∫_0 ^( ∞) ((tan^(−1) ((x/4))−tan^(−1) ((x/6)))/x) dx =?


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Question Number 117213 by bobhans last updated on 10/Oct/20

$\underset{0}{\int^{\infty}} \frac{\tan^{- 1} (\frac{x}{4}) - \tan^{- 1} (\frac{x}{6})}{x} dx = ?$
	Answered by TANMAY PANACEA last updated on 10/Oct/20

	$t a k i n g h e l p f r o m b o o k s$ $F r u l l a n i i n t r e g a l$ $\int_{0}^{\infty} \frac{f (a x) - f (b x)}{x} d x = (A - B) l n (\frac{b}{a})$ $A = \lim_{x \to 0} f (x) B = \lim_{x \to \infty} f (x)$ $h e r e f (x) = {t a n}^{- 1} (x)$ $\lim_{x \to 0} {t a n}^{- 1} (x) = 0 = A$ $\lim_{x \to \infty} {t a n}^{- 1} (x) = {t a n}^{- 1} (\infty) = \frac{π}{2} = B$ $a = \frac{1}{4} b = \frac{1}{6}$ $a n s w e r = (0 - \frac{π}{2}) l n (\frac{1 \times 4}{6 \times 1}) = \frac{π}{2} l n (\frac{3}{2})$
	Commented by bobhans last updated on 10/Oct/20

	$waw . . . . do you have that book ?$
	Commented by TANMAY PANACEA last updated on 10/Oct/20

	$y e s s i r . . . h a r d c o p y . . . i d o n o t k n o w h o w t o s h a r e i m a g e$ $o r d o c u m e n t s i n p d f f o r m$
	Commented by bemath last updated on 10/Oct/20

	$m a y b e s i r c a n s e n t b y g m a i l$
	Answered by AbduraufKodiriy last updated on 10/Oct/20

	$I = \int_{0}^{\infty} \frac{a r c t a n (\frac{x}{4}) - a r c t a n (\frac{x}{6})}{x} d x = \int_{0}^{\infty} \frac{a r c t a n (\frac{2 x}{x^{2} + 24})}{x} d x$ $I (t) = \int_{0}^{\infty} \frac{a r c t a n (\frac{2 t x}{x^{2} + 24})}{x} d x \Rightarrow I (0) = 0$ $I^{'} (t) = \int_{0}^{\infty} \partial_{t} \frac{a r c t a n (\frac{2 t x}{x^{2} + 24})}{x} d x = \int_{0}^{\infty} \frac{2 (x^{2} + 24)}{x^{4} + 48 x^{2} + 576 + 4 t^{2} x^{2}} d x =$ $= 2 \int_{0}^{\infty} \frac{1 + \frac{24}{x^{2}}}{x^{2} + \frac{576}{x^{2}} + 48 + 4 t^{2}} d x = 2 \int_{0}^{\infty} \frac{d (x - \frac{24}{x})}{{(x - \frac{24}{x})}^{2} + 96 + 4 t^{2}} =$ $= \frac{2}{2 \sqrt{24 + t^{2}}} a r c t a n (\frac{x - \frac{24}{x}}{2 \sqrt{24 + t^{2}}}) ∣_{0}^{\infty} + C = \frac{1}{\sqrt{24 + t^{2}}} a r c t a n (\frac{x^{2} - 24}{2 x \sqrt{24 + t^{2}}}) ∣_{0}^{\infty} =$ $= \frac{1}{\sqrt{24 + t^{2}}} (\frac{π}{2} - (- \frac{π}{2})) = \frac{π}{\sqrt{24 + t^{2}}} \Rightarrow$ $\Rightarrow I (t) = π l n (t + \sqrt{t^{2} + 24}) + C$ $I (0) = \frac{π}{2} l n 24 + C = 0 \Rightarrow C = - \frac{π}{2} l n 24$ $S o, I (t) = π l n (t + \sqrt{t^{2} + 24}) - \frac{π}{2} l n 24$ $I = I (1) = π l n (1 + 5) - \frac{π}{2} l n 24 = \frac{π}{2} l n \frac{36}{24} = \frac{π}{2} l n \frac{3}{2}$
	Answered by mathmax by abdo last updated on 10/Oct/20

	$I = \int_{0}^{\infty} \frac{\arctan (ax) - \arctan (bx)}{x} dx = limI (ξ) with I (ξ) = \int_{0}^{ξ} \frac{\arctan (ax) - \arctan (bx)}{x} dx = I$ $\int_{0}^{\infty} (. . .) dx = \lim_{ξ \to + \infty} I (ξ)$ $I (ξ) = \int_{0}^{ξ} \frac{\arctan (ax)}{x} dx (\to ax = u) - \int_{0}^{ξ} \frac{\arctan (bx)}{x} dx (bx = v)$ $= \int_{0}^{a ξ} \frac{arctanu}{u} du - \int_{0}^{b ξ} \frac{arctanv}{v} dv = \int_{b ξ}^{a ξ} \frac{\arctan (u)}{u} du$ $\exists c \in] b ξ, a ξ [/ \int_{b ξ}^{a ξ} \frac{\arctan (u)}{u} du = arctanc \int_{b ξ}^{a ξ} \frac{du}{u} = arctancln (\frac{a}{b})$ $\Rightarrow \lim_{ξ \to + \infty} I (ξ) = \frac{π}{2} \ln (\frac{a}{b}) with a = \frac{1}{4} and b = \frac{1}{6} \Rightarrow \frac{a}{b} = \frac{6}{4} = \frac{3}{2}$ $\Rightarrow \int_{0}^{\infty} \frac{\arctan (\frac{x}{4}) - \arctan (\frac{x}{6})}{x} dx = \frac{π}{2} \ln (\frac{3}{2})$
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