Show-that-0-lnx-x-2-1-2-dx-pi-4-

Question Number 136210 by Ar Brandon last updated on 19/Mar/21

Show that \int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} dx = - \frac{π}{4}

Commented by Ar Brandon last updated on 19/Mar/21

Oh! Thank You Sir

Commented by Ar Brandon last updated on 19/Mar/21

Ω = \int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} dx

= \frac{\partial}{\partial s} ∣_{s = 0} \int_{0}^{\infty} \frac{x^{s}}{{(x^{2} + 1)}^{2}} dx \overset{u = x^{2}}{=} \frac{\partial}{\partial s} ∣_{s = 0} \frac{1}{2} \int_{0}^{\infty} \frac{u^{\frac{s}{2} - \frac{1}{2}}}{{(u + 1)}^{2}} du

= \frac{\partial}{\partial s} ∣_{s = 0} \frac{1}{2} β (\frac{s}{2} + \frac{1}{2}, \frac{3}{2} - \frac{s}{2}) = \frac{\partial}{\partial s} ∣_{s = 0} \frac{1}{2} Γ (\frac{s}{2} + \frac{1}{2}) Γ (\frac{3}{2} - \frac{s}{2})

= \frac{1}{2} ∣_{s = 0} Γ (\frac{s}{2} + \frac{1}{2}) Γ^{'} (\frac{3}{2} - \frac{s}{2}) + Γ (\frac{3}{2} - \frac{s}{2}) Γ^{'} (\frac{s}{2} + \frac{1}{2})

= \frac{1}{2} \cdot \frac{1}{2} Γ (\frac{1}{2}) Γ (\frac{3}{2}) [ψ (\frac{1}{2}) - ψ (\frac{3}{2})] = \frac{π}{8} [ψ (\frac{1}{2}) - ψ (\frac{1}{2}) - 2] = - \frac{π}{4}

Commented by mnjuly1970 last updated on 19/Mar/21

e r r o r i n s i g n o f d r i v a t i v e

o f ψ (\frac{3}{2} - s) \dots \dots .^{'} = - ψ^{'} (\frac{3}{2} - s)

\dots g o o d l u c k m y b r o t h e r m r b r a n d o n \dots

Answered by mnjuly1970 last updated on 19/Mar/21

\boldsymbol ϕ \overset{x^{2} = y}{=} \frac{1}{4} \int_{0}^{\infty} \frac{y^{- \frac{1}{2}} l n (y)}{{(y + 1)}^{2}} d y

f (a) = \int_{0}^{\infty} \frac{y^{a - \frac{1}{2}}}{{(y + 1)}^{2}} d y

\boldsymbol ϕ = \frac{f^{'} (0)}{4} \dots ✓

f (a) = \int_{0}^{\infty} \frac{y^{a + \frac{1}{2} - 1}}{{(y + 1)}^{2}} d y = β (a + \frac{1}{2}, \frac{3}{2} - a)

f (a) = Γ (a + \frac{1}{2}) . Γ (\frac{3}{2} - a)

f^{'} (a) = Γ^{'} (a + \frac{1}{2}) Γ (\frac{3}{2} - a) - Γ^{'} (\frac{3}{2} - a) Γ (\frac{1}{2} + a)

\boldsymbol ϕ = \frac{1}{4} f^{'} (0) = \frac{1}{4} (\frac{1}{2} ψ (\frac{1}{2}) Γ^{2} (\frac{1}{2}) - \frac{1}{2} ψ (\frac{3}{2}) Γ^{2} (\frac{1}{2}))

= \frac{π}{8} (ψ (\frac{1}{2}) - ψ (\frac{3}{2}))

= \frac{π}{8} (ψ (\frac{1}{2}) - ψ (\frac{1}{2}) - 2) = \frac{- π}{4} . . ✓ ✓

Answered by Dwaipayan Shikari last updated on 19/Mar/21

\frac{1}{4} \int_{0}^{\infty} \frac{u^{- \frac{1}{2}} l o g (u)}{{(1 + u)}^{2}} d u = \frac{1}{4} τ^{'} (\frac{1}{2}) = \int_{0}^{\infty} \frac{l o g (x)}{{(x^{2} + 1)}^{2}} d x

τ (α) = \int_{0}^{\infty} \frac{u^{α - 1}}{{(1 + u)}^{2}} d u = \frac{Γ (α) Γ (2 - α)}{Γ (2)} = \frac{π (1 - α)}{s i n (π α)}

\Rightarrow τ^{'} (α) = \int_{0}^{\infty} \frac{u^{α - 1} l o g (u)}{{(1 + u)}^{2}} d u = - π^{2} c o s e c (π α) c o t (π α) - \frac{π}{s i n (π α)}

\frac{1}{4} τ^{'} (\frac{1}{2}) = - 0 - \frac{π}{4} = - \frac{π}{4}

Commented by Ar Brandon last updated on 19/Mar/21

Thanks bro

Commented by Dwaipayan Shikari last updated on 19/Mar/21

Answered by Ajetunmobi last updated on 19/Mar/21

\boldsymbol i \boldsymbol have \boldsymbol drop \boldsymbol the \boldsymbol solution \boldsymbol before \boldsymbol here

Commented by Ar Brandon last updated on 19/Mar/21

Alright!

Before posting I used the search option

to check if it was previously posted but

{couldn}^{'} t find it . Thank You Sir

Commented by Ajetunmobi last updated on 19/Mar/21

\boldsymbol ok

Answered by Ajetunmobi last updated on 19/Mar/21

Answered by mathmax by abdo last updated on 19/Mar/21

let f (a) = \int_{0}^{\infty} \frac{lnx}{x^{2} + a^{2}} dx with a > 0 \Rightarrow f^{^{'}} (a) = - 2 a \int_{0}^{\infty} \frac{lnx}{{(x^{2} + a^{2})}^{2}} dx \Rightarrow

f^{^{'}} (1) = - 2 \int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} \Rightarrow \int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} = - \frac{1}{2} f^{^{'}} (1)

f (a) =_{x = at} \int_{0}^{\infty} \frac{lna + lnt}{a^{2} (1 + t^{2})} adt = \frac{1}{a} \int_{0}^{\infty} \frac{lna + lnt}{(1 + t^{2})} dt

= \frac{lna}{a} \int_{0}^{\infty} \frac{dt}{1 + t^{2}} + \frac{1}{a} \int_{0}^{\infty} \frac{lnt}{1 + t^{2}} dt but \int_{0}^{\infty} \frac{lnt}{1 + t^{2}} dt = 0 (proved) \Rightarrow

f (a) = \frac{lna}{a} . \frac{π}{2} = \frac{π lna}{2 a} \Rightarrow f^{^{'}} (a) = \frac{\frac{π}{a} (2 a) - 2 π lna}{{4 a}^{2}} \Rightarrow f^{^{'}} (1) = \frac{2 π}{4} = \frac{π}{2} \Rightarrow

\int_{0}^{\infty} \frac{lnx}{{(x^{2} + 1)}^{2}} dx = - \frac{1}{2} . \frac{π}{2} = - \frac{π}{4}

Commented by mathmax by abdo last updated on 19/Mar/21

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Commented by Ajetunmobi last updated on 19/Mar/21

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