0-log-2-x-sin-x-2-dx-i-had-solved-that-already-and-answ-pi-2-2pi-32-

Question Number 130979 by mnjuly1970 last updated on 31/Jan/21

ϕ = \int_{0}^{\infty} {l o g}^{2} (x) s i n (x^{2}) d x = ?

i h a d s o l v e d t h a t a l r e a d y a n d :

a n s w : = - \frac{π^{2} \sqrt{2 π}}{32}

Commented by Dwaipayan Shikari last updated on 31/Jan/21

\int_{0}^{\infty} \frac{t a n x}{x} d x \Rightarrow D i v e r g e s

Commented by Dwaipayan Shikari last updated on 31/Jan/21

\int_{0}^{\infty} \frac{s i n x}{x} d x = \int_{0}^{\infty} \int_{0}^{\infty} e^{- x t} s i n x d t d x

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

= \int_{0}^{\infty} \frac{1}{t^{2} + 1} d t = \frac{π}{2}

Commented by mnjuly1970 last updated on 31/Jan/21

y e s i t i s c o n v e r g e n t

i w i l l p r e p a r e

o t h e r q u e s t i o n \dots

Answered by mnjuly1970 last updated on 31/Jan/21

Answered by mnjuly1970 last updated on 31/Jan/21

Answered by Dwaipayan Shikari last updated on 31/Jan/21

I (a) = \int_{0}^{\infty} x^{a} s i n (x^{2}) d x

I (a) = \frac{1}{2 i} \int_{0}^{\infty} x^{a} e^{{i x}^{2}} - x^{a} e^{- {i x}^{2}} d x x^{2} = i u x^{2} = - i j

I (a) = \frac{1}{4} \int_{0}^{\infty} x^{a - 1} e^{- u} d u - x^{a - 1} e^{- j} d j

I (a) = \frac{{(i)}^{\frac{a - 1}{2}}}{4} \int_{0}^{\infty} u^{\frac{a - 1}{2}} e^{- u} + {(- i)}^{a - 1} \frac{1}{4} \int_{0}^{\infty} j^{\frac{a - 1}{2}} e^{- j} d j

= \frac{Γ (\frac{a + 1}{2})}{2} c o s (\frac{π}{4} (a + 1))

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

I^{″} (a) = \frac{Γ^{″} (\frac{a + 1}{2})}{8} c o s (\frac{π}{4} (a + 1)) - \frac{π}{4} . \frac{Γ^{'} (\frac{a + 1}{2})}{4} s i n (\frac{π}{4} (a + 1)) - \frac{π}{16} Γ^{'} (\frac{a + 1}{2}) s i n (\frac{π}{4} (a + 1)) - \frac{π^{2}}{32} . Γ (\frac{a + 1}{2}) c o s (\frac{π}{4} (a + 1))

I^{″} (0) = \frac{Γ^{″} (\frac{1}{2})}{2 \sqrt{2}} - \frac{π Γ^{'} (\frac{1}{2})}{8 \sqrt{2}} - \frac{π^{\frac{5}{2}}}{32}

= \frac{π ψ (\frac{1}{2}) + ψ^{'} (\frac{1}{2}) \sqrt{π}}{2 \sqrt{2}} - \frac{π \sqrt{π}}{8 \sqrt{2}} ψ (\frac{1}{2}) - \frac{π^{\frac{5}{2}}}{32}

= \frac{π \sqrt{π}}{8 \sqrt{2}} (γ + l o g (4)) - \frac{π^{\frac{5}{2}}}{32} - \frac{π (γ + l o g (4)}{2 \sqrt{2}} + \frac{ψ^{'} (\frac{1}{2}) \sqrt{π}}{2 \sqrt{2}}

Answered by mathmax by abdo last updated on 31/Jan/21

Φ =_{x = \sqrt{t}} \int_{0}^{\infty} \ln^{2} (\sqrt{t}) \sin (t) (2 t) dt = \frac{1}{2} \int_{0}^{\infty} \ln^{2} (t) tsint dt

φ (a) = \int_{0}^{\infty} t^{a + 1} sint dt = \int_{0}^{\infty} e^{(a + 1) lnt} sint dt \Rightarrow φ^{^{'}} (a) = \int_{0}^{\infty} lnt . e^{(a + 1) lnt} sint dt

\Rightarrow φ^{(2)} (a) = \int_{0}^{\infty} \ln^{2} t . t^{a + 1} sint dt \Rightarrow φ^{(2)} (0) = \int_{0}^{\infty} \ln^{2} t . tsint dt = 2 Φ

φ (a) = - Im (\int_{0}^{\infty} t^{a + 1} e^{- it} dt) and

\int_{0}^{\infty} t^{a + 1} e^{- it} dt =_{it = u} \int_{0}^{\infty} {(\frac{u}{i})}^{a + 1} e^{- u} \frac{du}{i}

= \frac{1}{i^{a + 2}} \int_{0}^{\infty} u^{a + 2 - 1} e^{- u} du = \frac{1}{{(e^{\frac{i π}{2}})}^{a + 2}} \times Γ (a + 2)

= e^{- \frac{i π}{2} (a + 2)} . Γ (a + 2) = Γ (a + 2) {\cos (\frac{π}{2} (a + 2)) - isin (\frac{π}{2} (a + 2))} \Rightarrow

φ (a) = \sin (\frac{π a}{2} + π) . Γ (a + 2) = - \sin (\frac{π a}{2}) . Γ (a + 2) (witha > - 2)

Γ (a + 2) = Γ (a + 1 + 1) = (a + 1) Γ (a + 1) = a (a + 1) Γ (a) \Rightarrow

φ (a) = - a (a + 1) \sin (\frac{π a}{2}) . Γ (a) = w (a) . Γ (a) \Rightarrow

φ^{^{'}} (a) = W^{,} (a) . Γ (a) + w (a) . Γ^{^{'}} (a)

w^{^{'}} (a) = - {(2 a + 1) \sin (\frac{π a}{2}) + (a^{2} + a) \frac{π}{2} \cos (\frac{π a}{2})} \Rightarrow

φ^{,} (a) = - {(2 a + 1) \sin (\frac{π a}{2}) + \frac{π}{2} (a^{2} + a) \cos (\frac{π a}{2})} . Γ (a)

- (a^{2} + a) \sin (\frac{π a}{2}) . Γ^{^{'}} (a) \dots . . be continued \dots