Integration Questions

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

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

Answered by mathdave last updated on 03/Oct/20

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

Commented by mnjuly1970 last updated on 03/Oct/20

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

Commented by mindispower last updated on 03/Oct/20

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

Commented by Tawa11 last updated on 06/Sep/21

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