nice-calculus-calculate-0-arctan-1-x-sin-x-dx-

Question Number 127649 by mnjuly1970 last updated on 31/Dec/20

\dots n i c e c a l c u l u s \dots

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) = \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 \dots

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