Question Number 115558 by mnjuly1970 last updated on 26/Sep/20
$$\:\:\:\:\:\:\:…\:{advanced}\:\:\:{calculus}…\: \\ $$$$\:\:\:\:\:\:\:\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} {ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right){ln}\left(\mathrm{1}+\frac{{b}}{{x}^{\mathrm{2}} }\right){dx} \\ $$$$\:\:\:\:\:\:\:{m}.{n}.{july} \\ $$$$ \\ $$
Commented by sachin1221 last updated on 26/Sep/20
![show that lim_(n→∞) (1/n)[((cos^(2p) π)/(2n))+((cos^(2p) 2π)/(2n))+((cos^(2p) 3π)/(2n))+.....((cos^(2p) π)/2)]=Π_(r=1) ^p ((p+r)/(4r))](https://www.tinkutara.com/question/Q115574.png)
$$ \\ $$$$ \\ $$$${show}\:{that}\:\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\left[\frac{{cos}^{\mathrm{2}{p}} \pi}{\mathrm{2}{n}}+\frac{{cos}^{\mathrm{2}{p}} \mathrm{2}\pi}{\mathrm{2}{n}}+\frac{{cos}^{\mathrm{2}{p}} \mathrm{3}\pi}{\mathrm{2}{n}}+…..\frac{{cos}^{\mathrm{2}{p}} \pi}{\mathrm{2}}\right]=\underset{{r}=\mathrm{1}} {\overset{{p}} {\prod}}\frac{{p}+{r}}{\mathrm{4}{r}} \\ $$
Commented by TANMAY PANACEA last updated on 26/Sep/20
$${i}\:{think}\:{some}\:{error}\:{in}\:{problem} \\ $$
Answered by Olaf last updated on 26/Sep/20
![I = ∫_0 ^∞ ln(1+ax^2 )ln(1+(b/x^2 ))dx a > 0, b > 0 I = (π/( (√a)))(−2(√(ab))+((√(ab))+1)[ln(1−ab)+2arctan(√(ab))]) (Wolfram) I will try to find the demonstration but it may take time !](https://www.tinkutara.com/question/Q115570.png)
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \mathrm{ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{1}+\frac{{b}}{{x}^{\mathrm{2}} }\right){dx} \\ $$$$\mathrm{a}\:>\:\mathrm{0},\:\mathrm{b}\:>\:\mathrm{0} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\:\sqrt{{a}}}\left(−\mathrm{2}\sqrt{{ab}}+\left(\sqrt{{ab}}+\mathrm{1}\right)\left[\mathrm{ln}\left(\mathrm{1}−{ab}\right)+\mathrm{2arctan}\sqrt{\mathrm{ab}}\right]\right) \\ $$$$\left(\mathrm{Wolfram}\right) \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{try}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{demonstration} \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{may}\:\mathrm{take}\:\mathrm{time}\:! \\ $$
Commented by mnjuly1970 last updated on 26/Sep/20
$${thank}\:{you}\:{for}\:{yor} \\ $$$${attention}\:{an}\:{effort}.. \\ $$
Answered by maths mind last updated on 27/Sep/20
![∫ln(1+ax^2 )dx=xln(1+ax^2 )−∫((2ax^2 )/(1+ax^2 ))dx =xln(1+ax^2 )−∫((2ax^2 +2−2)/(1+ax^2 ))dx =xln(1+ax^2 )−2x+(2/( (√a)))∫(((√a)dx)/(1+((√a)x)^2 )) =xln(1+ax^2 )−2x+(2/( (√a)))arctan(x(√a)) ∫_0 ^∞ ln(1+ax^2 )ln(1+(b/x^2 ))dx=G(a,b)∫ =[(xln(1+ax^2 )−2x+(2/( (√a)))arctan(x(√a))]ln(1+(b/x^2 ))]_0 ^∞ −∫_0 ^∞ (xln(1+ax^2 )−2x+(2/( (√a)))arctan(x(√a)))((−2b)/(x(x^2 +b)))dx =2b{∫_0 ^∞ ((ln(1+ax^2 ))/((x^2 +b)))dx−∫_0 ^∞ (2/(x^2 +b))+(2/( (√a)))∫_0 ^∞ ((arcran(x(√a)))/(x(x^2 +b)))dx ∫_0 ^∞ (2/(x^2 +b))dx=(2/( (√b)))∫(1/( (√b))).(dx/(1+((x/( (√b))))^2 ))=(2/( (√b))).(π/2)=(π/( (√b))) ∫_0 ^∞ ((ln(1+ax^2 ))/((x^2 +b)))=f(a)⇒f′(a)=∫_0 ^∞ (x^2 /((1+ax^2 )(x^2 +b))) a,b>0 =∫_0 ^∞ ((b(1+ax^2 )−(b+x^2 ))/((ba−1)(1+ax^2 )(x^2 +b)))dx=(b/(ba−1))∫_0 ^∞ (dx/(b+x^2 ))−(1/(ba−1))∫_0 ^∞ (dx/(1+ax^2 )) =((√b)/( (ba−1)))(π/2)−(1/( (√a)(ba−1)))(π/2) f(a)=(π/(2(√b)))ln(ba−1)+(π/( (√b)))th^− ((√(ab))) ∫_0 ^∞ ((arctan(x(√a)))/(x(b+x^2 )))dx=s(a,b) let x=y(√(b )) s=∫_0 ^∞ ((arctan(y(√(ab))))/(b(1+y^2 )y)) ∫_0 ^∞ ((arcran(st))/(t(1+t^2 )))dt=f(s),f′(s)=∫_0 ^∞ (dt/((1+s^2 t^2 )(1+t^2 ))) .=(1/(1−s^2 ))∫_0 ^∞ ((1/(1+t^2 ))−(s^2 /(1+s^2 t^2 )))dt =(π/(2(1−s^2 )))−(s/(1−s^2 )).(π/2) f(s)=∫_0 ^s ((π/(2(1−s^2 ))−(π/2).(s/(1−s^2 )))ds=(π/2)th^− (s)+(π/2)ln((√(1−s^2 ))) ⇒∫_0 ^∞ ((arctan(x(√a)))/(x(x^2 +b)))=(1/b)((π/2)th^− ((√(ab)))+(π/2)ln((√(∣1−ab∣)))) ∫_0 ^∞ ln(1+ax^2 )ln(1+(b/x^2 ))=2b((π/(2(√b)))ln(∣ba−1∣)+(π/( (√b)))th^− ((√(ab)))) −2b.(π/( (√b)))+2b.(2/( (√a)))((1/b)((π/2)th^− ((√(ab)))+(π/4)ln(∣ba−1∣)) =π((√b)+(1/( (√a))))ln∣ba−1∣+π(2(√b)+(2/( (√a))))th^− ((√(ab)))−2π(√b) =((π(−2(√(ab))+((√(ab))+1)ln∣ab−1∣+(2(√(ab))+2)th^− ((√(ab)))))/( (√a)))](https://www.tinkutara.com/question/Q115618.png)
$$\int{ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right){dx}={xln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\int\frac{\mathrm{2}{ax}^{\mathrm{2}} }{\mathrm{1}+{ax}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\int\frac{\mathrm{2}{ax}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}}{\mathrm{1}+{ax}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\mathrm{2}{x}+\frac{\mathrm{2}}{\:\sqrt{{a}}}\int\frac{\sqrt{{a}}{dx}}{\mathrm{1}+\left(\sqrt{{a}}{x}\right)^{\mathrm{2}} } \\ $$$$={xln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\mathrm{2}{x}+\frac{\mathrm{2}}{\:\sqrt{{a}}}{arctan}\left({x}\sqrt{{a}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right){ln}\left(\mathrm{1}+\frac{{b}}{{x}^{\mathrm{2}} }\right){dx}={G}\left({a},{b}\right)\int \\ $$$$=\left[\left({xln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\mathrm{2}{x}+\frac{\mathrm{2}}{\:\sqrt{{a}}}{arctan}\left({x}\sqrt{{a}}\right)\right]{ln}\left(\mathrm{1}+\frac{{b}}{{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$−\int_{\mathrm{0}} ^{\infty} \left({xln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\mathrm{2}{x}+\frac{\mathrm{2}}{\:\sqrt{{a}}}{arctan}\left({x}\sqrt{{a}}\right)\right)\frac{−\mathrm{2}{b}}{{x}\left({x}^{\mathrm{2}} +{b}\right)}{dx} \\ $$$$=\mathrm{2}{b}\left\{\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +{b}\right)}{dx}−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}}{{x}^{\mathrm{2}} +{b}}+\frac{\mathrm{2}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} \frac{{arcran}\left({x}\sqrt{{a}}\right)}{{x}\left({x}^{\mathrm{2}} +{b}\right)}{dx}\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}}{{x}^{\mathrm{2}} +{b}}{dx}=\frac{\mathrm{2}}{\:\sqrt{{b}}}\int\frac{\mathrm{1}}{\:\sqrt{{b}}}.\frac{{dx}}{\mathrm{1}+\left(\frac{\mathrm{x}}{\:\sqrt{\mathrm{b}}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{\:\sqrt{\mathrm{b}}}.\frac{\pi}{\mathrm{2}}=\frac{\pi}{\:\sqrt{{b}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +{b}\right)}={f}\left({a}\right)\Rightarrow{f}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{b}\right)}\:\:{a},{b}>\mathrm{0} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{b}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)−\left({b}+{x}^{\mathrm{2}} \right)}{\left({ba}−\mathrm{1}\right)\left(\mathrm{1}+{ax}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{b}\right)}{dx}=\frac{{b}}{{ba}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{b}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{ba}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{ax}^{\mathrm{2}} } \\ $$$$=\frac{\sqrt{{b}}}{\:\left({ba}−\mathrm{1}\right)}\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{{a}}\left({ba}−\mathrm{1}\right)}\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({a}\right)=\frac{\pi}{\mathrm{2}\sqrt{{b}}}{ln}\left({ba}−\mathrm{1}\right)+\frac{\pi}{\:\sqrt{{b}}}{th}^{−} \left(\sqrt{{ab}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\sqrt{{a}}\right)}{{x}\left({b}+{x}^{\mathrm{2}} \right)}{dx}={s}\left({a},{b}\right)\:{let}\:{x}={y}\sqrt{{b}\:\:} \\ $$$${s}=\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({y}\sqrt{{ab}}\right)}{{b}\left(\mathrm{1}+{y}^{\mathrm{2}} \right){y}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{arcran}\left({st}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}={f}\left({s}\right),{f}'\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left(\mathrm{1}+{s}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$.=\frac{\mathrm{1}}{\mathrm{1}−{s}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{{s}^{\mathrm{2}} }{\mathrm{1}+{s}^{\mathrm{2}} {t}^{\mathrm{2}} }\right){dt} \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{s}^{\mathrm{2}} \right)}−\frac{{s}}{\mathrm{1}−{s}^{\mathrm{2}} }.\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({s}\right)=\int_{\mathrm{0}} ^{{s}} \left(\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{s}^{\mathrm{2}} \right.}−\frac{\pi}{\mathrm{2}}.\frac{{s}}{\mathrm{1}−{s}^{\mathrm{2}} }\right){ds}=\frac{\pi}{\mathrm{2}}{th}^{−} \left({s}\right)+\frac{\pi}{\mathrm{2}}{ln}\left(\sqrt{\left.\mathrm{1}−{s}^{\mathrm{2}} \right)}\right. \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\sqrt{{a}}\right)}{{x}\left({x}^{\mathrm{2}} +{b}\right)}=\frac{\mathrm{1}}{{b}}\left(\frac{\pi}{\mathrm{2}}{th}^{−} \left(\sqrt{{ab}}\right)+\frac{\pi}{\mathrm{2}}{ln}\left(\sqrt{\mid\mathrm{1}−{ab}\mid}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right){ln}\left(\mathrm{1}+\frac{{b}}{{x}^{\mathrm{2}} }\right)=\mathrm{2}{b}\left(\frac{\pi}{\mathrm{2}\sqrt{{b}}}{ln}\left(\mid{ba}−\mathrm{1}\mid\right)+\frac{\pi}{\:\sqrt{{b}}}{th}^{−} \left(\sqrt{{ab}}\right)\right) \\ $$$$−\mathrm{2}{b}.\frac{\pi}{\:\sqrt{{b}}}+\mathrm{2}{b}.\frac{\mathrm{2}}{\:\sqrt{{a}}}\left(\frac{\mathrm{1}}{{b}}\left(\frac{\pi}{\mathrm{2}}{th}^{−} \left(\sqrt{{ab}}\right)+\frac{\pi}{\mathrm{4}}{ln}\left(\mid{ba}−\mathrm{1}\mid\right)\right)\right. \\ $$$$=\pi\left(\sqrt{{b}}+\frac{\mathrm{1}}{\:\sqrt{{a}}}\right){ln}\mid{ba}−\mathrm{1}\mid+\pi\left(\mathrm{2}\sqrt{{b}}+\frac{\mathrm{2}}{\:\sqrt{{a}}}\right){th}^{−} \left(\sqrt{{ab}}\right)−\mathrm{2}\pi\sqrt{{b}} \\ $$$$=\frac{\pi\left(−\mathrm{2}\sqrt{{ab}}+\left(\sqrt{{ab}}+\mathrm{1}\right){ln}\mid{ab}−\mathrm{1}\mid+\left(\mathrm{2}\sqrt{{ab}}+\mathrm{2}\right){th}^{−} \left(\sqrt{{ab}}\right)\right)}{\:\sqrt{{a}}} \\ $$$$ \\ $$$$ \\ $$