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Question Number 132459 by mnjuly1970 last updated on 14/Feb/21

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

ϕ = \int_{0}^{\infty} \frac{s i n (x^{2}) l n (x)}{x^{\frac{3}{2}}} d x =

s o l u t i o n :

ϕ \overset{x^{2} = t}{=} \frac{1}{2} \int_{0}^{\infty} \frac{s i n (t) l n (\sqrt{t})}{t^{\frac{3}{4}}} \frac{d t}{t^{\frac{1}{2}}}

= \frac{1}{4} \int_{0}^{\infty} \frac{s i n (t) l n (t)}{t^{\frac{5}{4}}} d t

ψ (a) = \int_{0}^{\infty} \frac{s i n (t) . t^{a}}{t^{\frac{5}{4}}} d t = \int_{0}^{\infty} \frac{s i n (t)}{t^{\frac{5}{4} - a}} d t

ψ (a) = \frac{π}{2 Γ (\frac{5}{4} - a) s i n (\frac{5 π}{8} - \frac{π}{2} a)}

∴ ϕ = \frac{ψ^{'} (0)}{4} \dots \dots

ψ^{'} (a) = \frac{π}{2} [\frac{Γ^{'} (\frac{5}{4} - a) s i n (\frac{5 π}{8} - \frac{π a}{2}) + \frac{π}{2} Γ (\frac{5}{4} - a) c o s (\frac{5 π}{8} - \frac{π a}{2})}{Γ^{2} (\frac{5}{4} - a) {s i n}^{2} (\frac{5 π}{8} - \frac{π a}{2})}]

ψ^{'} (0) = \frac{π}{2} (\frac{Γ^{'} (\frac{5}{4}) c o s (\frac{π}{8}) - \frac{π}{2} Γ (\frac{5}{4}) s i n (\frac{π}{8})}{Γ^{2} (\frac{5}{4}) {c o s}^{2} (\frac{π}{8})})

= \frac{π}{2} (\frac{4 ψ (1 + \frac{1}{4}) Γ (\frac{1}{4}) c o s (\frac{π}{8}) - 2 π Γ (\frac{1}{4}) s i n (\frac{π}{8})}{Γ^{2} (\frac{1}{4}) {c o s}^{2} (\frac{π}{8})})

ϕ = \frac{π}{4} (\frac{(2 ψ (\frac{1}{4}) + 8) Γ (\frac{1}{4}) c o s (\frac{π}{8}) - π Γ (\frac{1}{4}) s i n (\frac{π}{8})}{Γ^{2} (\frac{1}{4}) {c o s}^{2} (\frac{π}{8})})