Calcul-0-lnt-t-1-t-2-dt-

Question Number 196408 by Erico last updated on 24/Aug/23

Calcul {\int_{0}}^{+ \infty} \frac{lnt}{\sqrt{t} (1 + t^{2})} dt

Answered by qaz last updated on 24/Aug/23

\int_{0}^{\infty} \frac{l n t}{\sqrt{t} (1 + t^{2})} d t = \frac{\partial}{\partial a} ∣_{a = - 1 / 2} \int_{0}^{\infty} \frac{t^{a}}{1 + t^{2}} d t

= - \frac{\partial}{\partial a} ∣_{a = - 1 / 2} \frac{π}{2} c s c (\frac{a + 1}{2} π) = \frac{π^{2}}{4} \cot (\frac{a + 1}{2} π) c s c (\frac{a + 1}{2} π) ∣_{a = - 1 / 2}

= \frac{\sqrt{2} π^{2}}{4}

Answered by Mathspace last updated on 25/Aug/23

I = \int_{0}^{\infty} \frac{l n t}{\sqrt{t} (1 + t^{2})} d t (\sqrt{t} = u)

= \int_{0}^{\infty} \frac{2 l n u}{u (1 + u^{4})} (2 u) d u

= 4 \int_{0}^{\infty} \frac{l n u}{1 + u^{4}} d u (u^{4} = x)

= 4 \int_{0}^{\infty} \frac{l n (x^{\frac{1}{4}})}{1 + x} \times \frac{1}{4} x^{\frac{1}{4} - 1} d x

= \frac{1}{4} \int_{0}^{\infty} \frac{x^{\frac{1}{4} - 1} l n x}{1 + x} d x

f (a) = \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x} d x - 1 < a < 1

f^{^{'}} (a) = \int_{0}^{\infty} \frac{x^{a - 1}}{1 + x} l n x d x \Rightarrow

4 I = f^{^{'}} (\frac{1}{4}) b u t f (a) = \frac{π}{s i n (π a)}

\Rightarrow f^{^{'}} (a) = π . \frac{- π c o s (π a)}{{s i n}^{2} (π a)}

= - π^{2} \frac{c o s (π a)}{{s i n}^{2} (π a)} \Rightarrow

f^{^{'}} (\frac{1}{4}) = - π^{2} \frac{c o s (\frac{π}{4})}{{s i n}^{2} (\frac{π}{4})}

= - π^{2} . \frac{1}{\sqrt{2} . (\frac{1}{2})} = - π^{2} \sqrt{2}

I = \frac{1}{4} f^{^{'}} (\frac{1}{4}) \Rightarrow I = - \frac{\sqrt{2}}{4} π^{2}