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

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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

**[Post your solution here]**

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

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":

Note by Aditya Kumar
2 years, 7 months ago

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Ohh!! I missed this season. I even didn't know that this happened. :(

Surya Prakash - 2 years, 3 months ago

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Maybe this time you can host it :)

Ishan Singh - 2 years, 3 months ago

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When is starting the season 4? :)

FDP DPF - 2 years, 2 months ago

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@Fdp Dpf I think in December 2017.

Ishan Singh - 2 years, 2 months ago

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Problem 50 :

Evaluate 0sin(2nx)cos(2arctan(ax))(1+a2x2)sinx dx \int_{0}^{\infty} \dfrac{\sin(2nx) \cos(2\arctan(ax))}{(1+a^2x^2) \sin x} \ \mathrm{d}x

where a>0a>0 , nZ+n \in \mathbb{Z^+}

Ishan Singh - 2 years, 6 months ago

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Let Cn1(a)C_{n-1}(a) be the previous integral. I suspect one can find a closed form for the ordinary generating function for the sequence (Cn(a))\left(C_n(a)\right) because U2n+1(cosx)=sin((2n+2)xsinxU_{2n+1}(\cos x)=\dfrac{\sin((2n+2)x}{\sin x}, UjU_j being the jj-nth Chebyshev polynomial of the second kind.

FDP DPF - 2 years, 6 months ago

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I looked hard on the web and could not find a closed expression for the Laplace transform of the UnU_n, which would have been really useful!

Mark Hennings - 2 years, 6 months ago

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Suppose that a>0a > 0 and nNn \in \mathbb{N}. Note that I  =  0sin(2nx)cos(2tan1(ax))(1+a2x2)sinxdx  =  01a2x2(1+a2x2)2sin2nxsinxdx  =  limδ0+Iδ I \; = \; \int_0^\infty \frac{\sin(2nx)\cos\big(2\tan^{-1}(ax)\big)}{(1 + a^2x^2)\sin x}\,dx \; = \; \int_0^\infty \frac{1-a^2x^2}{(1 + a^2x^2)^2}\frac{\sin 2nx}{\sin x}\,dx \; = \; \lim_{\delta \to 0+}I_\delta where Iδ  =  01a2(x+δ)2(1+a2(x+δ)2)2sin2nxsinxdx I_\delta \; = \; \int_0^\infty \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\,dx where we can use the Dominated Convergence Theorem to justify the limit, since 1a2(x+δ)2(1+a2(x+δ)2)2sin2nxsinx    11+a2(x+δ)2sin2nxsinx    Nn1+a2x2 \left| \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \left| \frac{1}{1 + a^2(x+\delta)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \frac{N_n}{1 + a^2x^2} for all x0x \ge 0 and δ>0\delta > 0, where sin2nxsinxNn\big|\tfrac{\sin 2nx}{\sin x}\big| \le N_n for all x0x \ge 0. It is trivial that 0tcos(ta)extdt  =  a2a2x21(1+a2x2)2a,x>0 \int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-xt}\,dt \; = \; a^2 \frac{a^2x^2 - 1}{(1 + a^2x^2)^2} \hspace{2cm} a,x > 0 and so Iδ=a20(0tcos(ta)e(x+δ)tdt)sin2nxsinxdx=a20tcos(ta)eδt(0extsin2nxsinxdx)dt\begin{aligned} I_\delta & = -a^{-2}\int_0^\infty\left(\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-(x+\delta)t}\,dt\right) \frac{\sin 2nx}{\sin x}\,dx \\ & = -a^{-2}\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx\right)\,dt \end{aligned} Since 0extsin(2m+2)xsinxdx0extsin2mxsinxdx  =  20extcos(2m+1)xdx  =  2tt2+(2m+1)2 \int_0^\infty e^{-xt} \frac{\sin(2m+2)x}{\sin x}\,dx - \int_0^\infty e^{-xt} \frac{\sin 2mx}{\sin x}\,dx \; = \; 2\int_0^\infty e^{-xt} \cos(2m+1)x\,dx \; = \; \frac{2t}{t^2 + (2m+1)^2} a simple induction shows that 0extsin2nxsinxdx  =  2tk=0n11t2+(2k+1)2t>0 \int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx \; = \; 2t\sum_{k=0}^{n-1} \frac{1}{t^2 + (2k+1)^2} \hspace{2cm} t > 0 and hence Iδ=2a20cos(ta)eδt(k=0n1t2t2+(2k+1)2)dt=2a20cos(ta)eδt(nk=0n1(2k+1)2t2+(2k+1)2)dt=2a2[nδδ2+a2k=0n1(2k+1)20cos(ta)eδtt2+(2k+1)2dt]\begin{aligned} I_\delta & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\sum_{k=0}^{n-1} \frac{t^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left( n - \sum_{k=0}^{n-1} \frac{(2k+1)^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2}\left[ \frac{n\delta}{\delta^2 + a^{-2}} - \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big) e^{-\delta t}}{t^2 + (2k+1)^2}\,dt \right] \end{aligned} and hence, letting δ0+\delta \to 0+, I=2a2k=0n1(2k+1)20cos(ta)t2+(2k+1)2dt=2a2k=0n1(2k+1)2×π2(2k+1)e2k+1a=πa2k=0n1(2k+1)e2k+1a\begin{aligned} I & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big)}{t^2 + (2k+1)^2}\,dt \\ & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \times \frac{\pi}{2(2k+1)}e^{-\frac{2k+1}{a}} \\ & = \frac{\pi}{a^2}\sum_{k=0}^{n-1} (2k+1)e^{-\frac{2k+1}{a}} \end{aligned} using a standard Fourier transform to finish off. This sum can be calculated, of course, but the end result is not particularly pretty, do I shall leave things here.

Mark Hennings - 2 years, 6 months ago

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Nice! I used the expansion sin(2nx)2sinx=k=1ncos((2k1)x) \dfrac{\sin(2nx)}{2\sin x} = \sum_{k=1}^{n} \cos((2k-1)x)

The result can also be generalized to 0sin(2nx)cos(parctan(ax))(1+a2x2)p2sinx dx ; nZ,a,p>0 \int_{0}^{\infty} \dfrac{\sin(2nx) \cos(p\arctan(ax))}{(1+a^2x^2)^{\frac{p}{2}} \sin x} \ \mathrm{d}x \ ; \ n \in \mathbb{Z}, a,p>0

Ishan Singh - 2 years, 6 months ago

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Problem 49:

Evaluate,

0π6xsin(2x)4sin4x2sin2x+1dx\int_{0}^{\frac{\pi }{6}}\frac{x \mathrm{sin}\left( 2 x\right) }{4 {{\mathrm{sin}^{4}x }}-2 {{\mathrm{sin^{2}}x }}+1}dx

FDP DPF - 2 years, 6 months ago

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J=0π6xsin(2x)12sin2x+4sin4xdx=0π62xsinxcosx(1sin2x)2+3sin4xdx=0π62xsinxcosxcos4+3sin4xdx=0π62xtanxcos2x(1+3tan4x)dx\begin{aligned}J&=\int_0^{\tfrac{\pi}{6}} \dfrac{x\sin(2x)}{1-2\sin^2 x+4\sin^4 x}dx\\ &= \int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{(1-\sin^2 x)^2+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{\cos^4+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\tan x}{\cos^2 x(1+3\tan^4 x)}dx\\ \end{aligned}

Perform the change of variable y=3tanxy=\sqrt{3}\tan x,

J=012xarctan(x3)3+x4dxJ=\int_0^1 \dfrac{2x\arctan\left(\tfrac{x}{\sqrt{3}}\right)}{3+x^4}dx

Since,

arctanu=01u1+t2u2dt\arctan u=\int_0^1 \dfrac{u}{1+t^2u^2}dt

J=230101x2(3+x4)(3+t2x2)dtdx=230101(t2+x2(t4+3)(x4+3)t2(t4+3)(t2x2+3))dtdx=430101t2(3+t4)(3+x4)dtdxJ\begin{aligned}J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{x^2}{(3+x^4)(3+t^2x^2)}dtdx\\ &=2\sqrt{3}\int_0^1 \int_0^1 \left(\frac{{{t}^{2}}+{{x}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{x}^{4}}+3\right) }-\frac{{{t}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{t}^{2}}{{x}^{2}}+3\right) }\right)dtdx\\ &=4\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx-J\\ \end{aligned}

therefore,

J=230101t2(3+t4)(3+x4)dtdx=23(01x23+x4dx)(0113+x4dx)\begin{aligned} J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx\\ &=2\sqrt{3}\left(\int_0^1 \dfrac{x^2}{3+x^4}dx \right)\left(\int_0^1 \dfrac{1}{3+x^4}dx\right)\\ \end{aligned}

Since,

0113+x4dx=[arctan(2314x3x2)232314log(32314x+x2x2+2314x+3)252314]01=1232334arctan(22314+22334)1252334log(1+323141+3+2314)\begin{aligned} \int_0^1 \dfrac{1}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}{{3}^{\frac{1}{4}}}}-\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot{{3}^{\frac{1}{4}}} x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}{{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(\dfrac{1+\sqrt{3}-\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{1+\sqrt{3}+\sqrt{2}\cdot 3^{\tfrac{1}{4}}}\right) \end{aligned}

1+3+33141+\sqrt{3}+\sqrt{3}\cdot 3^{\tfrac{1}{4}} is a root of the polynomial X44X316X+16X^4-4X^3-16X+16,

Therefore,

0113+x4dx=1232334arctan(231431)1252334log(1+32231422334)\int_0^1 \dfrac{1}{3+x^4}dx=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{\sqrt{3}-1}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right)

and,

01x23+x4dx=[arctan(2314x3x2)232314+log(32314x+x2x2+2314x+3)252314]01=1232314arctan(22314+22334)+1252314log(1+32231422334)\begin{aligned} \int_0^1 \dfrac{x^2}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{1}{4}}}}+\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}\cdot {{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{1}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)+\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{1}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) \end{aligned}

Therefore,

J=143arctan2(22314+22334)1163log2(1+32231422334)\boxed{J=\dfrac{1}{4\sqrt{3}}\arctan^2\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{16\sqrt{3}}\log^2\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) }

PS:

F(x,t)=x(a+x2)(a+tx)=ta+t21a+x2+1a+t2xa+x2t(a+t2)(a+tx)F(x,t)=\dfrac{x}{(a+x^2)(a+tx)}=\dfrac{t}{a+t^2}\cdot \dfrac{1}{a+x^2} +\dfrac{1}{a+t^2}\cdot \dfrac{x}{a+x^2}-\dfrac{t}{(a+t^2)(a+tx)}

is a nice function.

FDP DPF - 2 years, 6 months ago

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What was your method?

Ishan Singh - 2 years, 6 months ago

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I am interested. My solution used/generalised a lot of the tricks that FDP uses in his solutions, so it will be interesting if he has a different approach.

Mark Hennings - 2 years, 6 months ago

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Aliter,

Let.

I=0π6xsin(2x)4sin4x2sin2x+1 dx\text{I} = \int_{0}^{\frac{\pi}{6}} \dfrac{x \sin(2x)}{4\sin^4 x - 2\sin^2 x + 1} \ \mathrm{d}x

=14(αβ)0π6xsin(2x)(1cos(2x)1+β1cos(2x)1+α) = \dfrac{1}{4(\alpha - \beta)} \int_{0}^{\frac{\pi}{6}} x \sin (2x) \left(\dfrac{1}{\cos(2x) - 1 + \beta} - \dfrac{1}{\cos(2x) - 1 + \alpha}\right)

where α\alpha and β\beta are the roots of the equation t212t+14=0 t^2 - \dfrac{1}{2} t + \dfrac{1}{4} = 0

Now, from Problem 25, we have,

sinxp2+2pcosx+1=k=1(p)k1sin(kx) \dfrac{\sin x}{p^2 +2p \cos x +1} = \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)

Using this and interchanging, sum and integral ( and after a lot of straight forward simplification/calculations involving dilogarithm and logarithm) we get,

I=143[tan1(2314314)]21163ln2(314+2+3143142+314)\text{I} = \dfrac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right)

Ishan Singh - 2 years, 6 months ago

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Just pipped you! You might want to clarify what α\alpha and β\beta are, though.

Why don't you post the last integral!

Mark Hennings - 2 years, 6 months ago

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@Mark Hennings Thanks for letting me do the honors. I'll post it tomorrow, I'll need some time to create a nice one :)

Ishan Singh - 2 years, 6 months ago

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For any a>0a > 0 define A(a)=01dxa4+x4  =  1a30a1dz1+z4=12a32[tan1(1+2a1)tan1(12a1)]+14a32ln(a+2+a1a2+a1)B(a)=01x2dxa4+x4  =  1a0a1z2dz1+z4=12a2[tan1(1+2a1)tan1(12a1)]14a2ln(a+2+a1a2+a1)a4A(a)B(a)=18[tan1(1+2a1)tan1(12a1)]2132ln2(a+2+a1a2+a1)\begin{aligned} A(a) & = \int_0^1 \frac{dx}{a^4 + x^4} \; = \; \frac{1}{a^3}\int_0^{a^{-1}} \frac{dz}{1 + z^4} \\ & = \frac{1}{2a^3\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] + \frac{1}{4a^3\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ B(a) & = \int_0^1 \frac{x^2\,dx}{a^4 + x^4} \; = \; \frac{1}{a}\int_0^{a^{-1}} \frac{z^2\,dz}{1 + z^4} \\ & = \frac{1}{2a\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] - \frac{1}{4a\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ a^4A(a)B(a) & = \tfrac{1}{8}\left[ \tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \right]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{aligned} Thus we deduce that (putting z=axz = ax and playing symmetry games) I(a)=0a1xtan1(xa)1+x4dx  =  0a1xdx1+x401xady1+a2x2y2=a40101z2(a4+z4)(a4+y2z2)dydz=12a40101{z2(a4+z4)(a4+y2z2)+y2(a4+y4)(a4+y2z2)}dydz=12a40101y2+z2(a4+y4)(a4+z4)dydz  =  a4A(a)B(a)=18[tan1(1+2a1)tan1(12a1)]2132ln2(a+2+a1a2+a1)\begin{aligned} I(a) & = \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \; = \; \int_0^{a^{-1}} \frac{x\,dx}{1 + x^4} \int_0^1 \frac{\frac{x}{a}\,dy}{1 + a^{-2}x^2y^2} \\ & = a^4 \int_0^1 \int_0^1 \frac{z^2}{(a^4 + z^4)(a^4 + y^2 z^2)}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \left\{ \frac{z^2}{(a^4 + z^4)(a^4 + y^2z^2)} + \frac{y^2}{(a^4 + y^4)(a^4 + y^2z^2)}\right\}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \frac{y^2 + z^2}{(a^4 + y^4)(a^4 + z^4)}\,dy\,dz \; = \; a^4 A(a)B(a) \\ & = \tfrac{1}{8}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{aligned} Moreover J(a)=0a1tan1x2x2+a2dx  =  [1atan1x2tan1(xa)]0a12a0a1xtan1(xa)1+x4dx=1a(tan1a2)22aI(a)\begin{aligned} J(a) & = \int_0^{a^{-1}} \frac{\tan^{-1}x^2}{x^2 + a^2}\,dx \; = \; \Big[\frac{1}{a} \tan^{-1}x^2\,\tan^{-1}\big(\tfrac{x}{a}\big)\Big]_0^{a^{-1}} - \frac{2}{a} \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \\ & = \frac{1}{a}\big(\tan^{-1}a^{-2}\big)^2 - \frac{2}{a}I(a) \end{aligned} Thus, putting u=cos2xu = \cos2x and v=2u13v= \frac{2u-1}{\sqrt{3}}, we see that K=016πxsin2x4sin4x2sin2x+1dx  =  016πxsin2xcos22xcos2x+1dx=14121cos1uduu2u+1  =  14[23tan1(2u13)cos1u]121+1412123tan1(2u13)du1u2=123121tan1(2u13)du1u2  =  12×314013tan1vdv(13v)(3+v)\begin{aligned} K & = \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{4\sin^4x - 2\sin^2x + 1}\,dx \; = \; \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{\cos^22x - \cos2x + 1}\,dx \\ & = \tfrac14\int_{\frac12}^1 \frac{\cos^{-1}u\,du}{u^2 - u + 1} \; = \; \tfrac14\Big[\tfrac{2}{\sqrt{3}}\tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \cos^{-1}u\Big]_{\frac12}^1 + \tfrac14\int_{\frac12}^1 \tfrac{2}{\sqrt{3}} \tan^{-1}\big(\tfrac{2u-1}{\sqrt{3}}\big) \frac{du}{\sqrt{1-u^2}} \\ & = \tfrac{1}{2\sqrt{3}} \int_{\frac12}^1 \tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \frac{du}{\sqrt{1-u^2}} \; = \; \frac{1}{2 \times 3^{\frac14}} \int_0^{\frac{1}{\sqrt{3}}} \frac{\tan^{-1}v\,dv}{\sqrt{(1 - \sqrt{3}v)(\sqrt{3}+v)}} \end{aligned} The key substitution w  =  13v3+v w \; = \; \sqrt{\frac{1 - \sqrt{3}v}{\sqrt{3}+v}} yields K=13140314tan1(13w23+w2)dw3+w2  =  13140314[16πtan1w2]dw3+w2=π6×314[314tan1(w314)]0314314J(314)=π2363314[314(tan1312)22×314I(314)]  =  23I(314)=143[tan1(2314314)]21163ln2(314+2+3143142+314)\begin{aligned} K & = \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \tan^{-1}\left(\frac{1 - \sqrt{3}w^2}{\sqrt{3} + w^2}\right) \frac{dw}{\sqrt{3} + w^2} \; = \; \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \Big[\tfrac16\pi - \tan^{-1}w^2\Big] \frac{dw}{\sqrt{3} + w^2} \\ & = \frac{\pi}{6 \times 3^{\frac14}} \Big[3^{-\frac14} \tan^{-1}\big(\tfrac{w}{3^{\frac14}}\big)\Big]_0^{3^{-\frac14}} - 3^{-\frac14}J(3^{\frac14}) \\ & = \frac{\pi^2}{36\sqrt{3}} - 3^{-\frac14}\Big[3^{-\frac14}\left(\tan^{-1}3^{-\frac12}\right)^2 - 2 \times 3^{-\frac14} I\big(3^{\frac14}\big)\Big] \; = \; \tfrac{2}{\sqrt{3}} I\big(3^{\frac14}\big) \\ & = \frac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right) \end{aligned} noting that tan1(1+2a1)tan1(12a1)  =  tan1(2aa1)a>1 \tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \; =\; \tan^{-1}\left(\frac{\sqrt{2}}{a - a^{-1}}\right) \hspace{1cm} a > 1

Mark Hennings - 2 years, 6 months ago

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Actually my solution is the same than yours. The only difference is the way you manage to transform the trigonometric integral.

FDP DPF - 2 years, 6 months ago

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@Fdp Dpf Yes, but your method of transforming the original integral into what is, apart from a variable scaling, 23I(31/4)\tfrac{2}{\sqrt{3}} I(3^{1/4}) (in my notation), is elegant.

Mark Hennings - 2 years, 6 months ago

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@Mark Hennings if F(a,t,x)=x(tx+a)(x2+a)F(a,t,x)=\frac{x}{\left( t\cdot x+a\right) \cdot \left( {{x}^{2}}+a\right) }, compute 01F(1,t,x)dt\int_0^1 F(1,t,x)dt isn't nice?

Problem 47 relies on the evaluation of 0101F(1,t2,x2)dtdx\int_0^1 \int_0^1 F(1,t^2,x^2) dtdx and problem 49 relies on 0101F(3,t2,x2)dtdx\int_0^1 \int_0^1 F(3,t^2,x^2) dtdx

Who will post problem 50, the last one?

FDP DPF - 2 years, 6 months ago

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@Fdp Dpf Let me clarify. Both of our methods evaluate 23I(314)\tfrac{2}{\sqrt{3}}I(3^{\frac14}). What I was saying was that your technique for converting the original integral to 23I(314)\tfrac{2}{\sqrt{3}}I(3^{\frac14}) was more elegant than mine - in Mathematics, the word "elegant" is complimentary!

Mark Hennings - 2 years, 6 months ago

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@Fdp Dpf I'll post by evening, in 5 hours.

Ishan Singh - 2 years, 6 months ago

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My solution is close to yours. I will post it later (01 AM here) . This integral is based on the solution i have provided for problem 47. I have played around with it and i have build this one.

FDP DPF - 2 years, 6 months ago

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* PROBLEM 48: *

Prove that 0xμ12(x+r)μ(x+s)μdx  =  π(r+s)12μΓ(μ12)Γ(μ) \int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu (x+s)^\mu}\,dx \; = \; \sqrt{\pi}(\sqrt{r}+\sqrt{s})^{1-2\mu}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)} for all r,s>0r,s > 0 and μ>12\mu > \tfrac12.

Mark Hennings - 2 years, 6 months ago

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Someone else can post the next one...

For any a>0a > 0 and 0<b<c0 < b < c and x<1|x| < 1 we have Ia,b,c(x)=01tb1(1t)cb1(1xt)adt=01tb1(1t)cb1(n=0(a)(n)n!xntn)dt=n=0(a)(n)n!xn01tn+b1(1t)cb1dt  =  n=0(a)(n)n!xnB(n+b,cb)=n=0(a)(n)n!xnΓ(n+b)Γ(cb)Γ(n+c)  =  n=0(a)(n)n!xn(b)(n)Γ(b)Γ(cb)(c)(n)Γ(c)=B(b,cb)n=0(a)(n)(b)(n)(c)(n)xnn!  =  B(b,cb)(2F1)(a,b;c;x)\begin{aligned} I_{a,b,c}(x) & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - xt)^{-a}\,dt \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1} \left(\sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n t^n\right)\,dt \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \int_0^1 t^{n+b-1}(1-t)^{c-b-1}\,dt \; = \; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n B(n+b,c-b) \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{\Gamma(n+b) \Gamma(c-b)}{\Gamma(n+c)} \; =\; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{(b)^{(n)} \Gamma(b)\Gamma(c-b)}{(c)^{(n)} \Gamma(c)} \\ & = B(b,c-b)\sum_{n=0}^\infty \frac{(a)^{(n)} (b)^{(n)}}{(c)^{(n)}} \frac{x^n}{n!} \; = \; B(b,c-b) \big({}_2F_1\big)(a,b;c;x) \end{aligned}