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

$${\int }_0^{\frac{\pi }{2}}ln(x^2+{ln}^2\left(cos\left(x\right)\right))dx=\pi ln\left(ln\left(2\right)\right)$$$$postedQuation$$$$notsolvedyetihopsomeonGivideafor$$$$thisonethankyou$$

Answered by mathdave last updated on 03/Oct/20

$$solution$$$$letconsiderthiseasyproperty$$$${\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-zt}dt=\frac{\mathrm{\Gamma }\left(s\right)}{Z^s}soletZ=a+ibtoget$$$${\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-at}.e^{ibt}dt=\frac{\mathrm{\Gamma }\left(s\right)}{{(a+ib)}^s}..........\left(1\right)$$$$letdconjugate\left(Z\right)=\stackrel{-}{Z}=a-ibtoget$$$${\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-at}.e^{ibt}dt=\frac{\mathrm{\Gamma }\left(s\right)}{{(a-ib)}^s}........\left(2\right)$$$$adding\left(1\right)and\left(2\right)resultstove$$$${\int }_0^{\mathrm{\infty }}{t}^{s-1}{e}^{-at}(e^{ibt}+e^{-ibt})dt=\mathrm{\Gamma }\left(s\right)(\frac{1}{{(a+ib)}^s}+\frac{1}{{(a-ib)}^s})$$$$but\cos z=\frac{e^{iz}+e^{-iz}}{2}then$$$$2{\int }_0^{\mathrm{\infty }}{t}^{s-1}\cos \left(bt\right)e^{-at}dt=\frac{{(a-ib)}^s+{(a+ib)}^s}{{(a^2+b^2)}^s}(from{z}^n={\left({re}^{i\theta }\right)}^n,r^n=\mid z{\mid }^n={(a^2+b^2)}^n,\theta =Arg\left(z\right)orArg\left(\stackrel{-}{z}\right),n=s)$$$$butnote{z}^n=\mid z{\mid }^n{e}^{n(Arg\left(z\right)+2k\pi )i},k=0,\mid z{\mid }^n={(a^2+b^2)}^{\frac{s}{2}}$$$$Arg\left(z\right)orArg\left(\stackrel{-}{z}\right)={\tan }^{-1}\left(\frac{b}{a}\right)or{\tan }^{-1}(-\frac{b}{a})$$$$\because I={\int }_0^{\mathrm{\infty }}{t}^{s-1}\cos \left(bt\right)e^{-at}dt=\frac{1}{2}\mathrm{\Gamma }\left(s\right)\bullet \frac{{(a^2+b^2)}^{\frac{s}{2}}(e^{s.Arg\left(\stackrel{-}{z}\right)}+e^{s.Arg\left(z\right)})}{{(a^2+b^2)}^s}$$$$I=\frac{\mathrm{\Gamma }\left(s\right)}{{(a^2+b^2)}^{\frac{s}{2}}}.\frac{e^{-s{\tan }^{-1}\left(\frac{b}{a}\right)i}+e^{s{\tan }^{-1}\left(\frac{b}{a}\right)i}}{2}$$$$I=\frac{\mathrm{\Gamma }\left(s\right)}{{(a^2+b^2)}^{\frac{s}{2}}}.\cos \left(s{\tan }^{-1}\left(\frac{b}{a}\right)\right)$$$$I={\int }_0^{\mathrm{\infty }}{t}^{s-1}\cos \left(bt\right)e^{-at}dt=\frac{\mathrm{\Gamma }\left(s\right)\cos \left(s{\tan }^{-1}\left(\frac{b}{a}\right)\right)}{{(a^2+b^2)}^{\frac{s}{2}}}$$$$putb=x,a=-\ln \left(\cos x\right)andmultiplybothsideby{\int }_0^{\frac{\pi }{2}}dxtove$$$${\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}{t}^{s-1}\cos \left(tx\right)e^{t\ln \left(\cos x\right)}dtdx=\mathrm{\Gamma }\left(s\right){\int }_0^{\frac{\pi }{2}}\frac{\cos \left(s{\tan }^{-1}(-\frac{x}{\ln \left(\cos x\right)})\right)}{{(x^2+{\ln }^2\left(\cos x\right))}^{\frac{s}{2}}}dx$$$${\int }_0^{\frac{\pi }{2}}\frac{\cos \left(s{\tan }^{-1}(-\frac{x}{\ln \left(\cos x\right)})\right)}{{(x^2+{\ln }^2\left(\cos x\right))}^{\frac{s}{2}}}dx=\frac{1}{\mathrm{\Gamma }\left(s\right)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}{\int }_0^{\frac{\pi }{2}}{\cos }^{t}x\cos \left(tx\right)dtdx$$$$letsolve{\int }_0^{\frac{\pi }{2}}{\cos }^{t}x\cos \left(tx\right)dtdxfirst$$$$notefromdformular$$$${\int }_0^{\frac{\pi }{2}}{\cos }^{a-1}x\cos \left(bx\right)dx=\frac{\pi }{2^{a}a}\bullet \frac{1}{\beta (\frac{a+b+1}{2},\frac{a-b+1}{2})}then$$$${\int }_0^{\frac{\pi }{2}}{\cos }^{t}x\cos \left(tx\right)dx=\frac{\pi }{2^{t+1}}$$$$\because I={\int }_0^{\frac{\pi }{2}}\frac{\cos \left(s{\tan }^{-1}(-\frac{x}{\ln \left(\cos x\right)})\right)}{{(x^2+{\ln }^2\left(\cos x\right))}^{\frac{s}{2}}}dx=\frac{1}{\mathrm{\Gamma }\left(s\right)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}.\frac{\pi }{2^{t+1}}dt$$$$I=\frac{\pi }{2}\bullet \frac{1}{\mathrm{\Gamma }\left(s\right)}{\int }_0^{\mathrm{\infty }}{t}^{s-1}.e^{-t\ln 2}dt=\frac{\pi }{2}\bullet \frac{1}{\mathrm{\Gamma }\left(s\right)}\bullet \frac{\mathrm{\Gamma }\left(s\right)}{{\ln }^s\left(2\right)}=\frac{\pi }{{2\ln }^s\left(2\right)}$$$${\int }_0^{\frac{\pi }{2}}\frac{\cos \left(s{\tan }^{-1}(-\frac{x}{\ln \left(\cos x\right)})\right)}{{(x^2+{\ln }^2\left(\cos x\right))}^{\frac{s}{2}}}dx=\frac{\pi }{{2\ln }^s\left(2\right)}$$$$usingDuiswithrespecttosofbothside$$$$\frac{\partial }{\partial s}{\mid }_{s=0}{\int }_0^{\frac{\pi }{2}}\frac{\cos \left(s{\tan }^{-1}(-\frac{x}{\ln \left(\cos x\right)})\right)}{{(x^2+{\ln }^2\left(\cos x\right))}^{\frac{s}{2}}}dx=\frac{\partial }{\partial s}{\mid }_{s=0}\frac{\pi }{{2\ln }^s\left(2\right)}$$$${\int }_0^{\frac{\pi }{2}}(-\frac{1}{2}\ln (x^2+{\ln }^2\left(\cos x\right)))dx=-\frac{\pi }{2}\ln \left(\ln 2\right)$$$${\int }_0^{\frac{\pi }{2}}\ln (x^2+{\ln }^2\left(\cos x\right))dx=\pi \ln \left(\ln 2\right)Q.E.D$$$$bymathdave\left(03/10/2020\right)$$

Commented by mnjuly1970 last updated on 03/Oct/20

$$okay,niceverynicemrdave(good$$$$manbutalittlebitsensitive)$$

Commented by mindispower last updated on 03/Oct/20

$$thankyouihavenotthisidea$$$$goodjobsirthanxagain$$

Commented by Tawa11 last updated on 06/Sep/21

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

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