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Question Number 197461 by Erico last updated on 18/Sep/23

Prove that :

∙ {\int_{0}}^{x} \frac{lnt}{t^{2} - 1} dt = {\int_{0}}^{\frac{π}{2}} \arctan (xtan θ) d θ

∙ {\int_{\frac{1}{x}}}^{x} \frac{lnt}{t^{2} - 1} arctant dt = \frac{π}{8} {\int_{0}}^{π} \arctan (\frac{1}{2} (x - \frac{1}{x}) sint) dt

Answered by witcher3 last updated on 18/Sep/23

f (x) = \int_{0}^{x} \frac{\ln (t)}{t^{2} - 1} dt, f (0) = 0

f^{'} (x) = \frac{\ln (x)}{x^{2} - 1}

g (x) = \int_{0}^{\frac{π}{2}} \arctan (xtan (a)) da

g (0) = 0

g^{'} (x) = \int_{0}^{\frac{π}{2}} \frac{\tan (a)}{1 + {(xtan (a))}^{2}} da

= \int_{0}^{\frac{π}{2}} \frac{\sin (a) \cos (a)}{\cos^{2} (a) + x^{2} \sin^{2} (a)} dx

= \int_{0}^{\frac{π}{2}} \frac{\sin (2 a) da}{\cos (2 a) + 1 + x^{2} (1 - \cos (2 a)}, \cos (2 a) = y

= - \frac{1}{2} \int_{1}^{- 1} \frac{dy}{1 + x^{2} + (1 - x^{2}) y}

= \frac{1}{2 (1 - x^{2})} . {\ln [(1 + x^{2}) + (1 - x^{2}) y]}_{- 1}^{1}

\frac{1}{2 (1 - x^{2})} \ln (\frac{2}{{2 x}^{2}}) = \frac{\ln (x)}{x^{2} - 1} = f^{'} (x)

f^{'} (x) = g^{'} (x), f (0) = g (0) \Rightarrow f (x) = g (x)

(2) H (x) = \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} \tan^{- 1} (t) dt

t \to \frac{1}{t} \Rightarrow H (x) = \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} (\frac{π}{2} - \tan^{- 1} (t)) dt, \forall x \in R_{+}^{*}

H (x) = \frac{π}{2} \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} dt - H (x)

H (x) = \frac{π}{4} \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} dt = \frac{π}{4} (\int_{0}^{x} \frac{\ln (t)}{t^{2} - 1} dt - \int_{0}^{\frac{1}{x}} \frac{\ln (t)}{t^{2} - 1} dt)

= \frac{π}{4} (f (x) - f (\frac{1}{x})) = \frac{π}{4} (g (x) - g (\frac{1}{x}))

= \frac{π}{4} \tan^{- 1} (xtan (a)) - \frac{π}{4} \tan^{-} (\frac{\tan (a)}{x})

\tan^{- 1} (a) - \tan^{- 1} (b) = \tan^{- 1} (\frac{a - b}{1 + ab}),

\frac{π}{4} \int_{0}^{\frac{π}{2}} (\tan^{-} (\frac{(x - \frac{1}{x}) \tan (a)}{1 + \tan^{2} (a)})) da

= \frac{π}{4} \int_{0}^{\frac{π}{2}} \tan^{- 1} ((x - \frac{1}{x}) \frac{\sin (2 a)}{2}) da

2 a \to a

= \frac{π}{8} \int_{0}^{π} \tan^{- 1} (\frac{1}{2} (x - \frac{1}{x}) \sin (a)) da

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