0-ln-t-1-t-2-t-2-pi-2-2-

Question Number 197132 by Erico last updated on 08/Sep/23

{\int_{0}}^{+ \infty} {(\frac{\ln (t + \sqrt{1 + t^{2}})}{t})}^{2} = \frac{π^{2}}{2}

Commented by mnjuly1970 last updated on 09/Sep/23

t i t o k a g h a s a r d o o l . .

Answered by witcher3 last updated on 09/Sep/23

t = sh (x)

\int_{0}^{\infty} {(\frac{x}{sh (x)})}^{2} ch (x) dx

= \int_{0}^{\infty} \frac{x^{2}}{{sh}^{2} (x)} ch (x) = \lim_{(a, b) \to (0, \infty)} {[- \frac{x^{2}}{sh (x)}]}_{a}^{b} + \int_{0}^{\infty} \frac{2 x}{sh (x)} dx

= 2 \int_{0}^{\infty} \frac{2 x}{e^{x} - e^{- x}} dx

= \int_{0}^{\infty} \frac{{4 xe}^{- x}}{1 - e^{- 2 x}} dx

= 4 \sum_{n ⩾ 0} \int_{0}^{\infty} {xe}^{- (1 + 2 n) x} dx = \sum_{n ⩾ 0} \frac{4}{{(1 + 2 n)}^{2}}

= 4 (ζ (2) - \frac{1}{4} ζ (2)) = \frac{π^{2}}{2}

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