nice-calculus-prove-0-1-1-1-ln-2-x-dx-0-sin-x-1-x-dx-

Question Number 136497 by mnjuly1970 last updated on 22/Mar/21

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

p r o v e ::

\int_{0}^{1} \frac{1}{1 + {l n}^{2} (x)} d x = \int_{0}^{\infty} \frac{s i n (x)}{1 + x} d x

Commented by Dwaipayan Shikari last updated on 22/Mar/21

\int_{0}^{\infty} \frac{s i n (x)}{1 + x} d x

= \int_{0}^{\infty} \int_{0}^{\infty} e^{- (x + 1) t} s i n (x) d t d x

= \frac{1}{2 i} \int_{0}^{\infty} \int_{0}^{\infty} e^{- t} e^{- (t - i) x} - e^{- t} e^{- (t + i) x} d x d t

= \frac{1}{2 i} \int_{0}^{\infty} \frac{e^{- t}}{t - i} - \frac{e^{- t}}{t + i} d t

= \int_{0}^{\infty} \frac{e^{- t}}{t^{2} + 1} d t - t = l o g (u) \Rightarrow 1 = - \frac{1}{u} . \frac{d u}{d t}

= \int_{0}^{1} \frac{1}{1 + {l o g}^{2} (u)} d u

Commented by mnjuly1970 last updated on 22/Mar/21

g r a t e f u l m r p a y a n \dots

Answered by mindispower last updated on 22/Mar/21

t = - l n (x)

= \int_{0}^{\infty} \frac{e^{- t}}{1 + t^{2}} d t = \int_{0}^{\infty} \frac{(t + i) - (t - i)}{2 i (t + i) (t - i)} e^{- t}

= \frac{1}{2 i} (\int_{0}^{\infty} \frac{e^{- t}}{t - i} d t - \int_{0}^{\infty} \frac{e^{- t}}{t + i})

= I m \int_{0}^{\infty} \frac{e^{- t}}{t - i} d t

l e t z = i t

= I m \int_{0}^{i \infty} \frac{e^{i z}}{i (t + 1)} . \frac{d z}{i} = I m (- \int_{0}^{i \infty} \frac{e^{i z}}{t + 1} d z) \dots 2

l e t C_{R} = [0, R] \cup {({R e}^{i θ}, θ \in [0, \frac{π}{2}])}_{D} \cup [i R, 0]

\int_{C_{R}} \frac{e^{i z}}{z + 1} d z = 0 . . 1, z \to \frac{e^{i z}}{z + 1} h o l o m o r p h i c w i t h o u t p o l s

o v e r C_{R}

\int_{D} \frac{e^{i z}}{z + 1} = 0 \Rightarrow . . 2 (1) & (2) \Rightarrow

\int_{0}^{\infty} \frac{e^{i z}}{z + 1} d z = - \int_{0}^{i \infty} \frac{e^{i z}}{z + 1} d z

w e g e t I m \int_{0}^{\infty} \frac{e^{i z}}{z + 1} d z = \int_{0}^{\infty} \frac{s i n (z)}{z + 1} d z

Commented by mnjuly1970 last updated on 22/Mar/21

t h a n k y o u s o m u c h m r p o w e r \dots

Commented by mindispower last updated on 22/Mar/21

p l e a s u r