Math Question and Answers


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 75770 by Ajao yinka last updated on 16/Dec/19

Commented by ~blr237~ last updated on 17/Dec/19

$A = \int_{0}^{\infty} \frac{dx}{e^{x} \sqrt{\sinh 2 x}} = \sqrt{2} \int_{0}^{\infty} \frac{e^{- x} dx}{\sqrt{e^{2 x} - e^{- 2 x}}}$ $= \sqrt{2} \int_{0}^{\infty} \frac{e^{- 2 x} dx}{\sqrt{1 - {(e^{- 2 x})}^{2}}} = \frac{- 1}{\sqrt{2}} {[arsin (e^{- 2 x})]}_{0}^{\infty}$ $= \frac{π}{2 \sqrt{2}}$
Commented by ~blr237~ last updated on 17/Dec/19

$let consider for b \neq 0 and a \neq - 1, f (a, b) = \int_{0}^{\infty} \frac{x^{a}}{b + x} dx$ $f (a, b) = \int_{0}^{\infty} \frac{b^{a} {(\frac{x}{b})}^{a}}{1 + (\frac{x}{b})} d (\frac{x}{b}) = b^{a} \int_{0}^{\infty} \frac{t^{a}}{1 + t} dt$ $using the Mellin function \int_{0}^{\infty} \frac{t^{a}}{1 + t} dt = \frac{π}{\sin (π (a + 1))}$ $so f (a, b) = \frac{π b^{a}}{\sin (π (a + 1))}$ $Now study \frac{\partial f}{\partial a} (a, b) = \int_{0}^{\infty} \frac{x^{a} lnx}{b + x} dx$ $\frac{\partial^{n} f}{\partial b^{n - 1} \partial a} (a, b) = \int_{0}^{\infty} {(- 1)}^{n} (n - 1) \frac{x^{a} lnx}{{(b + x)}^{n}} dx$ $So \int_{0}^{\infty} \frac{lnx}{{(1 + x)}^{n}} dx = \frac{{(- 1)}^{n}}{n - 1} \frac{\partial^{n} f}{\partial b^{n - 1} \partial a} (0, 1)$ $\int_{0}^{\infty} \frac{(1 - 3 x)}{{(1 + x)}^{5}} {(logx)}^{2} dx = \int_{0}^{\infty} [\frac{4}{{(1 + x)}^{5}} {(logx)}^{2} - \frac{3}{{(1 + x)}^{4}} {(logx)}^{2}] dx$ $= - \frac{\partial^{6} f}{\partial b^{4} \partial a^{2}} (0, 1) - \frac{\partial^{5} f}{\partial b^{3} \partial a^{2}} (0, 1)$
Commented by ~blr237~ last updated on 17/Dec/19

$let consider for b \neq 0 and a \neq - 1, f (a, b) = \int_{0}^{\infty} \frac{x^{a}}{b + x} dx$ $f (a, b) = \int_{0}^{\infty} \frac{b^{a} {(\frac{x}{b})}^{a}}{1 + (\frac{x}{b})} d (\frac{x}{b}) = b^{a} \int_{0}^{\infty} \frac{t^{a}}{1 + t} dt$ $using the Mellin function \int_{0}^{\infty} \frac{t^{a}}{1 + t} dt = \frac{π}{\sin (π (a + 1))}$ $so f (a, b) = \frac{π b^{a}}{\sin (π (a + 1))}$ $Now study \frac{\partial f}{\partial a} (a, b) = \int_{0}^{\infty} \frac{x^{a} lnx}{b + x} dx$ $\frac{\partial^{n} f}{\partial b^{n - 1} \partial a} (a, b) = \int_{0}^{\infty} {(- 1)}^{n} (n - 1) \frac{x^{a} lnx}{{(b + x)}^{n}} dx$ $So \int_{0}^{\infty} \frac{lnx}{{(1 + x)}^{n}} dx = \frac{{(- 1)}^{n}}{n - 1} \frac{\partial^{n} f}{\partial b^{n - 1} \partial a} (0, 1)$ $\int_{0}^{\infty} \frac{(1 - 3 x)}{{(1 + x)}^{5}} {(logx)}^{2} dx = \int_{0}^{\infty} [\frac{4}{{(1 + x)}^{5}} {(logx)}^{2} - \frac{3}{{(1 + x)}^{4}} {(logx)}^{2}] dx$ $= - \frac{\partial^{6} f}{\partial b^{4} \partial a^{2}} (0, 1) - \frac{\partial^{5} f}{\partial b^{3} \partial a^{2}} (0, 1)$
Commented by Ajao yinka last updated on 17/Dec/19
Superb
Commented by Ajao yinka last updated on 17/Dec/19
So what's final answer?
Commented by ~blr237~ last updated on 17/Dec/19

$you find out the exprezsion of that derivate (it will be easy to start derivation on b)$ $after you just replace (a, b) by (0, 1)$
	Answered by ~blr237~ last updated on 17/Dec/19

	$B = \int_{\frac{- π}{4}}^{\frac{π}{4}} \log (sinx + cosx) dx$ $observe that sinx + cosx = \sqrt{2} \cos (x - \frac{π}{4})$ $B = \int_{- \frac{π}{2}}^{0} \log (\sqrt{2} cosu) du with u = x - \frac{π}{4}$ $B = - \frac{π}{4} \log 2 + \int_{0}^{\frac{π}{2}} \ln (cosu) du$ $secondly observe that$ $B = \int_{- \frac{π}{4}}^{\frac{π}{4}} \log (cost - sint) dt with t = - x$ $so 2 B = \int_{\frac{- π}{4}}^{\frac{π}{4}} [\log (cosx + sinx) + \log (cosx - sinx)] dx$ $= \int_{\frac{- π}{4}}^{\frac{π}{4}} \log (\cos 2 x) dx = 2 \int_{0}^{\frac{π}{4}} \log (\cos 2 x) dx$ $= \int_{0}^{\frac{π}{2}} \log (cosu) du with u = 2 x$ $Finally we have$ $B = - \frac{π}{4} \log 2 + 2 B$ $so B = \frac{π}{4} \log 2$
	Commented by Ajao yinka last updated on 17/Dec/19
	Nice
	Answered by mind is power last updated on 17/Dec/19

	$\int_{\frac{- π}{4}}^{\frac{π}{4}} \log (\sin (x) + \cos (x)) dx$ $= \int_{- \frac{π}{4}}^{\frac{π}{4}} \log (\sqrt{2} \sin (x + \frac{π}{4})) dx$ $= \int_{0}^{\frac{π}{2}} \log (\sqrt{2} \sin (u)) du$ $= \frac{π}{2} \log (\sqrt{2}) + \int_{0}^{\frac{π}{2}} \log (\sin (u)) du$ $\int_{0}^{\frac{π}{2}} \log (\sin (u)) du = \int_{0}^{\frac{π}{2}} \log (\cos (u)) du$ $\int_{0}^{π} \log (\sin (u)) = 2 \int_{0}^{\frac{π}{2}} \log (\sin (u))$ $\int_{0}^{\frac{π}{2}} \log (\sin (2 u)) du = \frac{1}{2} \int_{0}^{π} \log (\sin (u)) du = \int_{0}^{\frac{π}{2}} \log (\sin (u))$ $= \frac{π}{2} \log (2) + 2 \int_{0}^{\frac{π}{2}} \log (\sin (u)) du = \int_{0}^{\frac{π}{2}} \log (\sin (u)) du$ $\Rightarrow \int_{0}^{\frac{π}{2}} \log (\sin (u)) du = - \frac{π}{2} \log (2)$ $\Rightarrow \int_{- \frac{π}{4}}^{+ \frac{π}{4}} \log (\sin (u) + \cos (u)) du = \frac{π}{2} \log (\sqrt{2}) - \frac{π}{2} \log (2)$ $\frac{π}{2} \log (\frac{1}{\sqrt{2}})$
	Answered by mind is power last updated on 17/Dec/19
	$∫_0 ^(+∞) ((log(x))/((1+x)^n )) n≥2 if n =1 ,0diverge x=(1/y)⇒ =∫_0 ^(+∞) −((log(y))/((1+y)^n )).y^(n−2) dy if n=2 ⇒∫_0 ^(+∞) ((log(y))/((1+y)^2 ))dy=0 n≥3, by part U_n =∫_0 ^(+∞) ((log(x))/((1+x)^n ))dx ((log(x))/((1+x)^n ))=[((xlog(x)−x)/((1+x)^n ))]+n∫((xlog(x)−x)/((1+x)^(n+1) ))dx =n∫_0 ^(+∞) (((x+1)log(x)−log(x))/((1+x)^(n+1) ))dx−n∫_0 ^(+∞) {(1/((1+x)^n ))−(1/((1+x)^(n+1) ))}dx ==n∫_0 ^(+∞) ((log(x))/((1+x)^n ))−n∫_0 ^(+∞) ((log(x))/((1+x)^(n+1) ))dx−n[−(1/(n−1)).(1/((1+x)^(n−1) ))+(1/n).(1/((1+x)^n ))]_0 ^(+∞) ⇒U_n =nU_n −nU_(n+1) −n{−(1/(n−1))+(1/n)} ⇒(1−n)U_n =−nU_(n+1) +(1/((n−1))) ⇔U_(n+1) =((n−1)/n)U_n +(1/(n(n−1))) U_2 =0$
	$\int_{0}^{+ \infty} \frac{\log (x)}{{(1 + x)}^{n}}$ $n ⩾ 2 if n = 1, 0 diverge$ $x = \frac{1}{y} \Rightarrow$ $= \int_{0}^{+ \infty} - \frac{\log (y)}{{(1 + y)}^{n}} . y^{n - 2} dy$ $if n = 2 \Rightarrow \int_{0}^{+ \infty} \frac{\log (y)}{{(1 + y)}^{2}} dy = 0$ $n ⩾ 3,$ $by part U_{n} = \int_{0}^{+ \infty} \frac{\log (x)}{{(1 + x)}^{n}} dx$ $\frac{\log (x)}{{(1 + x)}^{n}} = [\frac{xlog (x) - x}{{(1 + x)}^{n}}] + n \int \frac{xlog (x) - x}{{(1 + x)}^{n + 1}} dx$ $= n \int_{0}^{+ \infty} \frac{(x + 1) \log (x) - \log (x)}{{(1 + x)}^{n + 1}} dx - n \int_{0}^{+ \infty} {\frac{1}{{(1 + x)}^{n}} - \frac{1}{{(1 + x)}^{n + 1}}} dx$ $== n \int_{0}^{+ \infty} \frac{\log (x)}{{(1 + x)}^{n}} - n \int_{0}^{+ \infty} \frac{\log (x)}{{(1 + x)}^{n + 1}} dx - n {[- \frac{1}{n - 1} . \frac{1}{{(1 + x)}^{n - 1}} + \frac{1}{n} . \frac{1}{{(1 + x)}^{n}}]}_{0}^{+ \infty}$ $\Rightarrow U_{n} = {nU}_{n} - {nU}_{n + 1} - n {- \frac{1}{n - 1} + \frac{1}{n}}$ $\Rightarrow (1 - n) U_{n} = - {nU}_{n + 1} + \frac{1}{(n - 1)}$ $\Leftrightarrow U_{n + 1} = \frac{n - 1}{n} U_{n} + \frac{1}{n (n - 1)}$ $U_{2} = 0$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum