The Log-Trig Integral

The following is the reposted version of Kartik Sharma's problem on logarithmo-trigonometric integral:

IfRn+=2π0π2x2+(lncosx)22n12+1212+..+1212+12(lncosx)2x2+(lncosx)2dx{ R }_{ n }^{ + }=\frac { 2 }{ \pi } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt [ { 2 }^{ n } ]{ { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } } \sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +..+\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { { \left( \ln { \cos { x } } \right) }^{ 2 } }{ { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } } } } } } } dxthen evaluate Rn+{ R }_{ n }^{ + }

Here is the solution to the problem:

In this case..first we observe that the nested radicals are of the form:12+a2\sqrt { \frac { 1 }{ 2 } +\frac { a }{ 2 } } which suggests that aa can be considered as cosθ\cos { \theta } for some θ\theta ..Thus the nested radical part of the integral reduces to 12+1212+1212+...+1212+cosθ2=cos(θ2n)\sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +...+\frac { 1 }{ 2 } \sqrt { \frac { 1 }{ 2 } +\frac { \cos { \theta } }{ 2 } } } } } =\cos { \left( \frac { \theta }{ { 2 }^{ n } } \right) } where cosθ=(lncosx)2x2+(lncosx)2sec2θ=x2+(lncosx)2(lncosx)2=1+x2(lncosx)2=1+tan2θtanθ=±xlncosx\cos { \theta } =\sqrt { \frac { { \left( \ln { \cos { x } } \right) }^{ 2 } }{ { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } } } \\ \Rightarrow \sec ^{ 2 }{ \theta } =\frac { { x }^{ 2 }+{ \left( \ln { \cos { x } } \right) }^{ 2 } }{ { \left( \ln { \cos { x } } \right) }^{ 2 } } =1+\frac { { x }^{ 2 } }{ { \left( \ln { \cos { x } } \right) }^{ 2 } } =1+\tan ^{ 2 }{ \theta } \\ \Rightarrow \tan { \theta } =\pm \frac { x }{ \ln { \cos { x } } } We choose θ\theta to be such that tanθ=xlncosx\tan { \theta } =- \frac { x }{ \ln { \cos { x } } }. Now, let r(0,1)r\in \left( 0,1 \right) . We use the two following results:0+yr1cos(ay)ebydy=Γ(r)cos{rtan1(ab)}(a2+b2)r2\int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\cos { \left( ay \right) } { e }^{ -by } } dy=\Gamma \left( r \right) \frac { \cos { \left\{ r\tan ^{ -1 }{ \left( \frac { a }{ b } \right) } \right\} } }{ { \left( { a }^{ 2 }+{ b }^{ 2 } \right) }^{ \frac { r }{ 2 } } } and 0π2cos(yx)cosyxdx=π2y+1\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( yx \right) } \cos ^{ y }{ x } } dx=\frac { \pi }{ { 2 }^{ y+1 } } \\ which are trivial.

Then we can write 0π2cos{rtan1(xlncosx)}{x2+(lnx)2}r2dx=1Γ(r)0π2{0+yr1cos(xy)eylncosxdy}dx=1Γ(r)0π20+yr1cos(xy)eylncosxdydx=1Γ(r)0+yr1{0π2cos(xy)eylncosxdx}dy=1Γ(r)0+yr1{0π2cos(xy)cosyxdx}dy=1Γ(r)0+yr1π2y+1dy=π2Γ(r)0+yr12ydy=π2Γ(r)0+yr1eyln2dy=π2Γ(r).Γ(r)(ln2)r=π2(ln2)r\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos { \left\{ r\tan ^{ -1 }{ \left( -\frac { x }{ \ln { \cos { x } } } \right) } \right\} } }{ { \left\{ { x }^{ 2 }+{ \left( \ln { x } \right) }^{ 2 } \right\} }^{ \frac { r }{ 2 } } } } dx=\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left\{ \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\cos { \left( xy \right) } { e }^{ y\ln { \cos { x } } } } dy \right\} } dx\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\cos { \left( xy \right) } { e }^{ y\ln { \cos { x } } } } dy } dx\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\left\{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( xy \right) } { e }^{ y\ln { \cos { x } } } } dx \right\} } dy\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\left\{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \cos { \left( xy \right) } \cos ^{ y }{ x } } dx \right\} } dy\\ =\frac { 1 }{ \Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }\frac { \pi }{ { 2 }^{ y+1 } } } dy\\ =\frac { \pi }{ 2\Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }{ 2 }^{ -y } } dy\\ =\frac { \pi }{ 2\Gamma \left( r \right) } \int _{ 0 }^{ +\infty }{ { y }^{ r-1 }{ e }^{ -y\ln { 2 } } } dy\\ =\frac { \pi }{ 2\Gamma \left( r \right) } .\frac { \Gamma \left( r \right) }{ { \left( \ln { 2 } \right) }^{ r } } =\frac { \pi }{ 2{ \left( \ln { 2 } \right) }^{ r } } \\ (clearly in steps 2 and 3 we have used Fubini's theorem and in step 2 I have used gamma function since r>0r>0.)

We can apply analytic continuation to the above result to extend the domain of rr from (0,1)(0,1) to (1,1)(-1,1) since the original domain (0,1)(0,1) is open in RR and the integrand function is analytic in rr. Now we simply put r=12nr=-\frac { 1 }{ { 2 }^{ n } } to get that Rn+=2π.π2(ln2)12n=(ln2)2n{ R }_{ n }^{ + }=\frac { 2 }{ \pi } .\frac { \pi }{ 2{ \left( \ln { 2 } \right) }^{ -\frac { 1 }{ { 2 }^{ n } } } } ={ \left( \ln { 2 } \right) }^{ { 2 }^{ -n } }.

Note by Kuldeep Guha Mazumder
3 years, 8 months ago

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@Surya Prakash

Pi Han Goh - 3 years, 8 months ago

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