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Question Number 99576 by mathmax by abdo last updated on 21/Jun/20

1) let f (a) = \int_{0}^{\infty} \frac{t^{a - 1} \ln (t)}{1 + t} dt with 0 < a < 1 prove that f (a) is convergent and determine

it value

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

3) calculate \int_{0}^{\infty} \frac{lnt}{(^{3} \sqrt{t^{2}}) (1 + t)} dt

Answered by maths mind last updated on 22/Jun/20

x = l n (t)

= \int_{- \infty}^{+ \infty} \frac{e^{a x} t}{e^{x} + 1} d x

= - \int_{0}^{+ \infty} \frac{e^{- a y} y}{1 + e^{- y}} d y + \int_{0}^{+ \infty} \frac{e^{a y} y}{1 + e^{y}} d y

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

l e t f (β) = \int_{0}^{+ \infty} \frac{{y e}^{- β y}}{1 + e^{- y}}

= \int_{0}^{\infty} {y e}^{- β y} (\sum_{k ⩾ 0} {(- e^{- y})}^{k} d y

= \int_{0}^{+ \infty} \sum_{k ⩾ 0} y {(- 1)}^{k} e^{(- β - k) y} d y

= \sum_{k ⩾ 0} {(- 1)}^{k} \int_{0}^{+ \infty} {y e}^{- (k + β) y}

= \sum_{k ⩾ 0} {(- 1)}^{k} \int_{0}^{+ \infty} \frac{{t e}^{- t}}{{(k + β)}^{2}} = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(k + β)}^{2}}

= \sum_{k ⩾ 0} \frac{1}{4 {(k + \frac{β}{2})}^{2}} - \sum_{k ⩾ 0} \frac{1}{4 {(k + \frac{1 + β}{2})}^{2}} = \frac{1}{4} (ζ (2, \frac{β}{2}) - ζ (2, \frac{1 + β}{2}))

ζ (s, q) = \sum_{n \in N} \frac{1}{{(n + q)}^{s}} H u r w i t z z e t a F u n c t i o n n + q \neq 0

f (a) = \frac{1}{4} (ζ (2, \frac{1 - a}{2}) - ζ (2, \frac{2 - a}{2}) - ζ (2, \frac{a}{2}) + ζ (2, \frac{1 + a}{2}))

f (a) = \frac{1}{4} (ζ (2, \frac{1 - a}{2}) + ζ (2, \frac{1 + a}{2}))

= \frac{1}{4} \sum_{n ⩾ 0} \frac{1}{{(n + \frac{1 - a}{2})}^{2}} + \frac{1}{{(n + \frac{1 + a}{2})}^{2} = G} - \frac{1}{4} Σ \frac{1}{{(n + \frac{2 - a}{2})}^{2}} + \frac{1}{{(n + \frac{a}{2})}^{2} = F}

G = \frac{1}{4} . \sum_{n ⩾ 1} \frac{1}{{(n - \frac{1 + a}{2})}^{2}} + \frac{1}{{(n + \frac{1 + a}{2})}^{2}} + \frac{1}{4} \frac{1}{{(\frac{1 + a}{2})}^{2}} .

π c o t (π x) = \frac{1}{x} + \sum_{n ⩾ 1} \frac{1}{(x + n)} + \frac{1}{(x - n)}

\Rightarrow π^{2} (- 1 - {c o t}^{2} (π x)) = - \frac{1}{x^{2}} + \sum_{n ⩾ 1} \frac{- 1}{{(x + n)}^{2}} - \frac{1}{{(x - n)}^{2}}

\Rightarrow G = - \frac{1}{4} (- π^{2} - π^{2} {c o t}^{2} (π (\frac{a + 1}{2}))

s a m w e f i n d F

\Rightarrow F = \frac{1}{4} (- π^{2} - π^{2} {c o t}^{2} (\frac{π a}{2})) \Rightarrow f (a) = - \frac{π^{2}}{4} ({t g}^{2} (\frac{π a}{2}) - {c o t}^{2} (\frac{π a}{2}))

{t g}^{2} (x) - {c o t}^{2} (x) = \frac{{s i n}^{4} (x) - {c o s}^{4} (x)}{{c o s}^{2} (x) {s i n}^{2} (x)} = - 4 \frac{c o s (2 x)}{{s i n}^{2} (2 x)}

w e g e t = \frac{π^{2} c o s (π a)}{{s i n}^{2} (π a)} = f (a)

2) \int \frac{l n (t)}{\sqrt{t} (t + 1)} d t = f (\frac{1}{2})

\int \frac{l n (t)}{t^{\frac{2}{3}} (1 + t)} d t = f (\frac{1}{3})

Answered by mathmax by abdo last updated on 22/Jun/20

1) let φ (a) = \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt with 0 < a < 1 its eazy to prove thst φ is derivable

and φ^{^{'}} (a) = \int_{0}^{\infty} \frac{\partial}{\partial a} {\frac{e^{(a - 1) lnt}}{1 + t}} dt = \int_{0}^{\infty} \frac{lnt t^{a - 1}}{1 + t} dt = f (a)

but φ (a) = \frac{π}{\sin (π a)} \Rightarrow φ^{^{'}} (a) = π (\frac{- π \cos (π a)}{\sin^{2} (π a)}) = - π^{2} \frac{\cos (π a)}{\sin^{2} (π a)} \Rightarrow

f (a) = - \frac{π^{2} \cos (π a)}{\sin^{2} (π a)}

2) \int_{0}^{\infty} \frac{\ln (t)}{(1 + t) \sqrt{t}} dt = \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} \ln (t)}{1 + t} dt = \int_{0}^{\infty} \frac{t^{\frac{1}{2} - 1} \ln (t)}{1 + t} dt

= f (\frac{1}{2}) = - \frac{π^{2} \cos (\frac{π}{2})}{\sin^{2} (\frac{π}{2})} = 0

another we do tbe changement \sqrt{t} = u \Rightarrow \int_{0}^{\infty} \frac{lnt}{(1 + t) \sqrt{t}} dt = \int_{0}^{\infty} \frac{2 lnu}{(1 + u^{2}) u} (2 u) du

= 4 \int_{0}^{\infty} \frac{lnu}{1 + u^{2}} du = 0

3) \int_{0}^{\infty} \frac{\ln (t)}{(^{3} \sqrt{t^{2}}) (1 + t)} dt = \int_{0}^{\infty} \frac{t^{- \frac{2}{3}} \ln (t)}{1 + t} dt = \int_{0}^{\infty} \frac{t^{\frac{1}{3} - 1} \ln (t)}{1 + t} dt

= f (\frac{1}{3}) = - \frac{π^{2} \cos (\frac{π}{3})}{\sin^{2} (\frac{π}{3})} = - \frac{π^{2}}{2} \times \frac{1}{{(\frac{\sqrt{3}}{2})}^{2}} = - \frac{π^{2}}{2} \times \frac{4}{3} = - \frac{2 π^{2}}{3}