Brilliant Integration Contest - Season 1 (Part 2)

This is Brilliant Integration Contest - Season 1 (Part 2) as a continuation of the previous contest (Part 1). There is a major change in the rules of contest, so please read all of them carefully before take part in this contest.

I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. 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

  1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
  2. 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.
  3. Please make a substantial comment.
  4. Make sure you know how to solve your own problem before posting it in case there is no one can answer it within a week, then you must post the solution and you have a right to post another problem.
  5. 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.
  6. The scope of questions is only computation of integrals either definite or indefinite integrals.
  7. You are NOT allowed to post a multiple integrals problem as well as a complex integral problem.
  8. You are also NOT allowed to post a solution using a contour integration or residue method.
  9. The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, and trigonometric integral.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

[Post your solution here]

PROBLEM xxx (number of problem) :

[Post your problem here]

Remember, put them separately.

Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌

Okay, let the contest part 2 begin!

P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 3).

Note by Anastasiya Romanova
4 years, 10 months ago

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1 vote

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Thanks for doing this. There is a lot to learn from these integration questions that you have shared.

Calvin Lin Staff - 4 years, 10 months ago

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Thank you for your help. You're too kind to me. I really appreciate it \quad ¨\ddot\smile

Anastasiya Romanova - 4 years, 10 months ago

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Problem 20

01sinh1(x)log[(21)x+1]xdx=log(2)log(1+2)π224 \displaystyle \int_0^1 \frac{\sinh ^{-1}(x)-\log \left[\left(\sqrt{2}-1\right) \sqrt{x}+1\right]}{x} \, dx = \log (2) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{24}

Ruben Doornenbal - 4 years, 10 months ago

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Solving this problem is really tedious job. Honestly, I'm unwilling to answer it too (even if I know how to solve it). I don't know what is the OP's motivation by posting two well-defined integrals in a single problem. If the OP has an elementary and a clever method than mine, please do share to us. Okay, here is an attempt using a cannon.

Let split the integral into two parts

IJ=01arcsinhxxdx01ln(1+(21)x)xdxI-J=\int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx-\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx

Perform integration by parts for II by taking u=arcsinhxu={\rm{arcsinh}\,} x and dv=dxxdv=\dfrac{dx}{x}.

01arcsinhxxdx=arcsinhxlnx0101lnx1+x2dxx=tantI=0π/4ln(cost)ln(sint)costdt=0π/4ln(cost)costdt0π/4ln(sint)costdt=120π/4ln(1sin2t)costdt0π/4ln(sint)costdt\begin{aligned} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&={\rm{arcsinh}\,} x\,\ln x\bigg|_0^1-\int_0^1\frac{\ln x}{\sqrt{1+x^2}}\,dx\qquad\Rightarrow\qquad x=\tan t\\ I&=\int_0^{\pi/4}\frac{\ln(\cos t)-\ln (\sin t)}{\cos t}\,dt\\ &=\int_0^{\pi/4}\frac{\ln(\cos t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ &=\frac{1}{2}\int_0^{\pi/4}\frac{\ln(1-\sin^2 t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ \end{aligned}

Putting y=sinty=\sin t and a=12a=\dfrac{1}{\sqrt{2}}, we get

01arcsinhxxdx=0aln(1y2)1y2dy0alny1y2dy=0aln(1y)+ln(1+y)(1y)(1+y)dy0alny(1y)(1+y)dy=I1+I2\begin{aligned} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&=\int_0^{a}\frac{\ln (1-y^2)}{1-y^2}\,dy-\int_0^{a}\frac{\ln y}{1-y^2}\,dy\\ &=\int_0^{a}\frac{\ln (1-y)+\ln(1+y)}{(1-y)(1+y)}\,dy-\int_0^{a}\frac{\ln y}{(1-y)(1+y)}\,dy\\ &=I_1+I_2 \end{aligned}

Performing partial fractions decomposition we get

I1=120aln(1+y)1+ydy+120aln(1y)1+ydy+120aln(1+y)1ydy+120aln(1y)1ydy=ln2(1+a)4+120aln(1y)1+ydy+120aln(1+y)1ydyln2(1a)4=14ln2(1+a1a)+120aln(1y)1+ydy+120aln(1+y)1ydy2z=1+y=ln2(1+2)+12cbln(22z)zdz+12cblnz1zdz=ln2(1+2)+ln22cbdzz+12cbln(1z)zdz+121c1bln(1z)zdz=ln2(1+2)+ln22ln(1+22)+12cbln(1z)zdz+121c1bln(1z)zdz\begin{aligned} I_1&=\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1-y}\,dy\\ &=\frac{\ln^2 (1+a)}{4}+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy-\frac{\ln^2 (1-a)}{4}\\ &=\frac{1}{4}\ln^2\left(\frac{1+a}{1-a}\right)+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy\qquad\Rightarrow\qquad 2z=1+y\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (2-2z)}{z}\,dz+\frac{1}{2}\int_c^{b}\frac{\ln z}{1-z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\int_c^{b}\frac{dz}{z}+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ \end{aligned}

and

I2=120alny1ydy+120alny1+ydy=1211aln(1y)ydy+lnyln(1+y)20a0aln(1+y)ydy=1211aln(1y)ydy+lnaln(1+a)20aln(1+y)ydy=1211aln(1y)ydy0aln(1+y)ydyln24ln(1+22)\begin{aligned} I_2&=\frac{1}{2}\int_0^{a}\frac{\ln y}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln y}{1+y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln y\ln(1+y)}{2}\bigg|_0^{a}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln a\ln(1+a)}{2}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy-\int_0^{a}\frac{\ln (1+y)}{y}\,dy-\frac{\ln2}{4}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)\\ \end{aligned}

where b=1+222b=\dfrac{1+\sqrt{2}}{2\sqrt{2}} and c=12c=\dfrac{1}{2}.

Now, let us evaluate JJ. Set x=t2x=t^2, we get

J=01ln(1+(21)x)xdx=201ln(1+(21)t)tdtJ=\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx=2\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)t\right)}{t}\,dt

Here is the cannon, recall a special function dilogarithm.

Li2(z)=0zln(1t)tdt\operatorname{Li}_2(z)=-\int_0^z\frac{\ln(1 - t)}{t}\,dt

Hence, the rest integrals can be easily evaluated by using dilogarithm and an elementary substitution, i.e. t=kxt=kx, where kk is a constant. We may also utilize these identities Li2(z)+Li2(z)=12Li2(z2)Li2(1z)+Li2(11z)=ln2z2Li2(z)+Li2(1z)=π26lnzln(1z)\begin{aligned} \operatorname{Li}_2(z)+\operatorname{Li}_2(-z)&=\frac{1}{2}\operatorname{Li}_2(z^2)\\ \operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)&=-\frac{\ln^2z}{2}\\ \operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)&=\frac{{\pi}^2}{6}-\ln z \cdot\ln(1-z) \end{aligned} and special values Li2(1)=π212Li2(0)=0Li2(12)=π212ln222Li2(1)=π26\begin{aligned} \operatorname{Li}_2(-1)&=-\frac{{\pi}^2}{12}\\ \operatorname{Li}_2(0)&=0\\ \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{{\pi}^2}{12}-\frac{\ln^2 2}{2}\\ \operatorname{Li}_2(1)&=\frac{{\pi}^2}{6} \end{aligned} Performing a cumbersome and a tedious calculation, we will get the announced result.

I hope you understand my feelings while trying to solve and to write it down. So, please do not ever post a problem like this again. LOL

Anastasiya Romanova - 4 years, 10 months ago

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The reason for taking the difference of the two integrals is that the result is much simpler than the two integrals separately (which involve dilogarithms), which I found beautiful. It only works if both terms are exactly as they are (including the weird factor (21) (\sqrt 2 -1) ).

Your solution can be significantly simplified. Both terms in the integral have a relatively simple antiderivative in terms of dilogarithms. After plugging in the limits, it then boils down to showing

Li2(21)Li2(12)=π2812log2(21) \displaystyle \operatorname{Li}_2 (\sqrt{2} - 1) - \operatorname{Li}_2 (1 - \sqrt 2) = \frac{\pi^ 2}{8} - \frac 1 2 \log^ 2(\sqrt 2 - 1)

using the dilogarithm identities you posted.

To find the antiderivative of the second term, just substitute (21)x=u (\sqrt 2 -1) \sqrt x = u . For the first term, substitute u=(x+1+x2)2 u = \left( x + \sqrt{1+x^2} \right)^2 . This will reduce the integral to

lnuduu21, \displaystyle \int \frac{\ln u \,du}{u^2 - 1},

which I am sure you can calculate in a few lines.

Ruben Doornenbal - 4 years, 10 months ago

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@Ruben Doornenbal I'm sorry, I'm a bit dizzy right now so I can't follow your comment. Could you elaborate? If I may ask, could you post your solution of this problem? Thanks.

Edit : Aha! I get it. Use this relation: arcsinhx=ln(x+x2+1){\rm{arcsinh}\,}x=\ln\left(x+\sqrt{x^2+1}\right). Very clever!

Anastasiya Romanova - 4 years, 10 months ago

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@Anastasiya Romanova Here's a summary of what I said, maybe it helps:

The term with arcsinhx/x \operatorname{arcsinh} x / x can be evaluated by substituting u=(x+1+x2)2 u = \left( x + \sqrt{1+x^2} \right)^2 . The result is something with a dilogarithm. The other term also gives a dilogarithm, but the dilogarithm terms cancel precisely, leading to an elementary result. I hope you can appreciate the beauty of the problem :)

Ruben Doornenbal - 4 years, 10 months ago

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PROBLEM 16 :

Prove

0π2cosv1xcosax dx=π2vv B(v+a+12,va+12)\large\int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx=\frac{\pi}{2^v v\ \operatorname{B}\left(\frac{v+a+1}{2},\frac{v-a+1}{2}\right)}

where B(x,y)\operatorname{B}\left(x,y\right) is the beta function.


PS : POST YOUR SOLUTION BELOW EACH PROBLEM THREAD AND POST YOUR PROPOSED PROBLEM AS A NEW THREAD. PUT THEM IN SEPARATED THREAD. SO THAT THE POSTS LOOK MORE ORGANIZED. THANKS.

Anastasiya Romanova - 4 years, 10 months ago

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Isn't this problem too difficult to high school students Anna? I decide to answer Problem 16 because I'm afraid if this continues till a week, this contest will lose its interest. IMHO, you should propose an easy problem so that this contest will be fun as the stated aims of it. So, here is a solution:

SOLUTION OF PROBLEM 16 :

Rewrite the integral as follows 0π2cosv1xcosax dx=12π2π2(eix+eix2)v1cosax dx=12vπ2π2(1+e2ix)v1ei(v1)xcosax dx=12vπ2π2n=0v1(v1n)e2inxei(v1)xcosax dx=12vn=0v1(v1n)π2π2ei(2nv+1)xcosax dx.(1)\begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac12\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{v-1}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(1+e^{2ix}\right)^{v-1}e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\sum_{n=0}^{v-1}\binom{v-1}{n} e^{2inx}\cdot e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(2n-v+1)x}\cos ax\ dx.\qquad\qquad\qquad\tag1 \end{aligned} Consider f(x)={eiωxfor π2<x<π20otherwise f(x)=\left\{ \begin{array}{l l} e^{i\omega x} & \quad \text{for}\ -\frac\pi2<x<\frac\pi2\\[12pt] 0 & \quad \text{otherwise} \end{array} \right. The Fourier transform of f(x)f(x) is F[f(x)]=f(x) eiαx dxπ2π2eiωxcosαx dxiπ2π2eiωxsinαx dx=π2π2eiωx eiαx dx=π2π2ei(ωα)x dx=[ei(ωα)xi(ωα)]x=π2π2(F[f(x)])=[sin(ωα)xωα]x=π2π2π2π2cosαx dx=2sin(ωα)π2ωα.(2)\begin{aligned} \mathscr{F}\left[f(x)\right]&=\int_{-\infty}^\infty f(x)\ e^{-i\alpha x}\ dx\\ \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\cos \alpha x\ dx-i\int_{-\large\frac\pi2}^{\large\frac\pi2}e^{i\omega x}\sin \alpha x\ dx&=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\ e^{-i\alpha x}\ dx\\ &=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(\omega-\alpha) x}\ dx\\ &=\left[\frac{e^{i(\omega-\alpha) x}}{i(\omega-\alpha)}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \Re\bigg(\mathscr{F}\left[f(x)\right]\bigg)&=\left[\frac{\sin(\omega-\alpha) x}{\omega-\alpha}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \int_{\large-\frac\pi2}^{\large\frac\pi2}\cos \alpha x\ dx&=\frac{2\sin(\omega-\alpha) \frac\pi2}{\omega-\alpha}.\qquad\qquad\qquad\tag2 \end{aligned} Using (2)(2), then (1)(1) turns out to be 0π2cosv1xcosax dx=12v1n=0v1(v1n)sin(2nv+1a)π22nv+1a=12vn=0v1(v1n)sin(nv1+a2)πnv1+a2.(3)\begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac1{2^{v-1}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin(2n-v+1-a) \frac\pi2}{2n-v+1-a}\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin\left(n-\frac{v-1+a}2\right) \pi}{n-\frac{v-1+a}2}.\qquad\qquad\qquad\tag3\\ \end{aligned} Now, let us express (yz)\dbinom{y}{z} in term of beta function that can be related to (3)(3). (yz)=y!z!(yz)!=y!Γ(1+z)Γ(1+yz)=y!zΓ(z)Γ(1z)(yz)(1z)=sin(πz)πzy!(yz)(1z)=sin(πz)πzn=0y(yn)(1)nnzn=n=0y(yn)sinπ(zn)π(zn).(4)\begin{aligned} \binom{y}{z}&=\frac{y!}{z!(y-z)!}\\ &=\frac{y!}{\Gamma(1+z)\Gamma(1+y-z)}\\ &=\frac{y!}{z\Gamma(z)\Gamma(1-z)(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\cdot\frac{y!}{(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\sum_{n=0}^{y}\binom{y}{n}(-1)^n\frac{n}{z-n}\\ &=\sum_{n=0}^{y}\binom{y}{n}\frac{\sin\pi(z-n)}{\pi(z-n)}.\qquad\qquad\qquad\tag4 \end{aligned} Using (4)(4), then (3)(3) turns out to be 0π2cosv1xcosax dx=π2v(v1v1+a2)=π2vΓ(v)Γ(v+a+12)Γ(va+12)=π2v v B(v+a+12,va+12)\begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac{\pi}{2^{v}}\binom{v-1}{\frac{v-1+a}2}\\ &=\frac{\pi}{2^{v}}\frac{\Gamma(v)}{\Gamma\left(\frac{v+a+1}2\right)\Gamma\left(\frac{v-a+1}2\right)}\\ &=\frac{\pi}{2^{v}\ v\ \operatorname{B}\left(\frac{v+a+1}2,\frac{v-a+1}2\right)}\qquad\qquad\qquad\blacksquare \end{aligned}

Tunk-Fey Ariawan - 4 years, 10 months ago

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OK, fine. I'll post high school integral problems from now. -_-"

Anastasiya Romanova - 4 years, 10 months ago

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@Anastasiya Romanova No Just keep posting those hard integrals, it's challenging but we learn a lot from it

Oussama Boussif - 4 years, 10 months ago

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Your solution is valid only if vv is an integer, whereas the identity holds in general also.

Ishan Singh - 2 years, 10 months ago

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PROBLEM 18

Show that

0π/4tan1/3xdx=16(π33log2) \displaystyle \int_0^{\pi/4} \tan^{1/3} x dx = \frac{1}{6} \left( \pi \sqrt{3} -3\log 2\right)

My bad, it should be 3log2 3 \log 2 indeed. Kinshuk's result is correct. Sorry for the confusion.

Ruben Doornenbal - 4 years, 10 months ago

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There should be 3log(2) instead of 2log(2)

Oussama Boussif - 4 years, 10 months ago

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Solutionofproblem18:I=0π4tan13x.dxputtan13x=tdx=3t21+t6.dtI=3t31+t6.dtputt2=u2t.dt=duI=32u.du1+u3usingpartialfraction,ourintegrationturnsouttobe:I=12du1+u+12(1+u)duu2u+1aftersolvingandapplyinglimits:I=0π4tan13x.dx=16(π33log2)Solution\quad of\quad problem\quad 18:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \\ put\quad { tan }^{ \frac { 1 }{ 3 } }x=t\\ dx\quad =\quad \frac { { 3t }^{ 2 } }{ 1+{ t }^{ 6 } } .dt\\ I\quad =\quad \int { \frac { { 3t }^{ 3 } }{ 1+{ t }^{ 6 } } .dt } \\ put\quad { t }^{ 2 }=u\\ 2t.dt=du\\ I\quad =\quad \frac { 3 }{ 2 } \int { \frac { u.du }{ 1+{ u }^{ 3 } } } \\ using\quad partial\quad fraction\quad ,\quad our\quad integration\\ turns\quad out\quad to\quad be\quad :\\ I\quad =\quad \frac { -1 }{ 2 } \int { \frac { du }{ 1+u } } \quad +\quad \frac { 1 }{ 2 } \int { \frac { (1+u)du }{ { u }^{ 2 }-u+1 } } \\ after\quad solving\quad and\quad applying\quad limits:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \quad =\quad \frac { 1 }{ 6 } (\pi \sqrt { 3 } -3\log { 2 } )

Kïñshük Sïñgh - 4 years, 10 months ago

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PROBLEM 21 :

Show that 0dxx4+2cos(2θ)x2+1=π4cosθ\int_0^\infty \frac{dx}{x^4+2\cos(2\theta)\,x^2+1}=\frac{\pi}{4\cos\theta}

Anastasiya Romanova - 4 years, 10 months ago

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We have

0dxx4+ax2+b2=π2b2b+a, \displaystyle \int_0^ {\infty} \frac{dx}{x^4 + a x^2 + b^2} = \frac{\pi}{2b \sqrt{2b+a}},

which I proved on MSE. Plugging in a=2cos2θ a = 2 \cos2 \theta and using that 2cos2(θ)=1+cos2θ2 \cos^2(\theta) = 1 + \cos 2\theta immediately gives the answer.

Ruben Doornenbal - 4 years, 10 months ago

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Oh I know you're now. You're user111187. I thought you're an old man. Haha

Nice to meet you here Ruben. It seems you'll be a tough opponent because you're a Math SE and I&S user. ¨\ddot\smile

Anastasiya Romanova - 4 years, 10 months ago

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@Anastasiya Romanova Yep, this will be good :)

Ruben Doornenbal - 4 years, 10 months ago

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Expecting a question from you @Ruben Doornenbal

Ronak Agarwal - 4 years, 10 months ago

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Problem23Problem\quad 23

Find 0π4ln(tan(x))dx\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(tan(x))dx }

Ronak Agarwal - 4 years, 10 months ago

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Solution 23

A well-known problem. Sub tanx=u \tan x = u to get

01dudulnu1+u2=k0(1)k01dulnuu2k=k0(1)k1(2k+1)2=G. \displaystyle \int_0^1 du \frac{du \ln u}{1+u^2} = \sum_{k \geq 0} (-1)^k \int_0^1 du \ln u \, u^{2k} = \sum_{k \geq 0} (-1)^k \frac{1}{(2k+1)^2} = G.

The penultimate equality follows from integration by parts.

Ruben Doornenbal - 4 years, 10 months ago

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@Ruben Doornenbal Can we have problem 24?

Samuel Jones - 4 years, 10 months ago

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Sir can you elaborate I did'nt understood this one @Ruben Doornenbal

U Z - 4 years, 9 months ago

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@U Z The idea is to expand the factor 11+u2 \displaystyle \frac{1}{1+u^2} in a geometric series and interchange summation and integration. The last equality is just the definition of Catalan's constant.

Ruben Doornenbal - 4 years, 9 months ago

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@Ronak Agarwal @Ruben Doornenbal The answer given is wrong!!! It should be -G!!! You must have forgotten the negative sign.......

Aaghaz Mahajan - 1 year ago

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Problem30Problem\quad 30

Find 1xx0.5xdx\displaystyle \int _{ 1 }^{ \infty }{ \frac { x-\left\lfloor x \right\rfloor -0.5 }{ x } dx }

Ronak Agarwal - 4 years, 10 months ago

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I=limnr=1r=n1rr+1xr12xdx I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} \int_{r}^{r+1} \frac{x-r-\frac{1}{2}}{x} dx I=limnr=1r=n11(r+12)ln(r+1r) I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 1-(r+\frac{1}{2})\ln(\frac{r+1}{r}) 2I=limnr=1r=n12(2r+1)ln(r+1r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+1)\ln(\frac{r+1}{r}) 2I=limnr=1r=n12(2r+2)ln(r+1r)+ln(r+1r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(\frac{r+1}{r}) + \ln(\frac{r+1}{r}) 2I=limnr=1r=n12(2r+2)ln(r+1)+(2r+2)ln(r)+ln(r+1r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(r+1) +(2r+2)\ln(r)+ \ln(\frac{r+1}{r}) 2I=limnr=1r=n122((r+1)ln(r+1)rln(r))+ln(r+1)+ln(r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r) 2I=limnr=1r=n122((r+1)ln(r+1)rln(r))+ln(r+1)+ln(r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r) 2I=limn2(n1)2nln(n)+2ln(n!)ln(n) 2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+2\ln(n!)-\ln(n) 2I=limn2(n1)2nln(n)+(2n+1)ln(n)2n+ln(2π)ln(n) 2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+(2n+1)\ln(n)-2n+\ln(2\pi)-\ln(n) 2I=ln(2π)2 2I = \ln(2\pi)-2 I=ln(2π)21 I =\frac{\ln(2\pi)}{2}-1

Shivang Jindal - 4 years, 10 months ago

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This one is easy too :) ln2π21\frac{\ln2\pi}{2}-1

@Shivang Jindal : Sorry, I was kidding & I am busy right now so I have no time to write down my answer. Could you elaborate yours then you're good to go (propose your problem). Sorry for the inconvenience...

Tunk-Fey Ariawan - 4 years, 10 months ago

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Expecting a question from you @Tunk-Fey Ariawan , also please give a proof of your answer.

Ronak Agarwal - 4 years, 10 months ago

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Trick , is to break the integral from (1,2),(2,3)...(n1,n) (1,2),(2,3)...(n-1,n) . Then, we compute the sum in terms of n n . and then use Stirling approximation :) .

Shivang Jindal - 4 years, 10 months ago

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Exactly, you got it perfectly right.

Ronak Agarwal - 4 years, 10 months ago

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PROBLEM 22

This one is particularly beautiful, in my opinion.

0axdxcosxcos(ax)=asinalnseca. \displaystyle \int_0^a \frac{x dx}{\cos x \cos(a-x)} = \frac{a}{\sin a} \ln \sec a.

Ruben Doornenbal - 4 years, 10 months ago

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SolutionofProblem22Solution\quad of\quad Problem\quad 22

I=0axdxcos(x)cos(ax)=0a(ax)dxcos(x)cos(ax)I=\displaystyle \int _{ 0 }^{ a }{ \frac { xdx }{ cos(x)cos(a-x) } } =\int _{ 0 }^{ a }{ \frac { (a-x)dx }{ cos(x)cos(a-x) } }

Adding these two forms we get :

I=a20adxcos(x)cos(ax)I=\displaystyle \frac { a }{ 2 } \int _{ 0 }^{ a }{ \frac { dx }{ cos(x)cos(a-x) } }

Multiplying and dividing by sin(a)sin(a) we get :

I=a2sin(a)0asin(x+(ax))dxcos(x)cos(ax)I=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ \frac { sin(x+(a-x))dx }{ cos(x)cos(a-x) } }

I=a2sin(a)0a(tan(x)+tan(ax))dxI=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ (tan(x)+tan(a-x))dx }

Also since 0atan(x)dx=0atan(ax)dx\displaystyle \int _{ 0 }^{ a }{ tan(x)dx } =\int _{ 0 }^{ a }{ tan(a-x)dx }

We get I=asin(a)0atan(x)dxI=\displaystyle \frac{a}{sin(a)}\int _{ 0 }^{ a }{ tan(x)dx }

I=asin(a)ln(sec(a))I=\frac { a }{ sin(a) } ln(sec(a))

Ronak Agarwal - 4 years, 10 months ago

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PROBLEM 24

01arcsechxarcsinxdx=π28ln2. \displaystyle \int_0^1 \operatorname{arcsech} x \operatorname{arcsin} xdx =\frac{\pi^2}{8} - \ln 2.

Ruben Doornenbal - 4 years, 10 months ago

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Solutionofproblem24Solution\quad of\quad problem\quad 24

First note that :

arcsech(x)=ln(1+1x2x)arcsech(x)=ln(\frac{1+\sqrt{1-{x}^{2}}}{x})

In our integral put x=sin(θ)x=sin(\theta)

I=0π2θcosθln(1+cosθsinθ)dθ\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta cos\theta ln(\frac { 1+cos\theta }{ sin\theta } ) } d\theta

Applying integration by parts we get , u=ln(1+cosθsinθ),dv=θcosθdθu=ln(\frac{1+cos\theta }{sin\theta}),dv=\theta cos\theta d\theta

I=(θsinθ+cosθ)ln(1+cosθsinθ)0π2+0π2(θsinθ+cosθ)cosecθdθ\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\theta sin\theta +cos\theta )cosec\theta d\theta }

I=(θsinθ+cosθ)ln(1+cosθsinθ)+θ22+ln(sinθ)0π2\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )+\frac { { \theta }^{ 2 } }{ 2 } +ln(sin\theta )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } }

Which on evaluating we get :

I=π28ln(2)I=\frac { { \pi }^{ 2 } }{ 8 } -ln(2)

Ronak Agarwal - 4 years, 10 months ago

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Can you elaborate the first line please , it will be great to learn from you

U Z - 4 years, 9 months ago

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PROBLEM 26 :

Prove

0lnxcoshxdx=π2ln(Γ4(34)π)\int_0^\infty\frac{\ln x}{\cosh x}\,dx=\frac{\pi}{2}\ln\left(\frac{\Gamma^4\left(\frac{3}{4}\right)}{\pi}\right)

P.S. You may use any well-known expressions.

Anastasiya Romanova - 4 years, 10 months ago

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SOLUTION 26

Consider

I(a)=0dxxa1coshx=2k0(1)k0dxxa1e(2k+1)x=2Γ(a)β(a). \displaystyle I(a) = \int_0^\infty dx \frac{x^{a-1}}{\cosh x} = 2 \sum \limits_{k \geq 0} (-1)^k \int_0^\infty dx x^{a-1} e^{-(2k+1)x} = 2 \Gamma(a) \beta(a).

Our integral is

I(1)=2Γ(1)β(1)+2Γ(1)β(1)=2(γ)(π/4)+2π4[γ+2ln2+3lnπ4lnΓ14]. \displaystyle I'(1) = 2 \Gamma'(1)\beta(1) + 2 \Gamma(1) \beta'(1) = 2(-\gamma)(\pi/4) + 2 \frac \pi 4 \left[\gamma + 2 \ln 2 + 3 \ln \pi - 4 \ln \Gamma \frac 1 4 \right].

Here we used a result from Mathworld. Using the Euler reflection formula,

Γ(1/4)=π2(Γ(3/4))1. \displaystyle \Gamma(1/4) = \pi \sqrt 2 (\Gamma(3/4))^{-1}.

Collecting all the terms gives I(1)=π2lnπ+2πlnΓ(3/4), \displaystyle I'(1) = -\frac \pi 2 \ln \pi + 2\pi \ln \Gamma(3/4),

which equals the stated result.

Ruben Doornenbal - 4 years, 10 months ago

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PROBLEM 27

My last two integrals were clearly too easy. By finding an antiderivative or otherwise, show that

0dxln2tanhx=74ζ(3). \int_0^{\infty} dx\, \ln^2 \tanh x = \frac{7}{4}\zeta{(3)}.

Ruben Doornenbal - 4 years, 10 months ago

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Solution of Problem 27

Set t=tanhxt=\tanh x, we have

0ln2(tanhx)dx=01ln2t1t2dt=01n=0t2nln2tdt=n=001t2nln2tdt=2n=01(2n+1)3see solution of Problem 13=2[n=11n3n=11(2n)3]=74n=11n3=74ζ(3)\begin{aligned} \int_0^\infty\ln^2(\tanh x)\,dx&=\int_0^1\frac{\ln^2t}{1-t^2}\,dt\\ &=\int_0^1\sum_{n=0}^\infty t^{2n}\ln^2t\,dt\\ &=\sum_{n=0}^\infty\int_0^1 t^{2n}\ln^2t\,dt\\ &=2\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\qquad\Rightarrow\qquad\text{see solution of Problem 13}\\ &=2\left[\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3}\right]\\ &=\frac{7}{4}\sum_{n=1}^\infty\frac{1}{n^3}\\ &=\frac{7}{4}\zeta(3) \end{aligned}

Anastasiya Romanova - 4 years, 10 months ago

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Problem 28

Prove

sinh2xcos2xsinhπx dx=sin2cos2+cosh2\begin{aligned} \int_{-\infty}^{\infty} \frac{\sinh 2x\cos 2x}{\sinh \pi x} \ dx = \frac{\sin 2}{\cos 2 + \cosh 2} \end{aligned}

Anastasiya Romanova - 4 years, 10 months ago

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Solution 28

The integral equals

I=dx(e2xe2x)eπxe2ixe2πx1.\displaystyle I = \Re \int_{-\infty}^\infty dx \frac{\left(e^{2x} - e^{- 2x} \right) e^{\pi x} e^{2 i x}}{e^{2 \pi x} - 1}.

Substitute eπx=u e^{\pi x} = u . We get

I=J+J, \displaystyle I = J_+ - J_-,

where

J±=1π0dxu2(i±1)/πu21=12cot[π2(2(i±1)/π+1)]=12tan(i±1).\displaystyle \begin{aligned} J_\pm &= \Re \frac 1 \pi \int_0^\infty dx \frac{u^{2( i\pm1)/\pi}}{u^2 - 1} \\&= -\Re \frac 1 2 \cot\left[\frac \pi 2 \left(2(i\pm1)/\pi + 1\right) \right] \\&= \Re\frac 1 2 \tan(i \pm 1). \end{aligned}

Here we made use of the well-known integral

PV0xa11xb=πbcotπab.\displaystyle PV \int_0^\infty \frac{x^{a-1}}{1-x^b} = \frac \pi b \cot \frac{\pi a}{b}.

Now using the identity

tanA+B2=sinA+sinBcosA+cosB\displaystyle \tan\frac{A+B}{2} = \frac{\sin A + \sin B}{\cos A + \cos B}

gives

J±=±12sin2cosh2+cos2,\displaystyle J_\pm = \pm \frac 1 2 \frac{\sin 2}{\cosh 2 + \cos 2},

which gives the desired result.

Ruben Doornenbal - 4 years, 10 months ago

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Wait!? For PV0xa11xbdx=πbcot(πab)PV\int_0^\infty \frac{x^{a-1}}{1-x^b}\,dx=\frac{\pi}{b}\cot\left(\frac{\pi a}{b}\right) could you prove it without using contour integration or residue method? See the rules.

Anastasiya Romanova - 4 years, 10 months ago

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@Anastasiya Romanova Of course, my dear. It is clear that we can take b=1 b = 1 in the proof. The general result follows from a substitution. Separate the integrals over (0,1) (0,1) and over (1,) (1,\infty) . Put u=1/x u = 1/x in the second integral. The result is

01dxxa1xa1x=ψ(1a)ψ(a)=πcotπa, \displaystyle \int_0^1 dx \frac{x^{a-1} - x^{-a}}{1-x} = \psi(1-a) - \psi(a) = \pi \cot \pi a,

as was to be proven. Here we used a result derived by real methods here.

You have sharp eye for integrals that I normally derive with residues :p

Ruben Doornenbal - 4 years, 10 months ago

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@Ruben Doornenbal OK, I accept your explanation. You may propose your problem but I have to sleep now. It's already late night here. Maybe someone else will answer your problem. So, in the mean time, you fight with them. Good night! 👋(>‿◠)