Brilliant Integration Contest - Season 3 (Part 2)

Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 3(Part 1)

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • The scope of questions is only computation of integrals either definite or indefinite integrals.

  • You are NOT allowed to post a multiple integrals problem.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

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**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

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Note by Aditya Kumar
2 years, 8 months ago

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Problem 26 :

Show that 0ln(as+xsbs+xs)dx  =  π(ab)cosecπsa,b>0,s>1  . \int_0^\infty \ln\left(\frac{a^s + x^s}{b^s + x^s}\right)\,dx \; =\; \pi(a-b)\mathrm{cosec}\tfrac{\pi}{s} \hspace{1cm} a,b > 0\,,\,s > 1 \;.

This problem has been solved by Fdp Dpf.

Mark Hennings - 2 years, 8 months ago

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Solution problem 26:

Let , F(a)=0+ln(as+xsbs+xs)dx\displaystyle F(a)=\int_0^{+\infty} \ln\left(\dfrac{a^s+x^s}{b^s+x^s}\right)dx

Observe that, F(b)=0F(b)=0

F(a)=sas10+1as+xsdx\displaystyle F^\prime(a)=sa^{s-1}\int_0^{+\infty} \dfrac{1}{a^s+x^s}dx

Perform the change of variable,

y=xay=\dfrac{x}{a},

F(a)=s0+11+xsdx\displaystyle F^\prime(a)=s\int_0^{+\infty} \dfrac{1}{1+x^s}dx

It's well known that,

0+11+xsdx=πscosec(πs)\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)

Therefore,

F(a)=πcosec(πs)\displaystyle F^\prime(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)

Therefore, F(a)=πcosec(πs)a+kF(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)a+k ,

k a real constant.

Since F(b)=0F(b)=0 then k=πcosec(πs)bk=-\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)b

Therefore,

F(a)=π(ab)cosec(πs)\boxed{\displaystyle F(a)=\pi (a-b)\mathrm{cosec}\left(\dfrac{\pi}{s}\right)}

Proof of:

0+11+xsdx=πscosec(πs)\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)

Perform the change of variable

y=xsy=x^s

0+11+xsdx=1s0+x1s11+xdx\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{1}{s} \int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx

0+x1s11+xdx=β(1s,11s)=Γ(1s)Γ(11s)Γ(1)=πsin(πs)=πcosec(πs) \displaystyle \begin{aligned}\int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx&=\beta\left(\dfrac{1}{s},1-\dfrac{1}{s}\right)\\ &=\dfrac{\Gamma\left(\dfrac{1}{s}\right)\Gamma\left(1-\dfrac{1}{s}\right)}{\Gamma(1)}\\ &= \dfrac{\pi}{\sin\left(\dfrac{\pi}{s}\right)}\\ &=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right) \end{aligned}

β\beta being the Beta Euler function. Third line is the use of Euler's reflection formula.

FDP DPF - 2 years, 8 months ago

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Problem 25 :

Prove That

1π0πxarctan(psinx1+pcosx) dx=Li2(p) p<1 \dfrac{1}{\pi} \int_{0}^{\pi} x \arctan\left( \dfrac{p \sin x}{1+p \cos x} \right) \ \mathrm{d}x = \operatorname{Li}_{2}(p) \quad \forall \ |p| <1

Notation : Li2(z)\operatorname{Li}_{2}(z) denotes the Dilogarithm Function.

This problem has been solved by Mark Hennings.

Ishan Singh - 2 years, 8 months ago

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If we define g(p)  =  0πln(1+2pcosx+p2)dxp<1 g(p) \; = \; \int_0^\pi \ln(1 + 2p \cos x + p^2)\,dx \hspace{1cm} |p| < 1 then g(0)=0g(0) = 0 and, using the complex subsitution z=eixz = e^{ix}, g(p)=0π2cosx+2p1+2pcosx+p2dx  =  12ππ2cosx+2p1+2pcosx+p2dx=12z=1z+z1+2p1+p(z+z1)+p2dziz  =  12iz=1z2+2pz+1z(pz+1)(z+p)dz=π(Resz=0+Resz=p)z2+2pz+1z(pz+1)(z+p)=π(1p1p2p(1p2))  =  0\begin{aligned} g'(p) & = \int_0^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \; = \; \frac12\int_{-\pi}^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{2}\int_{|z|=1} \frac{z + z^{-1} + 2p}{1 + p(z + z^{-1}) + p^2} \frac{dz}{iz} \; = \; \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 2pz + 1}{z(pz+1)(z+p)}\,dz \\ & = \pi\left(\mathrm{Res}_{z=0} + \mathrm{Res}_{z=-p}\right)\frac{z^2 + 2pz + 1}{z(pz+1)(z+p)} \\ & = \pi\left(\frac{1}{p} - \frac{1-p^2}{p(1-p^2)}\right) \; = \; 0 \end{aligned} so that g(p)=0g(p) = 0 for all p<1|p| < 1. If we now define f(p)  =  1π0πxtan1(psinx1+pcosx)dxp<1 f(p) \; = \; \frac{1}{\pi}\int_0^\pi x \tan^{-1}\left(\frac{p \sin x}{1 + p \cos x}\right)\,dx \hspace{1cm} |p| < 1 then f(0)=0f(0) = 0 and f(p)=1π0πx1+(psinx1+pcosx)2×sinx(1+pcosx)2dx=1π0πxsinx1+2pcosx+p2dx=1π[x2pln(1+2pcosx+p2)]0π+12πp0πln(1+2pcosx+p2)dx=ln(1p)p+g(p)  =  ln(1p)p\begin{aligned} f'(p) & = \frac{1}{\pi}\int_0^\pi \frac{x}{1 + \left(\frac{p \sin x}{1 + p\cos x}\right)^2} \times \frac{\sin x}{(1 + p\cos x)^2}\,dx \\ & = \frac{1}{\pi}\int_0^\pi \frac{x \sin x}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{\pi}\Big[-\frac{x}{2p}\ln(1 + 2p\cos x + p^2)\Big]_0^\pi + \frac{1}{2\pi p}\int_0^\pi \ln(1 + 2p\cos x + p^2)\,dx \\ & = -\frac{\ln(1-p)}{p} + g(p) \; = \; -\frac{\ln(1-p)}{p} \end{aligned} and hence it follows that f(p)=Li2(p)f(p) \,=\, \mathrm{Li}_2(p).

Mark Hennings - 2 years, 8 months ago

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Aliter,

Let f(p)=1π0πxarctan(psinx1+pcosx) dxf(p) = \dfrac{1}{\pi}\int_{0}^{\pi} x \arctan\left( \dfrac{p\sin x}{1+ p\cos x} \right) \ \mathrm{d}x

Firstly, we have f(0)=0f(0) = 0

Note that,

k=1(p)k1sin(kx)=1p(k=0(peix)k) \displaystyle \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx) = \dfrac{1}{p}\Im \left( \sum_{k=0}^{\infty} (-p e^{ix})^{k} \right)

=1p(11+peix)\displaystyle = \dfrac{1}{p} \Im \left( \dfrac{1}{1+p e^{ix}} \right)

=sinxp2+2pcosx+1\displaystyle = \dfrac{\sin x}{p^2 +2p \cos x +1}

where (z)\Im(z) denotes the imaginary part of zz.

So we have,

f(p)=1π0πxsinxp2+2pcosx+1 dx \displaystyle f'(p) = \dfrac{1}{\pi} \int_{0}^{\pi} \dfrac{x \sin x}{p^2 +2p \cos x +1} \ \mathrm{d}x

=1π0πx(k=1(p)k1sin(kx)) dx \displaystyle = \dfrac{1}{\pi} \int_{0}^{\pi} x \left(\sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)\right) \ \mathrm{d}x

=1πk=1(p)k10πxsin(kx) dx \displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \int_{0}^{\pi}x \sin (kx) \ \mathrm{d}x

=1πk=1(p)k1(sin(kπ)kπcos(kπ)k2) \displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \left( \dfrac{\sin(k \pi) - k \pi \cos(k \pi)}{k^2} \right)

Since sin(kπ)=0\sin (k \pi) = 0 and cos(kπ)=(1)k  kZ\cos (k \pi) = (-1)^k \ \forall \ k \in \mathbb{Z}, we get,

f(p)=k=1pk1k=ln(1p)p \displaystyle f'(p) = \sum_{k=1}^{\infty} \dfrac{p^{k-1}}{k} = -\dfrac{\ln(1-p)}{p}

f(p)=Li2(p)  \displaystyle \therefore f(p) = \operatorname{Li}_{2}(p) \ \square

Ishan Singh - 2 years, 8 months ago

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Problem 27:

Show that

062162+1lnx(x1)x22(15+83)x+1dx=23(23)G\displaystyle\int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}\, dx=\dfrac{2}{3}(2-\sqrt{3})G

This problem has been solved by Mark Hennings.

FDP DPF - 2 years, 8 months ago

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Put α=112π\alpha = \tfrac{1}{12}\pi. With the substitution x=1cosθ1+cosθx = \frac{1-\cos\theta}{1+\cos\theta}, we obtain after much manipulation that I  =  062162+1lnxdx(x1)x22(15+83)x+1)  =  12sinα0αln(1cosθ1+cosθ)sinθdθcosθcos2θcos2α I \; = \; \int_0^{\frac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \frac{\ln x\,dx}{(x-1)\sqrt{x^2 - 2(15+8\sqrt{3})x + 1)}} \; = \; -\tfrac12\sin\alpha \int_0^\alpha \frac{\ln\left(\frac{1-\cos\theta}{1+\cos\theta}\right) \sin\theta\,d\theta}{\cos\theta\sqrt{\cos^2\theta - \cos^2\alpha}} Putting u=cosθu = \cos\theta now gives I  =  12sinαcosα1ln(1u1+u)duuu2cos2α I \; = \; -\tfrac12\sin\alpha \int_{\cos\alpha}^1 \frac{\ln\left(\frac{1-u}{1+u}\right)\,du}{u\sqrt{u^2 - \cos^2\alpha}} Now putting u=cosαsecϕu = \cos\alpha \sec\phi yields I=12tanα0αln(cosϕcosαcosϕ+cosα)dϕ  =  12tanα0αln(tan(α+ϕ2)tan(αϕ2))dϕ=12tanα0α{ln(tan(α+ϕ2))+ln(tan(αϕ2))}dϕ=tanα{12ααln(tanϕ)dϕ12α0ln(tanϕ)dϕ}=tanα0αln(tanϕ)dϕ  =  tanα0112πln(tanϕ)dϕ=23tanαG  =  23(23)G\begin{aligned} I & = -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\frac{\cos\phi - \cos\alpha}{\cos\phi + \cos\alpha}\right)\,d\phi \; = \; -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big) \tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\,d\phi \\ & = -\tfrac12\tan\alpha \int_0^\alpha \left\{ \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big)\right) + \ln\left(\tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\right\}\,d\phi \\ & = -\tan\alpha \left\{ \int_{\frac12\alpha}^\alpha \ln(\tan \phi)\,d\phi - \int_{\frac12\alpha}^0 \ln(\tan\phi)\,d\phi\right\} \\ & = -\tan\alpha \int_0^\alpha \ln(\tan\phi)\,d\phi \; = \; -\tan\alpha \int_0^{\frac{1}{12}\pi}\ln(\tan\phi)\,d\phi \\ & = \tfrac23\tan\alpha\, G \; = \; \tfrac23(2 - \sqrt{3})G \end{aligned} using a standard integral representation of GG at the very last stage.

Mark Hennings - 2 years, 8 months ago

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Problem 29:

Show that 0etxx(x2+1)dx=π[2C(2tπ)sint2S(2tπ)costsin(tπ/4)]\displaystyle \int\limits_{0}^{\infty} \frac{e^{-tx}}{\sqrt x (x^2+1)}dx=\pi \left[ \sqrt2 C\left( \sqrt{\frac{2t}{\pi}}\right)\sin t - \sqrt2S\left( \sqrt{\frac{2t}{\pi}} \right) \cos t - \sin(t-\pi/4) \right]

where CC and SS are Fresnel cosine integral and Fresnel sine integral, respectively.

This problem has been solved by Mark Hennings.

Aman Rajput - 2 years, 8 months ago

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In this question, we are using the Fresnel integrals C(x)  =  0xcos(12πt2)dtS(x)  =  0xsin(12πt2)dt C(x) \; = \; \int_0^x \cos\big(\tfrac12\pi t^2\big)\,dt \hspace{2cm} S(x) \; = \; \int_0^x \sin\big(\tfrac12\pi t^2\big)\,dt If we define A(t)  =  cost0etxxx2+1dxsint0etxx(x2+1)dx A(t) \; = \; \cos t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx for t>0t > 0 then A(t)=[t]lsint0etxxx2+1dxcost0etxxxx2+1dxcost0etxx(x2+1)dx+sint0etxxx2+1dx=cost0etxxdx  =  πtcost\begin{aligned} A'(t) & = \begin{array}{c}[t]{l}\displaystyle-\sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \cos t \int_0^\infty \frac{e^{-tx}x\sqrt{x}}{x^2+1}\,dx \\\displaystyle - \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx + \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx\end{array} \\ & = -\cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; = \; -\sqrt{\tfrac{\pi}{t}}\cos t \end{aligned} for any t>0t > 0, so that ddt[A(t)+π2C(2tπ)]  =  0 \frac{d}{dt}\Big[ A(t) + \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\Big] \; = \; 0 Since A(t)0A(t) \,\to\, 0 as tt \to \infty, we deduce that A(t)+π2C(2tπ)  =  12πt>0 A(t) + \pi\sqrt{2}C\Big(\sqrt{\frac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0 Similarly, if we define B(t)  =  sint0etxxx2+1dx+cost0etxx(x2+1)dx B(t) \; = \; \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx + \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx for t>0t > 0, then we can show that B(t)  =  sint0etxxdx  =  πtsint B'(t) \; =\; -\sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; =\; - \sqrt{\tfrac{\pi}{t}}\sin t and hence B(t)+π2S(2tπ)  =  12πt>0 B(t) + \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0 Thus we deduce that 0etxx(x2+1)dx=B(t)costA(t)sint=cost[12ππ2S(2tπ)]sint[12ππ2C(2tπ)]=π2C(2tπ)π2S(2tπ)πsin(t14π)\begin{aligned} \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx & = B(t)\cos t - A(t) \sin t \\ & = \cos t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] - \sin t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] \\ & = \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sin\big(t - \tfrac14\pi\big) \end{aligned} as required.

Mark Hennings - 2 years, 7 months ago

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Problem 30:

Show that 01ln(1+14x2+12x)dxx  =  120π2 \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\,\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2

This problem has been solved by Fdp Dpf.

Mark Hennings - 2 years, 7 months ago

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I=01ln(1+x24+x2)xdx\displaystyle I=\int_0^1 \dfrac{\ln(\sqrt{1+\dfrac{x^2}{4}}+\dfrac{x}{2})}{x}dx

Perform the change of variable y=x2y=\dfrac{x}{2},

I=012ln(1+x2+x)xdx\displaystyle I=\int_0^{\tfrac{1}{2}} \dfrac{\ln\left(\sqrt{1+x^2}+x\right)}{x}dx

Perform the change of variable y=arsinh(x)y=\mathrm{arsinh}(x),

I=0arsinh(12)xcosh(x)sinh(x)dx=0arsinh(12)x(1+e2x)1e2xdx=0arsinh(12)(x(1+e2x)n=0+e2nx)dx=(arsinh(12))22+20arsinh(12)(xn=1+e2nx)dx=(arsinh(12))22+2n=1+(0arsinh(12)xe2nx)dx=(arsinh(12))22+12n=1+(1n2(1+2arsinh(12)n)e2arsinh(12)nn2)=(arsinh(12))22+12ζ(2)12Li2(e2asinh(12))+arsinh(12)ln(1e2arsinh(12))\begin{aligned}\displaystyle I&=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\cosh(x)}{\sinh(x)}dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\left(1+\mathrm{e}^{-2x}\right)}{1-\mathrm{e}^{-2x}} dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\left(1+\mathrm{e}^{-2x}\right)\sum_{n=0}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\sum_{n=1}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\sum_{n=1}^{+\infty} \left(\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} x\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\sum_{n=1}^{+\infty}\left(\dfrac{1}{n^2}-\dfrac{\left(1+2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n\right)\mathrm{e}^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n}}{n^2}\right)\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\zeta(2)-\dfrac{1}{2}\mathrm{Li}_2\left(e^{-2\mathrm{asinh}\left(\tfrac{1}{2}\right)}\right)+\mathrm{arsinh}\left(\tfrac{1}{2}\right)\ln\left(1-e^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)}\right) \end{aligned}

Since, arsinh(12)=ln(12+52)\mathrm{arsinh}\left(\dfrac{1}{2}\right)=\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right) then,

I=π21212Li2(3252)12(ln(12+52))2I=\dfrac{\pi^2}{12}-\dfrac{1}{2}\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)-\dfrac{1}{2}\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2

Since , Li2(3252)=π215(ln(12+52))2\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=\dfrac{\pi^2}{15}-\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2 then,

I=π212π230=π220I=\dfrac{\pi^2}{12}-\dfrac{\pi^2}{30}=\boxed{\dfrac{\pi^2}{20}}

FDP DPF - 2 years, 7 months ago

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Nice. A bit of integration by parts makes the earlier stage a little clearer, perhaps...

Note that ddxln(1+14x2+12x)  =  12x121+14x2+121+14x2+12x  =  14+x2 \frac{d}{dx}\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right) \; = \; \frac{\frac{\frac12x}{\frac12\sqrt{1+\frac14x^2}} + \frac12}{\sqrt{1 + \frac14x^2} + \frac12x} \; = \; \frac{1}{\sqrt{4+x^2}} and so, integrating by parts, 01ln(1+14x2+12x)dxx=[(lnx)ln(1+14x2+12x)]0101lnx4+x2dx=01lnx4+x2dx\begin{aligned} \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} & = \Big[(\ln x)\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\Big]_0^1 - \int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \\ & = -\int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \end{aligned} Since sinh112=ln(12(5+1))\sinh^{-1}\tfrac12 = \ln\big(\tfrac12(\sqrt{5}+1)\big), the substitution x=2sinhux = 2\sinh u yields 01lnx4+x2dx=0sinh112ln(2sinhu)du  =  0sinh112[u+ln(1e2u)]du=12ln2(5+12)+0ln(5+12)ln(1e2u)e2ue2udu=12ln2(5+12)121352ln(1v)vdv\begin{aligned} \int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx & = \int_0^{\sinh^{-1}\frac12} \ln\big(2\sinh u\big)\,du \; = \; \int_0^{\sinh^{-1}\frac12}\Big[u + \ln\big(1 - e^{-2u}\big)\Big]\,du \\ & = \tfrac12 \ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \int_0^{\ln\big(\frac{\sqrt{5}+1}{2}\big)} \frac{\ln\big(1 - e^{-2u}\big)}{e^{-2u}}\,e^{-2u}\,du \\ & = \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) - \frac12\int_1^{\frac{3-\sqrt{5}}{2}} \frac{\ln(1-v)}{v}\,dv \end{aligned} using the substitution v=e2uv = e^{-2u}. Thus 01lnx4+x2dx  =  12ln2(5+12)+12Li2(352)12Li2(1)  =  130π2112π2  =  120π2 \int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx \; = \; \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \tfrac12\mathrm{Li}_2\left(\tfrac{3 - \sqrt{5}}{2}\right) - \tfrac12\mathrm{Li}_2(1) \; = \; \tfrac{1}{30}\pi^2 - \tfrac{1}{12}\pi^2 \; = \; -\tfrac{1}{20}\pi^2 and so 01ln(1+14x2+12x)dxx  =  120π2 \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2

Mark Hennings - 2 years, 7 months ago

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@Mark Hennings You're right, it's easiest this way. By the way, ln(1+14x2+12x)dx=xln(1+14x2+12x)x2+4\displaystyle \int \ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)dx=x\ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)-\sqrt{x^2+4}

FDP DPF - 2 years, 7 months ago

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Problem 31:

Show that

01x1(x+1)lnxdx=ln(π2)\displaystyle \int_0^1 \dfrac{x-1}{(x+1)\ln x}dx=\ln\left(\dfrac{\pi}{2}\right)

This problem has been solved by Ishan Singh.

FDP DPF - 2 years, 7 months ago

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Let

f(p)=01xp1(x+1)lnx dx;p>0f(p) = \int_{0}^{1} \dfrac{x^p - 1}{(x+1) \ln x} \ \mathrm{d}x \quad ; \quad p > 0

    f(p)=01xpx+1dx \displaystyle \implies f'(p) = \int_{0}^{1} \dfrac{x^p}{x+1} \mathrm{d}x

=01xp(x1)x21dx\displaystyle = \int_{0}^{1} \dfrac{x^p(x-1)}{x^2 -1} \mathrm{d}x

Substitute x2xx^2 \mapsto x and simplify to get,

f(p)=1201(xp21x1xp121x1)dx \displaystyle f'(p) = \dfrac{1}{2} \int_{0}^{1} \left( \dfrac{x^{\frac{p}{2}} - 1}{x - 1} - \dfrac{x^{\frac{p-1}{2}} - 1}{x - 1} \right) \mathrm{d}x

=12(ψ(p2+1)ψ(p12+1))\displaystyle = \dfrac{1}{2} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right)

Note that f(0)=0f(0) = 0

    f(1)=01f(p) dp\displaystyle \implies f(1) = \int_{0}^{1} f'(p) \ \mathrm{d}p

=1201(ψ(p2+1)ψ(p12+1))dp\displaystyle = \dfrac{1}{2} \int_{0}^{1} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right) \mathrm{d}p

=log(π2)\displaystyle = \log \left(\dfrac{\pi}{2}\right)

where the last equality follows since ψ(z)=ddzlog(Γ(z)) \psi (z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \log(\Gamma(z))

Ishan Singh - 2 years, 7 months ago

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Problem 37:

Evaluate 012πln(cosx+sinx)dx\int_0^{\frac12\pi}\, \ln(\cos x + \sin x)\,dx

This problem has been solved by Fdp Dpf.

Mark Hennings - 2 years, 7 months ago

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Solution to problem 37.

Since,

sin(x+π4)=cos(π4)sinx+sin(π4)cosx=22(sinx+cosx)\begin{aligned}\sin\left(x+\dfrac{\pi}{4}\right)&=\cos\left(\dfrac{\pi}{4}\right)\sin x+\sin \left(\dfrac{\pi}{4}\right)\cos x\\ &=\dfrac{\sqrt{2}}{2}(\sin x+\cos x)\end{aligned}

then,

0π2ln(cosx+sinx)dx=0π2ln(2sin(x+π4))dx=0π4ln(2sin(x+π4))dx+π4π2ln(2sin(x+π4))dx=0π4ln(2sin((π4x)+π4))dx+0π4ln(2sin((x+π4)+π4))dx=20π4ln(2cosx)dx=πln24+20π4ln(cosx)dx=πln240π4ln(1(cosx)2)dx=πln240π4ln(1+(tanx)2)dx\begin{aligned} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\int_0^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(\dfrac{\pi}{4}-x\right)+\dfrac{\pi}{4}\right)\right)dx+\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(x+\dfrac{\pi}{4}\right)+\dfrac{\pi}{4}\right)\right)dx\\ &=2\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}+2\int_0^{\tfrac{\pi}{4}} \ln\left(\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(\dfrac{1}{(\cos x)^2}\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(1+(\tan x)^2\right)dx\\ \end{aligned}

In the latter integral perform the change of variable y=tanxy=\tan x, therefore, 0π2ln(cosx+sinx)dx=πln2401ln(1+x2)1+x2dx\begin{aligned} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx \end{aligned}

0+ln(1+x2)1+x2dx=01ln(1+x2)1+x2dx+0+ln(1+x2)1+x2dx\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx

In the latter integral perform the change of variable y=1xy=\dfrac{1}{x}, therefore,

0+ln(1+x2)1+x2dx=01ln(1+x2)1+x2dx+01ln(1+1x2)1+x2dx=201ln(1+x2)1+x2dx201lnx1+x2dx=201ln(1+x2)1+x2dx+2G\begin{aligned}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx&=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^1 \dfrac{\ln\left(1+\dfrac{1}{x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx-2\int_0^1 \dfrac{\ln x}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+2\text{G}\\ \end{aligned}

G\text{G} being the Catalan constant. But, 0+ln(1+x2)1+x2dx=πln2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\pi\ln 2

(see my answer to problem 36)

Therefore, 01ln(1+x2)1+x2dx=πln22G\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{2}-\text{G}

Therefore,

0π2ln(cosx+sinx)dx=πln24(πln22G)=Gπln24\begin{aligned} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\left(\dfrac{\pi\ln 2}{2}-\text{G}\right)\\ &=\boxed{\text{G}-\dfrac{\pi\ln 2}{4}} \end{aligned}

FDP DPF - 2 years, 7 months ago

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Aliter,

The integral 0π4log(cosx) dx \displaystyle \int_{0}^{\frac{\pi}{4}} \log(\cos x) \ \mathrm{d}x can also be done using the identity log(cosx)=log2k=1cos(2kx)k \log(\cos x) = -\log 2 - \sum_{k=1}^{\infty} \dfrac{\cos (2kx)}{k}.