Prove that: •∫^( x) _( 0) ((lnt)/(t^2 −1))dt=∫^( (π/2)) _( 0) arctan(xtanθ)dθ • ∫^( x) _( (1/x)) ((lnt)/(t^2 −1))arctant dt=(π/8)∫^( π)


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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
	$f(x)=∫_0 ^x ((ln(t))/(t^2 −1))dt,f(0)=0 f′(x)=((ln(x))/(x^2 −1)) g(x)=∫_0 ^(π/2) arctan(xtan (a))da g(0)=0 g′(x)=∫_0 ^(π/2) ((tan(a))/(1+(xtan(a))^2 ))da =∫_0 ^(π/2) ((sin(a)cos(a))/(cos^2 (a)+x^2 sin^2 (a)))dx =∫_0 ^(π/2) ((sin(2a)da)/(cos(2a)+1+x^2 (1−cos(2a))),cos(2a)=y =−(1/2)∫_1 ^(−1) (dy/(1+x^2 +(1−x^2 )y)) =(1/(2(1−x^2 ))).{ln [(1+x^2 )+(1−x^2 )y]}_(−1) ^1 (1/(2(1−x^2 )))ln((2/(2x^2 )))=((ln(x))/(x^2 −1))=f′(x) f′(x)=g′(x) ,f(0)=g(0)⇒f(x)=g(x) (2)H(x)=∫_(1/x) ^x ((ln(t))/(t^2 −1))tan^(−1) (t)dt t→(1/t)⇒H(x)=∫_(1/x) ^x ((ln(t))/(t^2 −1))((π/2)−tan^(−1) (t))dt,∀x∈R_+ ^∗ H(x)=(π/2)∫_(1/x) ^x ((ln(t))/(t^2 −1))dt−H(x) H(x)=(π/4)∫_(1/x) ^x ((ln(t))/(t^2 −1))dt=(π/4)(∫_0 ^x ((ln(t))/(t^2 −1))dt−∫_0 ^(1/x) ((ln(t))/(t^2 −1))dt) =(π/4)(f(x)−f((1/x)))=(π/4)(g(x)−g((1/x))) =(π/4)tan^(−1) (xtan(a))−(π/4)tan^− (((tan(a))/x)) tan^(−1) (a)−tan^(−1) (b)=tan^(−1) (((a−b)/(1+ab))), (π/4)∫_0 ^(π/2) (tan^− ((((x−(1/x))tan (a))/(1+tan^2 (a)))))da =(π/4)∫_0 ^(π/2) tan^(−1) ((x−(1/x))((sin(2a))/2))da 2a→a =(π/8)∫_0 ^π tan^(−1) ((1/2)(x−(1/x))sin(a))da$
	$f (x) = \int_{0}^{x} \frac{\ln (t)}{t^{2} - 1} dt, f (0) = 0$ $f^{'} (x) = \frac{\ln (x)}{x^{2} - 1}$ $g (x) = \int_{0}^{\frac{π}{2}} \arctan (xtan (a)) da$ $g (0) = 0$ $g^{'} (x) = \int_{0}^{\frac{π}{2}} \frac{\tan (a)}{1 + {(xtan (a))}^{2}} da$ $= \int_{0}^{\frac{π}{2}} \frac{\sin (a) \cos (a)}{\cos^{2} (a) + x^{2} \sin^{2} (a)} dx$ $= \int_{0}^{\frac{π}{2}} \frac{\sin (2 a) da}{\cos (2 a) + 1 + x^{2} (1 - \cos (2 a)}, \cos (2 a) = y$ $= - \frac{1}{2} \int_{1}^{- 1} \frac{dy}{1 + x^{2} + (1 - x^{2}) y}$ $= \frac{1}{2 (1 - x^{2})} . {\ln [(1 + x^{2}) + (1 - x^{2}) y]}_{- 1}^{1}$ $\frac{1}{2 (1 - x^{2})} \ln (\frac{2}{{2 x}^{2}}) = \frac{\ln (x)}{x^{2} - 1} = f^{'} (x)$ $f^{'} (x) = g^{'} (x), f (0) = g (0) \Rightarrow f (x) = g (x)$ $(2) H (x) = \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} \tan^{- 1} (t) dt$ $t \to \frac{1}{t} \Rightarrow H (x) = \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} (\frac{π}{2} - \tan^{- 1} (t)) dt, \forall x \in R_{+}^{*}$ $H (x) = \frac{π}{2} \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} dt - H (x)$ $H (x) = \frac{π}{4} \int_{\frac{1}{x}}^{x} \frac{\ln (t)}{t^{2} - 1} dt = \frac{π}{4} (\int_{0}^{x} \frac{\ln (t)}{t^{2} - 1} dt - \int_{0}^{\frac{1}{x}} \frac{\ln (t)}{t^{2} - 1} dt)$ $= \frac{π}{4} (f (x) - f (\frac{1}{x})) = \frac{π}{4} (g (x) - g (\frac{1}{x}))$ $= \frac{π}{4} \tan^{- 1} (xtan (a)) - \frac{π}{4} \tan^{-} (\frac{\tan (a)}{x})$ $\tan^{- 1} (a) - \tan^{- 1} (b) = \tan^{- 1} (\frac{a - b}{1 + ab}),$ $\frac{π}{4} \int_{0}^{\frac{π}{2}} (\tan^{-} (\frac{(x - \frac{1}{x}) \tan (a)}{1 + \tan^{2} (a)})) da$ $= \frac{π}{4} \int_{0}^{\frac{π}{2}} \tan^{- 1} ((x - \frac{1}{x}) \frac{\sin (2 a)}{2}) da$ $2 a \to a$ $= \frac{π}{8} \int_{0}^{π} \tan^{- 1} (\frac{1}{2} (x - \frac{1}{x}) \sin (a)) da$
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