{n=0}^{\infty} (-1)^k\beta(2k+1) = \sum_{k=0}^{\infty} \frac{E_{2k}}{2(2k)!}\cdot (\frac{\pi}{2})^{2k} \cdot \frac{\pi}{2} = \frac{\pi}{4} \cdot \sum_{k=0}^{\infty}\frac{E_{2k}}{(2k)!}\cdot (\frac{\pi}{2})^{2k} = \frac{\pi}{4} \operatorname{sech(\frac{\pi}{2})}$$

I also noticed that if we rearrange $\ (-1)^k\beta(2k+1)$ a little, it turns out to be a sum of infinite GPs. Then it becomes equal to the sum:

$$\sum_{k=0}^{\infty}(-1)^k \beta(2k+1) = \sum_{k=0}^{\infty} \frac{(-1)^k(2k+1)}{1+(2k+1)^2}$$

which does indeed converge to $\frac{\pi}{4}\operatorname{sech(\frac{\pi}{2})}$

But the doubt regarding whether the sum should converge, if at all, remains unresolved. I cannot think of a reason to justify its convergence if the limit as n approaches infinity of $\beta(2n+1)$ is $1$.

Answer 2

Making the story short, if you enjoy the Gaussian hypergeometric function $$ I = \int \cos(\ln(\tan x))dx$$ $$I=\frac{1+i}{4}\, \tan ^{1-i}(x) \,\, _2F_1\left(\frac{1-i}{2},1;\frac{3-i}{2};- \tan ^2(x)\right)+$$ $$\frac{1-i}{4} \tan ^{1+i}(x) \,\, _2F_1\left(\frac{1+i}{2},1;\frac{3+i}{2};- \tan ^2(x)\right)$$ which is defined for $(0 \leq x \lt \frac \pi 2)$

$$ J = \int_0^{\frac{\pi}{4}} \cos(\ln(\tan x))dx$$ $$J=\frac 1 8\Bigg(\left(\psi\left(\frac{3+i}{4}\right)-\psi \left(\frac{3-i}{4}\right)\right)- \left(\psi \left(\frac{1+i}{4}\right)+\psi \left(\frac{1-i}{4}\right)\right) \Bigg)$$ $$J=\frac{\pi}{4} \ \text{sech}\left(\frac{\pi }{2}\right)$$