nice-calculus-prove-that-0-e-x-ln-x-cos-x-dx-1-8-4-pi-2ln-2-

Question Number 130433 by mnjuly1970 last updated on 25/Jan/21

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

p r o v e t h a t ::

Ψ = \int_{0}^{\infty} e^{- x} l n (x) c o s (x) d x \overset{?}{=} \frac{1}{8} (- 4 γ - π - 2 l n (2))

Answered by Dwaipayan Shikari last updated on 25/Jan/21

I (a) = \int_{0}^{\infty} x^{a} e^{- x} c o s x d x = \frac{Γ (a + 1)}{2^{\frac{a + 1}{2}}} c o s (\frac{π}{4} (a + 1))

I^{'} (a) = \frac{Γ^{'} (a + 1)}{2^{\frac{a + 1}{2}}} c o s (\frac{π}{4} (a + 1)) - \frac{π}{4} . \frac{Γ (a + 1)}{2^{\frac{a + 1}{2}}} s i n (\frac{π}{4} (a + 1)) - \frac{1}{2} . \frac{Γ (a + 1)}{2^{\frac{a + 1}{2}}} c o s (\frac{π}{4} (a + 1)) l o g (2)

I^{'} (0) = \frac{- γ}{2} - \frac{π}{8} - \frac{l o g (2)}{4} = - \frac{1}{8} (4 γ + π + l o g (4))

Commented by mnjuly1970 last updated on 25/Jan/21

v e r y n i c e .

A l l a h k e e p y o u \dots

Answered by mnjuly1970 last updated on 26/Jan/21

Ψ = R e (\int_{0}^{\infty} e^{- x} e^{- i x} l n (x) d x)

= R e (\frac{d}{d a} (\int_{0}^{\infty} e^{- x (1 + i)} x^{a} d x = Ω) ∣_{a = 0}

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

= \frac{1}{2^{\frac{a + 1}{2}} e^{\frac{i π (a + 1}{4}}} Γ (a + 1)

∴ \frac{d (Ω)}{d a} = Γ^{'} (a + 1) 2^{\frac{- (a + 1)}{2}} * e^{\frac{- i π (a + 1)}{4}}

- (\frac{l n (2)}{2} 2^{\frac{- (a + 1)}{2}} * e^{\frac{- i π (a + 1)}{4}} + \frac{i π}{4} e^{\frac{- i π (a + 1)}{4}} * 2^{\frac{- (a + 1)}{2}}) Γ (a + 1)

∴ (\frac{d (Ω)}{d a}) ∣_{a = 0} = \frac{Γ^{'} (1)}{\sqrt{2}} (c o s (\frac{π}{4}) - i s i n (\frac{π}{4}))

- [\frac{l n (2)}{2 \sqrt{2}} (c o s (\frac{π}{4}) - i s i n (\frac{π}{4})) + \frac{i π}{4 \sqrt{2}} (c o s (\frac{π}{4}) - i s i n (\frac{π}{4}))] Γ (1)

∴ R e (\frac{d Ω}{d a}) ∣_{a = 0} = \frac{- γ}{2} - \frac{l n (2)}{4} - \frac{π}{8}

Ψ = - \frac{1}{8} (π + 4 γ + 2 l n (2))