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Question Number 116272 by mindispower last updated on 02/Oct/20

∫_0 ^(π/2) ln(x^2 +ln^2 (cos(x)))dx=πln(ln(2)) posted Quation not solved yet i hop someon Giv idea for this one thank you

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({x}^{\mathrm{2}} +{ln}^{\mathrm{2}} \left({cos}\left({x}\right)\right)\right){dx}=\pi{ln}\left({ln}\left(\mathrm{2}\right)\right) \\ $$ $${posted}\:{Quation}\: \\ $$ $${not}\:{solved}\:{yet}\:{i}\:{hop}\:{someon}\:{Giv}\:{idea}\:{for} \\ $$ $${this}\:{one}\:{thank}\:{you} \\ $$

Answered by mathdave last updated on 03/Oct/20

solution let consider this easy property ∫_0 ^∞ t^(s−1) e^(−zt) dt=((Γ(s))/Z^s ) so let Z=a+ib to get ∫_0 ^∞ t^(s−1) e^(−at) .e^(ibt) dt=((Γ(s))/((a+ib)^s ))..........(1) let d conjugate(Z)=Z^− =a−ib to get ∫_0 ^∞ t^(s−1) e^(−at) .e^(ibt) dt=((Γ(s))/((a−ib)^s ))........(2) adding (1) and (2) results to ve ∫_0 ^∞ t^(s−1) e^(−at) (e^(ibt) +e^(−ibt) )dt=Γ(s)((1/((a+ib)^s ))+(1/((a−ib)^s ))) but cosz=((e^(iz) +e^(−iz) )/2) then 2∫_0 ^∞ t^(s−1) cos(bt)e^(−at) dt=(((a−ib)^s +(a+ib)^s )/((a^2 +b^2 )^s )) (from z^n =(re^(iθ) )^n ,r^n =∣z∣^n =(a^2 +b^2 )^n ,θ=Arg(z) or Arg(z^− ),n=s) but note z^n =∣z∣^n e^(n(Arg(z)+2kπ)i) ,k=0,∣z∣^n =(a^2 +b^2 )^(s/2) Arg(z) or Arg(z^− )=tan^(−1) ((b/a)) or tan^(−1) (−(b/a)) ∵I=∫_0 ^∞ t^(s−1) cos(bt)e^(−at) dt=(1/2)Γ(s)•(((a^2 +b^2 )^(s/2) (e^(s.Arg(z^− )) +e^(s.Arg(z)) ))/((a^2 +b^2 )^s )) I=((Γ(s))/((a^2 +b^2 )^(s/2) )).((e^(−stan^(−1) ((b/a))i) +e^(stan^(−1) ((b/a))i) )/2) I=((Γ(s))/((a^2 +b^2 )^(s/2) )).cos(stan^(−1) ((b/a))) I=∫_0 ^∞ t^(s−1) cos(bt)e^(−at) dt=((Γ(s)cos(stan^(−1) ((b/a))))/((a^2 +b^2 )^(s/2) )) put b=x,a=−ln(cosx) and multiply both side by ∫_0 ^(π/2) dx to ve ∫_0 ^(π/2) ∫_0 ^∞ t^(s−1) cos(tx)e^(tln(cosx)) dtdx=Γ(s)∫_0 ^(π/2) ((cos(stan^(−1) (−(x/(ln(cosx))))))/((x^2 +ln^2 (cosx))^(s/2) ))dx ∫_0 ^(π/2) ((cos(stan^(−1) (−(x/(ln(cosx))))))/((x^2 +ln^2 (cosx))^(s/2) ))dx=(1/(Γ(s)))∫_0 ^∞ t^(s−1) ∫_0 ^(π/2) cos^t xcos(tx)dtdx let solve ∫_0 ^(π/2) cos^t xcos(tx)dtdx first note from d formular ∫_0 ^(π/2) cos^(a−1) xcos(bx)dx=(π/(2^a a))•(1/(β(((a+b+1)/2),((a−b+1)/2)))) then ∫_0 ^(π/2) cos^t xcos(tx)dx=(π/2^(t+1) ) ∵I=∫_0 ^(π/2) ((cos(stan^(−1) (−(x/(ln(cosx))))))/((x^2 +ln^2 (cosx))^(s/2) ))dx=(1/(Γ(s)))∫_0 ^∞ t^(s−1) .(π/2^(t+1) )dt I=(π/2)•(1/(Γ(s)))∫_0 ^∞ t^(s−1) .e^(−tln2) dt=(π/2)•(1/(Γ(s)))•((Γ(s))/(ln^s (2)))=(π/(2ln^s (2))) ∫_0 ^(π/2) ((cos(stan^(−1) (−(x/(ln(cosx))))))/((x^2 +ln^2 (cosx))^(s/2) ))dx=(π/(2ln^s (2))) using Duis with respect to s of both side (∂/∂s)∣_(s=0) ∫_0 ^(π/2) ((cos(stan^(−1) (−(x/(ln(cosx))))))/((x^2 +ln^2 (cosx))^(s/2) ))dx=(∂/∂s)∣_(s=0) (π/(2ln^s (2))) ∫_0 ^(π/2) (−(1/2)ln(x^2 +ln^2 (cosx)))dx=−(π/2)ln(ln2) ∫_0 ^(π/2) ln(x^2 +ln^2 (cosx))dx=πln(ln2) Q.E.D by mathdave(03/10/2020)

$${solution} \\ $$ $${let}\:{consider}\:{this}\:{easy}\:{property} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{zt}} {dt}=\frac{\Gamma\left({s}\right)}{{Z}^{{s}} }\:\:\:{so}\:{let}\:\:{Z}={a}+{ib}\:\:{to}\:{get} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{at}} .{e}^{{ibt}} {dt}=\frac{\Gamma\left({s}\right)}{\left({a}+{ib}\right)^{{s}} }..........\left(\mathrm{1}\right) \\ $$ $${let}\:{d}\:{conjugate}\left({Z}\right)=\overset{−} {{Z}}={a}−{ib}\:\:{to}\:{get} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{at}} .{e}^{{ibt}} {dt}=\frac{\Gamma\left({s}\right)}{\left({a}−{ib}\right)^{{s}} }........\left(\mathrm{2}\right) \\ $$ $${adding}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\:{results}\:{to}\:{ve} \\ $$ $$\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{at}} \left({e}^{{ibt}} +{e}^{−{ibt}} \right){dt}=\Gamma\left({s}\right)\left(\frac{\mathrm{1}}{\left({a}+{ib}\right)^{{s}} }+\frac{\mathrm{1}}{\left({a}−{ib}\right)^{{s}} }\right) \\ $$ $${but}\:\mathrm{cos}{z}=\frac{{e}^{{iz}} +{e}^{−{iz}} }{\mathrm{2}}\:\:\:{then} \\ $$ $$\mathrm{2}\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} \mathrm{cos}\left({bt}\right){e}^{−{at}} {dt}=\frac{\left({a}−{ib}\right)^{{s}} +\left({a}+{ib}\right)^{{s}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{{s}} }\:\:\:\:\:\:\:\:\:\left({from}\:{z}^{{n}} =\left({re}^{{i}\theta} \right)^{{n}} \:,{r}^{{n}} =\mid{z}\mid^{{n}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{{n}} ,\theta={Arg}\left({z}\right)\:{or}\:{Arg}\left(\overset{−} {{z}}\right),{n}={s}\right)\: \\ $$ $${but}\:{note}\:{z}^{{n}} =\mid{z}\mid^{{n}} {e}^{{n}\left({Arg}\left({z}\right)+\mathrm{2}{k}\pi\right){i}} \:\:,{k}=\mathrm{0},\mid{z}\mid^{{n}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{{s}}{\mathrm{2}}} \\ $$ $${Arg}\left({z}\right)\:{or}\:{Arg}\left(\overset{−} {{z}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\:{or}\:\mathrm{tan}^{−\mathrm{1}} \:\left(−\frac{{b}}{{a}}\right) \\ $$ $$\because{I}=\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} \mathrm{cos}\left({bt}\right){e}^{−{at}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left({s}\right)\bullet\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{{s}}{\mathrm{2}}} \left({e}^{{s}.{Arg}\left(\overset{−} {{z}}\right)} +{e}^{{s}.{Arg}\left({z}\right)} \right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{{s}} } \\ $$ $${I}=\frac{\Gamma\left({s}\right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{{s}}{\mathrm{2}}} }.\frac{{e}^{−{s}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right){i}} +{e}^{{s}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right){i}} }{\mathrm{2}} \\ $$ $${I}=\frac{\Gamma\left({s}\right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{{s}}{\mathrm{2}}} }.\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right) \\ $$ $${I}=\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} \mathrm{cos}\left({bt}\right){e}^{−{at}} {dt}=\frac{\Gamma\left({s}\right)\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{b}}{{a}}\right)\right)}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{{s}}{\mathrm{2}}} } \\ $$ $${put}\:{b}={x},{a}=−\mathrm{ln}\left(\mathrm{cos}{x}\right)\:{and}\:{multiply}\:{both}\:{side}\:{by}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\:\:{to}\:\:{ve} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} \mathrm{cos}\left({tx}\right){e}^{{t}\mathrm{ln}\left(\mathrm{cos}{x}\right)} {dtdx}=\Gamma\left({s}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{x}}{\mathrm{ln}\left(\mathrm{cos}{x}\right)}\right)\right)}{\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right)^{\frac{{s}}{\mathrm{2}}} }{dx} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{x}}{\mathrm{ln}\left(\mathrm{cos}{x}\right)}\right)\right)}{\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right)^{\frac{{s}}{\mathrm{2}}} }{dx}=\frac{\mathrm{1}}{\Gamma\left({s}\right)}\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{{t}} {x}\mathrm{cos}\left({tx}\right){dtdx} \\ $$ $${let}\:{solve}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{{t}} {x}\mathrm{cos}\left({tx}\right){dtdx}\:\:{first} \\ $$ $${note}\:{from}\:{d}\:{formular} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{{a}−\mathrm{1}} {x}\mathrm{cos}\left({bx}\right){dx}=\frac{\pi}{\mathrm{2}^{{a}} {a}}\bullet\frac{\mathrm{1}}{\beta\left(\frac{{a}+{b}+\mathrm{1}}{\mathrm{2}},\frac{{a}−{b}+\mathrm{1}}{\mathrm{2}}\right)}\:\:{then} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{{t}} {x}\mathrm{cos}\left({tx}\right){dx}=\frac{\pi}{\mathrm{2}^{{t}+\mathrm{1}} } \\ $$ $$\because{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{x}}{\mathrm{ln}\left(\mathrm{cos}{x}\right)}\right)\right)}{\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right)^{\frac{{s}}{\mathrm{2}}} }{dx}=\frac{\mathrm{1}}{\Gamma\left({s}\right)}\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} .\frac{\pi}{\mathrm{2}^{{t}+\mathrm{1}} }{dt} \\ $$ $$\:\:\:\:\:\:\:\:{I}=\frac{\pi}{\mathrm{2}}\bullet\frac{\mathrm{1}}{\Gamma\left({s}\right)}\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} .{e}^{−{t}\mathrm{ln2}} {dt}=\frac{\pi}{\mathrm{2}}\bullet\frac{\mathrm{1}}{\Gamma\left({s}\right)}\bullet\frac{\Gamma\left({s}\right)}{\mathrm{ln}^{{s}} \left(\mathrm{2}\right)}=\frac{\pi}{\mathrm{2ln}^{{s}} \left(\mathrm{2}\right)} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{x}}{\mathrm{ln}\left(\mathrm{cos}{x}\right)}\right)\right)}{\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right)^{\frac{{s}}{\mathrm{2}}} }{dx}=\frac{\pi}{\mathrm{2ln}^{{s}} \left(\mathrm{2}\right)} \\ $$ $${using}\:{Duis}\:{with}\:{respect}\:{to}\:{s}\:{of}\:{both}\:{side} \\ $$ $$\frac{\partial}{\partial{s}}\mid_{{s}=\mathrm{0}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\left({s}\mathrm{tan}^{−\mathrm{1}} \left(−\frac{{x}}{\mathrm{ln}\left(\mathrm{cos}{x}\right)}\right)\right)}{\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right)^{\frac{{s}}{\mathrm{2}}} }{dx}=\frac{\partial}{\partial{s}}\mid_{{s}=\mathrm{0}} \frac{\pi}{\mathrm{2ln}^{{s}} \left(\mathrm{2}\right)} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right)\right){dx}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{ln2}\right) \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{ln}^{\mathrm{2}} \left(\mathrm{cos}{x}\right)\right){dx}=\pi\mathrm{ln}\left(\mathrm{ln2}\right)\:\:\:\:{Q}.{E}.{D} \\ $$ $${by}\:{mathdave}\left(\mathrm{03}/\mathrm{10}/\mathrm{2020}\right) \\ $$

Commented bymnjuly1970 last updated on 03/Oct/20

okay, nice very nice mr dave (good man but a little bit sensitive)

$${okay},\:{nice}\:\:{very}\:{nice}\:{mr}\:{dave}\:\left({good}\right. \\ $$ $$\left.{man}\:{but}\:\:{a}\:{little}\:{bit}\:{sensitive}\right) \\ $$

Commented bymindispower last updated on 03/Oct/20

thank you i have not this idea good job sir thanx again

$${thank}\:{you}\:{i}\:{have}\:{not}\:{this}\:{idea} \\ $$ $${good}\:{job}\:{sir}\:{thanx}\:{again} \\ $$

Commented byTawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

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