Resolved: "Existence of the integral ∫αβarctan⁡h(r)rdrFor −1≤α≤β≤1"

Erik Cantu

Answered question

2022-04-02

Existence of the integral

\int_{α}^{β} \sqrt{\frac{\arctan h (r)}{r}} d r

For

- 1 \leq α \leq β \leq 1

undodaonePvopxl24

Beginner2022-04-03Added 13 answers

\frac{arctanh r}{r}

is an even function, so it is enough to show that

\int_{0}^{t} \sqrt{\frac{arctanh r}{r}} d r

is finite for any

t \in [0, 1]

. This can be done through the Cauchy-Schwarz inequality, since

\int_{0}^{1} \frac{arctanh r}{r} d r = \frac{π^{2}}{8}

ensures:

\int_{0}^{t} \sqrt{\frac{arctanh r}{r}} d r \leq \sqrt{\int_{0}^{t} 1 d r \int_{0}^{t} \frac{arctanh r}{r} d r} \leq π \sqrt{\frac{t}{8}}

We may also state

\int_{0}^{t} \sqrt{\frac{arctanh r}{r}} d r \leq \frac{π t}{2 \sqrt{2}}

by exploiting convexity, but the above (quite crude) inequality is already enough to prove that

\int_{α}^{β} \sqrt{\frac{arctanh r}{r}} d r

is finite for any

α, β \in [- 1, 1]

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