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Question Number 121862 by Bird last updated on 12/Nov/20

1)explicite f(a)=∫_0 ^∞ ((t^(a−1) lnt)/(1+t))dt with 0<a<1 2)calculate ∫_0 ^∞ ((lnt)/((1+t)(√t)))dt

$$\left.\mathrm{1}\right){explicite}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} {lnt}}{\mathrm{1}+{t}}{dt} \\ $$ $${with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$ $$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}{dt} \\ $$

Answered by mnjuly1970 last updated on 12/Nov/20

solution:1 :: g(b)=∫_0 ^( ∞) (t^(a+b−1) /(1+t))dt f(a)=g′(0) g(b)=Γ(a+b)Γ(1−a−b)=(π/(sin(π(a+b)))) g′(b)=((−π^2 cos(π(a+b)))/(sin^2 (π(a+b)))) g′(0)=((−π^2 cos(πa))/(sin^2 (πa)))=f(a)^ f(a)=−π^2 cot(πa)csc(πa) ... 2:: f((1/2))=0 we know that :: ∫_0 ^( ∞) ((ln(x))/(1+x^2 ))dx=^(easy) 0

$${solution}:\mathrm{1}\:::\:{g}\left({b}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{{a}+{b}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$ $$\:\:\:\:\:\:\:\:\:\:{f}\left({a}\right)={g}'\left(\mathrm{0}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:{g}\left({b}\right)=\Gamma\left({a}+{b}\right)\Gamma\left(\mathrm{1}−{a}−{b}\right)=\frac{\pi}{{sin}\left(\pi\left({a}+{b}\right)\right)} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:{g}'\left({b}\right)=\frac{−\pi^{\mathrm{2}} {cos}\left(\pi\left({a}+{b}\right)\right)}{{sin}^{\mathrm{2}} \left(\pi\left({a}+{b}\right)\right)} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}'\left(\mathrm{0}\right)=\frac{−\pi^{\mathrm{2}} {cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}={f}\left({a}\overset{} {\right)} \\ $$ $$\:\:\:\:\:\:{f}\left({a}\right)=−\pi^{\mathrm{2}} {cot}\left(\pi{a}\right){csc}\left(\pi{a}\right)\:... \\ $$ $$\:\:\:\:\:\mathrm{2}::\:\:\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:{we}\:{know}\:{that}\::: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\overset{{easy}} {=}\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:\:\: \\ $$

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