... nice calculus... calculate:: ∅=∫_0 ^( ∞) arctan((1/x)).sin(x)dx=?


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 127649 by mnjuly1970 last updated on 31/Dec/20

$. . . n i c e c a l c u l u s . . .$ $c a l c u l a t e ::$ $\emptyset = \int_{0}^{\infty} a r c t a n (\frac{1}{x}) . s i n (x) d x = ?$
	Answered by mindispower last updated on 31/Dec/20
	$f(t)=∫_0 ^∞ arctan((t/x))sin(x)dx t∈]0,1] f(0)=0 f′(t)=∫_0 ^∞ ((xsin(x))/(x^2 +t^2 ))dx let f(z)=((zsin(z))/(z^2 +t^2 )),Im(z)>0 pol(f)=z∈{it} ∫_(−∞) ^∞ f(z)dz=2∫_0 ^∞ f(z)dz= ∫_(−∞) ^∞ f(z)dz=Im∫_(−∞) ^∞ ((ze^(iz) )/(z^2 +t^2 ))dz=Im(2iπRes(((ze^(iz) )/(z^2 +t^2 ))),z=it} =Im(2iπ((ite^(−t) )/(2it)))=πe^(−t) f′(t)=(π/2)e^(−t) f(t)=−((πe^(−t) )/2)+c f(0)=0⇒c=(π/2) f(t)=(π/2)(1−e^(−t) ) ∅=f(1)=πe^(−(1/2)) (((e^(1/2) −e^(−(1/2)) )/2))=((πsh((1/2)))/( (√e)))≈0.9923$
	$f (t) = \int_{0}^{\infty} a r c t a n (\frac{t}{x}) s i n (x) d x$ $t \in] 0, 1]$ $f (0) = 0$ $f^{'} (t) = \int_{0}^{\infty} \frac{x s i n (x)}{x^{2} + t^{2}} d x$ $l e t f (z) = \frac{z s i n (z)}{z^{2} + t^{2}}, I m (z) > 0$ $p o l (f) = z \in {i t}$ $\int_{- \infty}^{\infty} f (z) d z = 2 \int_{0}^{\infty} f (z) d z =$ $\int_{- \infty}^{\infty} f (z) d z = I m \int_{- \infty}^{\infty} \frac{{z e}^{i z}}{z^{2} + t^{2}} d z = I m (2 i π R e s (\frac{{z e}^{i z}}{z^{2} + t^{2}}), z = i t}$ $= I m (2 i π \frac{{i t e}^{- t}}{2 i t}) = π e^{- t}$ $f^{'} (t) = \frac{π}{2} e^{- t}$ $f (t) = - \frac{π e^{- t}}{2} + c$ $f (0) = 0 \Rightarrow c = \frac{π}{2}$ $f (t) = \frac{π}{2} (1 - e^{- t})$ $\emptyset = f (1) = π e^{- \frac{1}{2}} (\frac{e^{\frac{1}{2}} - e^{- \frac{1}{2}}}{2}) = \frac{π s h (\frac{1}{2})}{\sqrt{e}} \approx 0.9923$
	Commented by mnjuly1970 last updated on 31/Dec/20

	$] t h a k y o u s i r p o w e r$ $e x t r a o r i n a r y . . a s a l w a y s . . .$
	Answered by mathmax by abdo last updated on 01/Jan/21

	$Φ = \int_{0}^{\infty} \arctan (\frac{1}{x}) sinx dx \Rightarrow Φ = \int_{0}^{\infty} (\frac{π}{2} - \arctan (x)) sinxdx$ $letf (a) = \int_{0}^{\infty} (\frac{π}{2} - \arctan (ax)) sinx dx with a > 0$ $f^{^{'}} (a) = \int_{0}^{\infty} - \frac{xsinx}{1 + a^{2} x^{2}} dx =_{ax = t} - \frac{1}{a} \int_{0}^{\infty} \frac{tsin (\frac{t}{a})}{t^{2} + 1} \frac{dt}{a}$ $= - \frac{1}{a^{2}} \int_{0}^{\infty} \frac{tsin (\frac{t}{a})}{t^{2} + 1} dt = - \frac{1}{{2 a}^{2}} \int_{- \infty}^{+ \infty} \frac{tsin (\frac{t}{a})}{t^{2} + 1} dt$ $= - \frac{1}{{2 a}^{2}} Im (\int_{- \infty}^{+ \infty} \frac{{ze}^{\frac{iz}{a}}}{z^{2} + 1} dz) we have$ $\int_{- \infty}^{+ \infty} \frac{{ze}^{\frac{iz}{a}}}{z^{2} + 1} dz = 2 i π Res (f, i) = 2 i π \times \frac{{ie}^{- \frac{1}{a}}}{2 i} = i π e^{- \frac{1}{a}} \Rightarrow$ $f^{^{'}} (a) = - \frac{π}{{2 a}^{2}} e^{- \frac{1}{a}} \Rightarrow f (a) = - \frac{π}{2} \int \frac{e^{- \frac{1}{a}}}{a^{2}} da + K$ $= K - \frac{π}{2} e^{- \frac{1}{a}} we have \lim_{a \to + \infty} f (a) = 0 = K - \frac{π}{2} \Rightarrow K = \frac{π}{2} \Rightarrow$ $f (a) = \frac{π}{2} (1 - e^{- \frac{1}{a}}) so$ $Φ = \int_{0}^{\infty} \arctan (\frac{1}{x}) sinx dx = f (1) = \frac{π}{2} (1 - e^{- 1})$ $= \frac{π}{2} (1 - \frac{1}{e})$
	Commented by mnjuly1970 last updated on 01/Jan/21

	$t h a n k s a l o t s i r m a x . .$
	Commented by mathmax by abdo last updated on 01/Jan/21

	$you are welcome sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum