Proving that $\lim_{s \to 0^-} \int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} x^{s} \, dx = \int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, dx $

In an attempt to show that $$\int_0^\infty \frac{\arcsin(\sin x)}{x} \, \mathrm dx =2G,$$ where $G$ is Catalan's constant, I first evaluated the Mellin transform $$I(s)=\int_0^\infty \frac{\arcsin(\sin x)}{x} x^s\, \mathrm dx, \quad -1 < s< 0,$$ by using the fact that $\arcsin(\sin x)$ can be represented by the Fourier series $$\arcsin(\sin x) = \frac{4}{\pi} \sum_{n=0}^\infty \frac{(-1)^n \sin \left[(2n+1)x \right]}{(2n+1)^2}. $$

Replacing $\arcsin(\sin x)$ with this representation and then switching the order of summation and integration, which is justified by Fubini's theorem, I got $$ I(s) = \frac{4}{\pi} \beta(2+s) \Gamma(s) \sin \left(\frac{\pi s}{2} \right),$$ where $\beta(s)$ is the Dirichlet beta function.

Now I want to argue that $$\lim_{s \to 0^-} I(s) = \int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm dx $$ so that I can conclude that$$\begin{align} \int_0^\infty \frac{\arcsin(\sin x)}{x} \, \mathrm dx &= \frac{4}{\pi} \lim_{s \to 0^-} \beta(2+s) \Gamma(s) \sin \left(\frac{\pi s}{2} \right) \\ &= \frac{4}{\pi} \beta(2) \lim_{s \to 0^-} \left( \frac{1}{s} + \mathcal{O}(1) \right) \sin \left(\frac{\pi s}{2} \right) \\ &= 2 G. \end{align} $$

The issue is that $\int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm dx$ doesn't converge absolutely, so we can't appeal immediately to the dominated convergence theorem.

In the case of the Laplace transform, it turns out that $$\lim_{s \to 0^+} \int_0^\infty \frac{f(x)}{x} e^{-sx} \, \mathrm dx = \int_0^\infty \frac{f(x)}{x} \, \mathrm dx$$ if $\int_0^\infty \frac{f(x)}{x} \, \mathrm dx$ exists as an improper Riemann integral. This has to do with the fact that the improper integrals $\int_0^\infty \frac{f(x)}{x} e^{-sx} \, \mathrm dx$ converge uniformly on $[0, \infty)$.

(See here and here.)

I don't know if the Mellin transform has a similar property, but I imagine it does since the two transforms are related.

Solution 1:

Here is a general statement:

Proposition. Let $g : (0, \infty) \to \mathbb{R}$ be a locally integrable function such that $$ I(s) = \int_{0}^{\infty} \frac{g(x)}{x^s} \, \mathrm{d}x = \lim_{\substack{a \to 0^+ \\ b \to \infty}} \int_{a}^{b} \frac{g(x)}{x^s} \, \mathrm{d}x $$ exists and is finite at $s \in \{0, \delta\}$ for some $\delta > 0$. Then $I(s)$ exists for any $s \in [0, \delta]$ and $$ \lim_{s \to 0^+} I(s) = I(0). $$

We first illustrate a quick proof. If we write

$$ \int_{0}^{\infty} \frac{g(x)}{x^s} \, \mathrm{d}x = \int_{0}^{1} \frac{g(x)}{x^{\delta}}x^{\delta-s} \, \mathrm{d}x + \int_{1}^{\infty} g(x)x^{-s} \, \mathrm{d}x, $$

then the right-hand side can be handled by the abelian theorem for the Laplace transform via suitable substitution.

Next we provide a more self-contained proof:

Step 1. Since $g$ is locally integrable, it admits an antiderivative $G : (0, \infty) \to \mathbb{R}$. Also, by noting that

$$ \int_{a}^{b} g(x) \, \mathrm{d}x = G(b) - G(a) $$

and it converges as $a\to0^+$ and $b\to\infty$, we find that both

$$ G(0^+) = \lim_{a\to0^+} G(a) \qquad\text{and}\qquad G(\infty) = \lim_{b\to\infty} G(b) $$

exist and are finite. Similarly, the map $x \mapsto g(x)x^{-\delta}$ admits an antiderivative $H : (0, \infty) \to \mathbb{R}$ with finite limits at both endpoints of $(0, \infty)$.

Step 2. Now let $s > 0$ and assume that $1 < b$. Then

\begin{align*} \int_{1}^{b} \frac{g(x)}{x^s} \, \mathrm{d}x &= \int_{1}^{b} g(x) \biggl( \int_{x}^{\infty} \frac{s}{u^{1+s}} \, \mathrm{d}u \biggr) \, \mathrm{d}x \\ &= \int_{1}^{\infty} \frac{s}{u^{1+s}} \biggl( \int_{1}^{b\wedge u} g(x) \, \mathrm{d}x \biggr) \, \mathrm{d}u \\ &= \int_{1}^{\infty} \frac{s}{u^{1+s}} \bigl( G(b\wedge u) - G(1) \bigr) \, \mathrm{d}u. \end{align*}

Since $G$ is bounded, letting $ b \to \infty$ and applying the dominated convergence theorem shows that the both sides converge to

\begin{align*} \int_{1}^{\infty} \frac{g(x)}{x^s} \, \mathrm{d}x &=\int_{1}^{\infty} \frac{s}{u^{1+s}} \bigl( G(u) - G(1) \bigr) \, \mathrm{d}u. \end{align*}

Then substituting $u = y^{1/s}$,

\begin{align*} \int_{1}^{\infty} \frac{g(x)}{x^s} \, \mathrm{d}x &= \int_{1}^{\infty} \frac{1}{y^{2}} \bigl( G(y^{1/s}) - G(1) \bigr) \, \mathrm{d}u. \end{align*}

So by the dominated convergence theorem again, it follows that

\begin{align*} \lim_{s\to0^+} \int_{1}^{\infty} \frac{g(x)}{x^s} \, \mathrm{d}x = \int_{1}^{\infty} \frac{1}{y^{2}} \bigl( G(\infty) - G(1) \bigr) \, \mathrm{d}u = G(\infty) - G(1). \end{align*}

Step 3. Similarly, for $s < \delta$ and $0 < a < 1$,

\begin{align*} \int_{a}^{1} \frac{g(x)}{x^s} \, \mathrm{d}x &= \int_{a}^{1} \frac{g(x)}{x^{\delta}} x^{\delta-s} \, \mathrm{d}x \\ &= \int_{a}^{1} \frac{g(x)}{x^{\delta}} \biggl( \int_{0}^{x} (\delta - s) u^{\delta-s-1} \, \mathrm{d}u \biggr) \, \mathrm{d}x \\ &= \int_{0}^{1} (\delta - s) u^{\delta-s-1} \biggl( \int_{a \vee u}^{1} \frac{g(x)}{x^{\delta}} \, \mathrm{d}x \biggr) \, \mathrm{d}u \\ &= \int_{0}^{1} (\delta - s) u^{\delta-s-1} \bigl( H(1) - H(a\vee u) \bigr) \, \mathrm{d}u \end{align*}

and letting $ a \to 0^+$ shows that both sides converge to

\begin{align*} \int_{0}^{1} \frac{g(x)}{x^s} \, \mathrm{d}x &= \int_{0}^{1} (\delta - s) u^{\delta-s-1} \bigl( H(1) - H(u) \bigr) \, \mathrm{d}u. \end{align*}

Also, appealing to the dominated convergence theorem again, the right-hand side is continuous for $s < \delta$ and hence the same is true for the integral in the left-hand side.

Conclusion. Altogether, we have shown that $I(s)$ exists for $s \in (0, \delta)$ and that

\begin{align*} \lim_{s \to 0^+} I(s) &= \lim_{s\to0^+} \int_{0}^{1} \frac{g(x)}{x^s} \, \mathrm{d}x + \lim_{s\to0^+} \int_{1}^{\infty} \frac{g(x)}{x^s} \, \mathrm{d}x \\ &= \int_{0}^{1} g(x) \, \mathrm{d}x + \left( G(\infty) - G(1) \right) \\ &= I(0). \end{align*}