Question Number 122159 by mnjuly1970 last updated on 14/Nov/20
$$\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:{prove}\:\:{that}:: \\ $$$$\Omega=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left\{{tan}^{−\mathrm{1}} \left({ptan}\left({x}\right)\right)−{tan}^{−\mathrm{1}} \left({qtan}\left({x}\right)\right)\right\}\left({tan}\left({x}\right)+{cot}\left({x}\right)\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\:{log}\left(\frac{{p}}{{q}}\right)\:\:\:\left(\:\:\:\:{p}\:,\:{q}\:>\mathrm{0}\:\:\:\right) \\ $$$$\:\:\:\:{m}.{n}. \\ $$
Answered by mathmax by abdo last updated on 14/Nov/20
![Ω =∫_0 ^(π/2) (arctan(ptanx)dx−∫_0 ^(π/2) arctan(qtanx)dx =A_p −A_q A_p =∫_0 ^(π/2) arctan(ptanx)dx =_(tanx=t) ∫_0 ^∞ ((arctan(pt))/(1+t^2 ))dt =f(p) f^′ (p) =∫_0 ^∞ (t/((1+p^2 t^2 )(1+t^2 )))dt =_(pt=x) ∫_0 ^∞ (x/(p(1+x^2 )(1+(x^2 /p^2 ))))(dx/p) =∫_0 ^∞ ((xdx)/((x^2 +1)(x^2 +p^2 ))) let decompose F(x)=(x/((x^2 +1)(x^2 +p^2 ))) F(x)=((ax+b)/(x^2 +1)) +((cx +d)/(x^2 +p^2 )) F(−x)=−f(x) ⇒((−ax+b)/(x^2 +1)) +((−cx+d)/(x^2 +p^2 )) =((−ax−b)/(x^2 +1))+((−cx−d)/(x^2 +p^2 )) ⇒b=d=0 ⇒F(x) =((ax)/(x^2 +1))+((cx)/(x^2 +p^2 )) lim_(x→+∞) xF(x)=0 =a+c ⇒c=−a ⇒ F(x)=((ax)/(x^2 +1))−((ax)/(x^2 +p^2 )) F(1)=(1/(2(p^2 +1))) =(a/2)−(a/(p^2 +1)) ⇒((1/2)−(1/(p^2 +1)))a =(1/(2(p^2 +1))) ⇒ ((p^2 −1)/(2(p^2 +1)))a =(1/(2(p^2 +1))) ⇒a =(1/(p^2 −1)) ⇒F(x)=(1/(p^2 −1)){(x/(x^2 +1))−(x/(x^2 +p^2 ))} ⇒ f^, (p) =(1/(p^2 −1))∫_0 ^∞ {(x/(x^2 +1))−(x/(x^2 +p^2 ))}dx =(1/(p^2 −1))×(1/2)[ln∣((x^2 +1)/(x^2 +p^2 ))∣]_0 ^∞ =(1/(2(p^2 −1)))(−2lnp) =−((lnp)/(p^2 −1)) =((lnp)/(1−p^2 )) ⇒ f(p) =∫_1 ^p ((lnt)/(1−t^2 ))dt +c c =f(1) =∫_0 ^∞ ((arctant)/(1+t^2 )) =[arctan^2 t]_0 ^∞ −∫_0 ^∞ ((arctant)/(1+t^2 ))dt ⇒ 2c =((π/2))^2 =(π^2 /4) ⇒c =(π^2 /8) ⇒f(p) =∫_1 ^p ((lnt)/(1−t^2 ))dt +(π^2 /8) if p<1 ⇒∫_1 ^p ((lnt)/(1−t^2 ))dt =−∫_p ^1 ((lnt)/(1−t^2 )) =−∫_p ^1 lntΣ_(n=0) ^∞ t^(2n) dt =−Σ_(n=0) ^∞ ∫_p ^1 t^(2n) lnt dt =−Σ_(n=0) ^∞ u_n u_n =[(t^(2n+1) /(2n+1))lnt]_p ^1 −∫_p ^1 (t^(2n+1) /(2n+1))(dt/t) =−(p^(2n+1) /(2n+1))−(1/(2n+1))∫_p ^1 t^(2n) dt =−(p^(2n+1) /(2n+1))−(1/((2n+1)^2 ))[t^(2n+1) ]_p ^1 =−(p^(2n+1) /(2n+1))−(1/((2n+1)^2 ))(1−p^(2n+1) ) ⇒ ∫_1 ^p ((lnt)/(1−t^2 ))dt =Σ_(n=0) ^∞ (p^(2n+1) /(2n+1)) +Σ_(n=0) ^∞ ((1−p^(2n+1) )/((2n+1)^2 ))....becontinued...](https://www.tinkutara.com/question/Q122165.png)
$$\Omega\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{arctan}\left(\mathrm{ptanx}\right)\mathrm{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{arctan}\left(\mathrm{qtanx}\right)\mathrm{dx}\right. \\ $$$$=\mathrm{A}_{\mathrm{p}} −\mathrm{A}_{\mathrm{q}} \\ $$$$\mathrm{A}_{\mathrm{p}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{arctan}\left(\mathrm{ptanx}\right)\mathrm{dx}\:=_{\mathrm{tanx}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{pt}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\mathrm{f}\left(\mathrm{p}\right) \\ $$$$\mathrm{f}^{'} \left(\mathrm{p}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}}{\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt}\:=_{\mathrm{pt}=\mathrm{x}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{x}}{\mathrm{p}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{p}^{\mathrm{2}} }\right)}\frac{\mathrm{dx}}{\mathrm{p}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{xdx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{p}^{\mathrm{2}} \right)}\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{p}^{\mathrm{2}} \right)} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{cx}\:+\mathrm{d}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{p}^{\mathrm{2}} } \\ $$$$\mathrm{F}\left(−\mathrm{x}\right)=−\mathrm{f}\left(\mathrm{x}\right)\:\Rightarrow\frac{−\mathrm{ax}+\mathrm{b}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:+\frac{−\mathrm{cx}+\mathrm{d}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{p}^{\mathrm{2}} }\:=\frac{−\mathrm{ax}−\mathrm{b}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}+\frac{−\mathrm{cx}−\mathrm{d}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{p}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{b}=\mathrm{d}=\mathrm{0}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{ax}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{cx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{p}^{\mathrm{2}} } \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{xF}\left(\mathrm{x}\right)=\mathrm{0}\:=\mathrm{a}+\mathrm{c}\:\Rightarrow\mathrm{c}=−\mathrm{a}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{ax}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{ax}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{p}^{\mathrm{2}} } \\ $$$$\mathrm{F}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{a}}{\mathrm{p}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{a}\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\frac{\mathrm{p}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{a}\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} +\mathrm{1}\right)}\:\Rightarrow\mathrm{a}\:=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{2}} −\mathrm{1}}\left\{\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{p}^{\mathrm{2}} }\right\}\:\Rightarrow \\ $$$$\mathrm{f}^{,} \left(\mathrm{p}\right)\:=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left\{\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{p}^{\mathrm{2}} }\right\}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{p}^{\mathrm{2}} −\mathrm{1}}×\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\mid\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{p}^{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{\infty} \:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} −\mathrm{1}\right)}\left(−\mathrm{2lnp}\right)\:=−\frac{\mathrm{lnp}}{\mathrm{p}^{\mathrm{2}} −\mathrm{1}}\:=\frac{\mathrm{lnp}}{\mathrm{1}−\mathrm{p}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{p}\right)\:=\int_{\mathrm{1}} ^{\mathrm{p}} \:\:\frac{\mathrm{lnt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:+\mathrm{c} \\ $$$$\mathrm{c}\:=\mathrm{f}\left(\mathrm{1}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctant}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:=\left[\mathrm{arctan}^{\mathrm{2}} \mathrm{t}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctant}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{2c}\:=\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\mathrm{c}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\mathrm{f}\left(\mathrm{p}\right)\:=\int_{\mathrm{1}} ^{\mathrm{p}} \:\frac{\mathrm{lnt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{if}\:\mathrm{p}<\mathrm{1}\:\Rightarrow\int_{\mathrm{1}} ^{\mathrm{p}} \:\frac{\mathrm{lnt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=−\int_{\mathrm{p}} ^{\mathrm{1}} \:\frac{\mathrm{lnt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:=−\int_{\mathrm{p}} ^{\mathrm{1}} \mathrm{lnt}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{2n}} \:\mathrm{dt} \\ $$$$=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{p}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{2n}} \:\mathrm{lnt}\:\mathrm{dt}\:\:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{u}_{\mathrm{n}} =\left[\frac{\mathrm{t}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\mathrm{lnt}\right]_{\mathrm{p}} ^{\mathrm{1}} −\int_{\mathrm{p}} ^{\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\frac{\mathrm{dt}}{\mathrm{t}}\:=−\frac{\mathrm{p}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\int_{\mathrm{p}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{2n}} \mathrm{dt} \\ $$$$=−\frac{\mathrm{p}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\left[\mathrm{t}^{\mathrm{2n}+\mathrm{1}} \right]_{\mathrm{p}} ^{\mathrm{1}} \:=−\frac{\mathrm{p}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{p}^{\mathrm{2n}+\mathrm{1}} \right)\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\mathrm{p}} \:\frac{\mathrm{lnt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{p}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\:+\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−\mathrm{p}^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }….\mathrm{becontinued}… \\ $$
Answered by mindispower last updated on 14/Nov/20
$${not}\:{true}\:{sir} \\ $$$${q}=\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}^{−} \left({ptan}\left({x}\right)\right)−{xdx}=\frac{\pi}{\mathrm{2}}{ln}\left({p}\right) \\ $$$${tak}\:{p}\rightarrow\infty\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}^{−} \left({ptan}\left({x}\right)\right)−{x}\:\:{dx}={I}\rightarrow+\infty \\ $$$${but}\:{tan}^{−} \left({z}\right)<\frac{\pi}{\mathrm{2}}\Rightarrow \\ $$$${I}\leqslant\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\pi}{\mathrm{2}}−{x}\right){dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 14/Nov/20
$${thank}\:{you} \\ $$$$\:{corrected} \\ $$
Answered by mindispower last updated on 14/Nov/20
![∫_0 ^(π/2) {tan^− (t.tan(x))}{tan(x)+cot(x)}dx=f(t) =∫_0 ^(π/2) {tan^− (t.tan(x)){((1+tan^2 (x))/(tan(x)))}dx,tan(x)=s ⇔ =∫_0 ^∞ tan^− (ts).(ds/s) f(p)−f(q)=∫_0 ^∞ (tan^− (ps)−tan^− (qs))(ds/s) =[(tan^− (ps)−tan^− (qs))ln(x)]_0 ^∞ −∫_0 ^∞ {p((ln(s))/(1+p^2 s2))−q((ln(s))/(1+q^2 s^2 ))}ds tan^− (ps)−tan^− (qs)=((π/2)−(1/(ps))+o((1/s))−((π/2)−(1/(qs))+o((1/s))) =(1/s)((1/q)−(1/p))+o((1/s))⇒lim_(x→∞) ln(x){tan^− (ps)−tan^− (qs)}=0 ⇔ f(p)−f(q)=−∫_0 ^∞ ((pln(s))/(1+p^2 s2))−((qln(s))/(1+q^2 s^2 ))ds=−(g(p)−g(q)) g(z)=∫_0 ^∞ ((zln(x))/(1+z^2 x^2 ))dx,z>0 withe Quation llet zx=t ⇔g(z)=∫_0 ^∞ ((ln(t)−ln(z))/(1+t^2 ))dt =∫_0 ^∞ ((ln(t))/(1+t^2 ))dt_(=0) −ln(z)∫_0 ^∞ (dt/(1+t^2 )) =−ln(z)[tan^(−1) (t)]_0 ^∞ =−(π/2)ln(z) ∫_0 ^∞ {tan^(−1) (ptan(x))−tan^− (qtan(x)}(tan(x)+cot(x))dx =−g(p)+g(q)=(π/2)ln(p)−(π/2)ln(q)=(π/2)ln((p/q))](https://www.tinkutara.com/question/Q122215.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{{tan}^{−} \left({t}.{tan}\left({x}\right)\right)\right\}\left\{{tan}\left({x}\right)+{cot}\left({x}\right)\right\}{dx}={f}\left({t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{{tan}^{−} \left({t}.{tan}\left({x}\right)\right)\left\{\frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{{tan}\left({x}\right)}\right\}{dx},{tan}\left({x}\right)={s}\right. \\ $$$$\Leftrightarrow \\ $$$$=\int_{\mathrm{0}} ^{\infty} {tan}^{−} \left({ts}\right).\frac{{ds}}{{s}} \\ $$$${f}\left({p}\right)−{f}\left({q}\right)=\int_{\mathrm{0}} ^{\infty} \left({tan}^{−} \left({ps}\right)−{tan}^{−} \left({qs}\right)\right)\frac{{ds}}{{s}} \\ $$$$=\left[\left({tan}^{−} \left({ps}\right)−{tan}^{−} \left({qs}\right)\right){ln}\left({x}\right)\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \left\{{p}\frac{{ln}\left({s}\right)}{\mathrm{1}+{p}^{\mathrm{2}} {s}\mathrm{2}}−{q}\frac{{ln}\left({s}\right)}{\mathrm{1}+{q}^{\mathrm{2}} {s}^{\mathrm{2}} }\right\}{ds} \\ $$$${tan}^{−} \left({ps}\right)−{tan}^{−} \left({qs}\right)=\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{ps}}+{o}\left(\frac{\mathrm{1}}{{s}}\right)−\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{{qs}}+{o}\left(\frac{\mathrm{1}}{{s}}\right)\right)\right. \\ $$$$=\frac{\mathrm{1}}{{s}}\left(\frac{\mathrm{1}}{{q}}−\frac{\mathrm{1}}{{p}}\right)+{o}\left(\frac{\mathrm{1}}{{s}}\right)\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}{ln}\left({x}\right)\left\{{tan}^{−} \left({ps}\right)−{tan}^{−} \left({qs}\right)\right\}=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${f}\left({p}\right)−{f}\left({q}\right)=−\int_{\mathrm{0}} ^{\infty} \frac{{pln}\left({s}\right)}{\mathrm{1}+{p}^{\mathrm{2}} {s}\mathrm{2}}−\frac{{qln}\left({s}\right)}{\mathrm{1}+{q}^{\mathrm{2}} {s}^{\mathrm{2}} }{ds}=−\left({g}\left({p}\right)−{g}\left({q}\right)\right) \\ $$$${g}\left({z}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{zln}\left({x}\right)}{\mathrm{1}+{z}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx},{z}>\mathrm{0}\:{withe}\:{Quation} \\ $$$$\mathrm{l}{let}\:{zx}={t} \\ $$$$\Leftrightarrow{g}\left({z}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({t}\right)−{ln}\left({z}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}_{=\mathrm{0}} −{ln}\left({z}\right)\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=−{ln}\left({z}\right)\left[\mathrm{tan}^{−\mathrm{1}} \left({t}\right)\right]_{\mathrm{0}} ^{\infty} =−\frac{\pi}{\mathrm{2}}{ln}\left({z}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \left\{\mathrm{tan}^{−\mathrm{1}} \left({ptan}\left({x}\right)\right)−{tan}^{−} \left({qtan}\left({x}\right)\right\}\left({tan}\left({x}\right)+{cot}\left({x}\right)\right){dx}\right. \\ $$$$=−{g}\left({p}\right)+{g}\left({q}\right)=\frac{\pi}{\mathrm{2}}{ln}\left({p}\right)−\frac{\pi}{\mathrm{2}}{ln}\left({q}\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\frac{{p}}{{q}}\right) \\ $$$$ \\ $$$$ \\ $$