Brilliant Integration Contest - Season 1

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.

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

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]

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

Okay, let the contest begin! Here is the first problem:

PROBLEM 1 :

For a>0a>0, show that

0alnxaxx2dx=πln(a4) \int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,dx=\pi\ln\left(\frac{a}{4}\right)

P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 2) and Brilliant Integration Contest - Season 1 (Part 3).

Note by Anastasiya Romanova
5 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Ans:ln4a/2a^2

Shreyas Dighe - 5 years, 3 months ago

Log in to reply

Solution of Problem 15 :

We have 01(11x+1lnx)dx=01lnx+1x(1x)lnxdx \int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx

Proposition :

I(s)=01slnx+1xs(1x)lnxdx=lnΓ(s+1)+γs I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s

Proof :

Let

I(s)=01slnx+1xs(1x)lnxdx I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx

then

I(s)=011xs1xdxI(s)=01xslnx1xdx=01n=0xn+slnxdx=n=0s01xn+sdx=n=0s[1n+s+1]=n=01(n+s+1)2=ψ1(s+1)I(s)=ψ1(s+1)dsI(s)=s[ψ(s+1)]dsI(s)=ψ(s+1)+C\begin{aligned} I'(s)&=\int_0^1 \frac{1-x^s}{1-x}\,dx\\ I''(s)&=-\int_0^1 \frac{x^s\ln x}{1-x} \,dx\\ &=-\int_0^1\sum_{n=0}^\infty x^{n+s}\ln x\,dx\\ &=-\sum_{n=0}^\infty\partial_s\int_0^1 x^{n+s}\,dx\\ &=-\sum_{n=0}^\infty\partial_s\left[\frac{1}{n+s+1}\right]\\ &=\sum_{n=0}^\infty\frac{1}{(n+s+1)^2}\\ &=\psi_1(s+1)\\ I'(s)&=\int\psi_1(s+1)\,ds\\ I'(s)&=\int\frac{\partial}{\partial s}\bigg[\psi(s+1)\bigg]\,ds\\ I'(s)&=\psi(s+1)+C\\ \end{aligned} For s=0s=0, we have I(0)=0I'(0)=0. Implying C=ψ(1)=γC=-\psi(1)=\gamma, then I(s)=ψ(s+1)+γI(s)=ψ(s+1)ds+γs+C=s[lnΓ(s+1)]ds+γs+C=lnΓ(s+1)+γs+C\begin{aligned} I'(s)&=\psi(s+1)+\gamma\\ I(s)&=\int \psi(s+1)\,ds +\gamma s+C\\ &=\int \frac{\partial}{\partial s}\bigg[\ln\Gamma(s+1)\bigg]\,ds +\gamma s+C\\ &=\ln\Gamma(s+1) +\gamma s+C\\ \end{aligned}

where ψ(z)\psi(z) is the digamma function. For s=0s=0, we have I(0)=0I(0)=0. Implying C=0C=0, then

I(s)=01slnx+1xs(1x)lnxdx=lnΓ(s+1)+γs I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s\qquad\qquad\square


For s=1s=1, we have

I(1)=01(11x+1lnx)dx=01lnx+1x(1x)lnxdx=γ I(1)=\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx=\gamma

where γ\gamma is The Euler–Mascheroni constant.

Problem 16 :

Prove

0π2cosn1xcosax dx=π2nn B(n+a+12,na+12)\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-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 - 5 years, 3 months ago

Log in to reply

Can you please post a all together new note of the integration contest it's getting messy and messier, Please move this contest to a new note @Anastasiya Romanova

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

OK

The Brilliant Integration Contest - Season 1 (Part 1) moves to The Brilliant Integration Contest - Season 1 (Part 2). Enjoy it!! (>‿◠)✌

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

_ Solution of problem 14 _

I=01ln(3+x3x)1x(1x2)dxI = \displaystyle \int_{0}^{1} \ln \bigg(\dfrac{3+x}{3-x} \bigg) \dfrac{1}{\sqrt{x(1-x^2)}} {\mathrm dx}

= 01ln(1+x/3)1x(1x)dx\displaystyle \int_{0}^{1} \ln(1 + x/3) \dfrac{1}{\sqrt{x(1-x)}} {\mathrm dx}

= 0π/22ln(1+13sin2θ)dθ0π/22ln(113sin2θ)dθ\displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 + \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta - \displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 - \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta

Lemma 0π/2ln(1+asin2θ)=πln(1+a+12)\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)

Proof

Let I(a)=0π/2ln(1+asin2x)dxI(a) = \displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 x) {\mathrm dx}

I(a)=0π/2sin2x1+asin2xdxI'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 x}{1+ a \sin^2 x} {\mathrm dx}

I(a)=0π/21a(1sec2x1+(a+1)tan2xdx)I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{a}\bigg(1 - \dfrac{\sec^2 x}{1+(a+1) \tan^2x} {\mathrm dx}\bigg)

= π2a1a(0dz1+(a+1)z2) \dfrac{\pi}{2a} - \dfrac{1}{a} \bigg(\displaystyle \int_{0}^{\infty} \dfrac{dz}{1+(a+1)z^2}\bigg)

=π2a(11a+1)\dfrac{\pi}{2a} \bigg(1 - \dfrac{1}{\sqrt{a+1}}\bigg)

Hence, I(a)I(0)=0aπ2(1x1xx+1)dxI(a) - I(0) = \displaystyle \int_{0}^{a} \dfrac{\pi}{2} \bigg(\dfrac{1}{x} - \dfrac{1}{x\sqrt{x+1}}\bigg) {\mathrm dx}

= πln(x+1+1)0a=πln(1+a+12)\pi \ln(\sqrt{x+1} +1)\bigg|_{0}^{a} = \pi\ln\bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg) (Note that I(0)=0I(0) = 0)

Put a=1/3a=1/3 , and 1/3-1/3, then substitute back in original integral to get:

I=2πln(1+4/31+2/3)=πln(7+435+26)I = 2 \pi \ln \bigg(\dfrac{1+\sqrt{4/3}}{1+\sqrt{2/3}} \bigg) = \boxed{\pi \ln \bigg(\dfrac{7 + 4\sqrt{3}}{5+2\sqrt{6}}\bigg)}

jatin yadav - 5 years, 3 months ago

Log in to reply

Nice solution Jatin, +1! But you're a bit late about 10 min than Ronak, so the right to propose a new problem goes to Ronak.

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.

Sudeep Salgia - 5 years, 3 months ago

Log in to reply

Solutionofproblem14Solution\quad of\quad problem\quad 14

Lemma

π2π2ln(a2cos2x+b2sin2x)dx=2πln(a+b2)\displaystyle\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx } =2\pi ln(\frac { a+b }{ 2 } )

Proof :

I(a)=π2π2ln(a2cos2x+b2sin2x)dx\displaystyle I(a)=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx }

Differentiating with respect to aa we get :

I(a)=2aπ2π2cos2xa2cos2x+b2sin2xdx\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 2 }x }{ { a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x } dx }

Put tan(x)=ttan(x)=t to get our integral as :

I(a)=2aπ2π2dt(a2+b2t2)(1+t2)\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } }

=2aπ2π2dt(a2+b2t2)(1+t2)=2ab2a2(π2π2b2dta2+b2t2π2π2dt1+t2)\displaystyle =2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2a }{ { b }^{ 2 }-{ a }^{ 2 } } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { b }^{ 2 }dt }{ { a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 } } } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ 1+{ t }^{ 2 } } } )

Solving it we get :

I(a)=2πa+b\displaystyle {I}^{'}(a)=\frac { 2\pi }{ a+b }

I(a)=2πln(a+b)+C\Rightarrow \displaystyle I(a)=2\pi ln(a+b)+C

Put a=b=1a=b=1 to get C=2πln(2)C=-2\pi ln(2)

Hence I(a)=2πln(a+b2)\displaystyle I(a)=2\pi ln(\frac{a+b}{2})

In our integral in the given question put x=1sinθ2x=\frac{1-sin\theta }{2} to get the integral as :

I=π2π2ln(7sin(θ))dxπ2π2ln(5+sin(θ))dxI= \displaystyle \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7-sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5+sin(\theta))dx }

Using the property abf(x)dx=abf(a+bx)dx\displaystyle \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx } we get :

I=π2π2ln(7+sin(θ))dxπ2π2ln(5sin(θ))dx\displaystyle I=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7+sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5-sin(\theta))dx }

Adding these two we get :

I=12(π2π2ln(49sin2θ)dxπ2π2ln(25sin2(θ))dx)\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49-{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25-{ sin }^{ 2 }(\theta))dx } )

I=12(π2π2ln(49cos2θ+48sin2θ)dxπ2π2ln(25cos2θ+24sin2θ)dx)\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49{ cos }^{ 2 }\theta+48{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25{ cos }^{ 2 }\theta+24{ sin }^{ 2 }\theta)dx } )

Using our given lemma we get :

I=12(2πln(7+432)2πln(5+262))=πln(7+435+26)\displaystyle I=\frac { 1 }{ 2 } (2\pi ln(\frac{7+4\sqrt { 3 }}{2} )-2\pi ln(\frac{5+2\sqrt { 6 }}{2} ))=\pi ln(\frac { 7+4\sqrt { 3 } }{ 5+2\sqrt { 6 } } )

Problem15Problem\quad 15

Evaluate I=011ln(x)+11xdx\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx }

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

According to timeline, you're faster than @jatin yadav to post your solution, then the right to propose a new problem is yours.

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

Solution of Problem 13 :

First we prove the following proposition:

Proposition :

01xalnkx dx=(1)kk!(a+1)k+1for  k=0,1,2,\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}} \qquad\text{for }\ k=0,1,2,\ldots

Proof :

Note that 01xa dx=1a+1for  α>1. \int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1. Differentiating equation above kk times w.r.t. aa we have 01kak(xa) dx=01xalnkx dx=(1)kk!(a+1)k+1Q.E.D. \int_0^1 \frac{\partial^k}{\partial a^k}\left(x^a\right)\ dx=\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}}\qquad\text{Q.E.D.}

Now, we will evaluate the general case of

0xa1ebx1dx=0xa1ebx1ebxdxfor  a,b>0\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\int_0^\infty\frac{x^{a-1}\,e^{-bx}}{1-e^{-bx}}\,dx\qquad\text{for }\ a,b>0

Set y=ebxy=e^{-bx}, then

0xa1ebx1dx=(1)a1ba01lna1y1ydy\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\frac{(-1)^{a-1}}{b^a}\int_0^1\frac{\ln^{a-1}y}{1-y}\,dy

Use a geometric series for 11y\dfrac{1}{1-y} then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have

0xa1ebx1dx=(1)a1ba01k=0yklna1ydy=(1)a1bak=001yklna1ydy=(1)a1bak=0(1)a1(a1)!(k+1)a=(a1)!bak=11ka=Γ(a)ζ(a)ba\begin{aligned} \int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx&=\frac{(-1)^{a-1}}{b^a}\int_0^1\sum_{k=0}^\infty y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty\int_0^1 y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty \frac{(-1)^{a-1}\, (a-1)!}{(k+1)^{a}}\\ &=\frac{(a-1)!}{b^a}\sum_{k=1}^\infty \frac{1}{k^{a}}\\ &=\frac{\Gamma(a)\,\zeta(a)}{b^a} \end{aligned}

where Γ(a)\Gamma(a) is the gamma function and ζ(a)\zeta(a) is the Riemann zeta function.

Hence, by setting a=n+1a=n+1 and b=1b=1, we have

0xnex1dx=Γ(n+1)ζ(n+1)\int_0^\infty\frac{x^{n}}{e^{x}-1}\,dx=\Gamma(n+1)\,\zeta(n+1)

Problem 14 :

Prove that

01ln(3+x3x)dxx(1x)=πln(7+435+26)\begin{aligned} \int_{0}^1 \ln\left(\frac{3+x}{3-x}\right)\,\frac{dx}{\sqrt{x(1-x)}}=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\ \end{aligned}

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

SolutionofProblem12Solution\quad of\quad Problem\quad 12

It's a very good disguised integral.

First note that the function is an even function hence our integral can be written as :

I=20cos(arctan(2x))(1+x2)1+4x2dx\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { cos(arctan(2x)) }{ (1+{ x }^{ 2 })\sqrt { 1+4{ x }^{ 2 } } } dx }

Now we know that cos(arctan(2x))=11+4x2\displaystyle cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}. Using this our integral becomes :

I=201(1+x2)(1+4x2)dx\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+{ x }^{ 2 })(1+4{ x }^{ 2 }) } dx }

Splliting it by partial fractions we get :

I=23(0dx(14+x2)0dx1+x2)\displaystyle I=\frac { 2 }{ 3 } (\int _{ 0 }^{ \infty }{ \frac { dx }{ (\frac { 1 }{ 4 } +{ x }^{ 2 }) } } -\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } )

I=23(2tan1(2x)0tan1(x)0)\displaystyle I=\frac { 2 }{ 3 } (2{ tan }^{ -1 }(2x)\overset { \infty }{ \underset { 0 }{ | } } -{ tan }^{ -1 }(x)\overset { \infty }{ \underset { 0 }{ | } } )

I=π3\displaystyle I=\frac { \pi }{ 3 }

Problem13Problem\quad 13

Find closed form of I=0xnex1dx\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx }

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

Wait a sec!? Could you rectify your solution, because the original problem is written as arctan(2x)\arctan(2x). And one thing, what did you mean by "Now we know that cos(arctanx)=11+4x2\cos(\arctan x)=\frac{1}{\sqrt{1+4x^2}}"? Could you elaborate?

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

That was my typing mistake, also Letarctan(2x)=θ arctan(2x)=\theta, hence 2x=tan(θ)2x=tan(\theta), hence cos(θ)=11+4x2cos(\theta)=\frac{1}{\sqrt{1+4{x}^{2}}}

cos(arctan(2x))=11+4x2\Rightarrow cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

Solution of Problem 11 :

Using substitution u=α2x    x=α2u    dx=α2u2 duu=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du, then

0lnxx2+α2 dx=0ln(α2u)(α2u)2+α2α2u2 duI(α)=02lnαlnuα2+u2 du=2lnα01α2+u2 du0lnuu2+α2 du=2lnα01α2+u2 duI(α)I(α)=lnα01α2+u2 du.\begin{aligned} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\int_0^\infty\frac{\ln \left(\dfrac{\alpha^2}{u}\right)}{\left(\dfrac{\alpha^2}{u}\right)^2+\alpha^2}\cdot \dfrac{\alpha^2}{u^2}\ du\\ I(\alpha)&=\int_0^\infty\frac{2\ln \alpha-\ln u}{\alpha^2+u^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+\alpha^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-I(\alpha)\\ I(\alpha)&=\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du. \end{aligned}

The last integral can easily be evaluated by using substitution u=αtanθu=\alpha\tan\theta, then

0lnxx2+α2 dx=lnαα0π2 dθ=πlnα2α\begin{aligned} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\frac{\ln \alpha}{\alpha}\int_0^{\Large\frac\pi2} \ d\theta\\ &=\frac{\pi\ln \alpha}{2\alpha} \end{aligned}

Problem 12 :

Prove

cos(arctan2x)(1+x2)1+4x2dx=π3 \int_{-\infty}^{\infty}\frac{\cos \left(\, \arctan 2x\right)}{(1+x^2)\sqrt{1+4x^2}}\,dx=\frac{\pi}{3}

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

The last substitution should be u=αtanθu=\alpha \tan \theta :)

Pratik Shastri - 5 years, 3 months ago

Log in to reply

This is a simple kind of NCERT problem. It can be easily done by substituting t=tan1xα t = tan^{-1}\dfrac{x}{\alpha} and then applying by parts

U Z - 5 years, 1 month ago

Log in to reply

Fixed it. Sorry, I'm too hasty. Thanks... :)

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

Solution to problem 10

First integrate by parts -

I=0sin3xx2dx=03sin2xcosxxdxI= \int_{0}^{\infty} \dfrac{\sin^3{x}}{x^2} dx =\int_{0}^{\infty} \dfrac{3\sin^2{x} \cos{x}}{x} dx

Then, use the property of the laplace transform that L{f(t)t}(s)=sF(p)dp\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}(s)=\displaystyle\int_{s}^{\infty} F(p) dp (in our case, s0s \rightarrow 0).

I=30L{sin2tcost}(p) dpI=3\int_{0}^{\infty} \mathcal{L} \{\sin^2{t} \cos{t}\}(p) \ dp

I=302s(s2+1)(s2+9)ds=3log34I=3\int_{0}^{\infty} \dfrac{2s}{(s^2+1)(s^2+9)} ds=\boxed{\dfrac{3\log {3}}{4}} The last integral can be found by using partial fractions.

Note : L{f(t)}(s)=0estf(t)dt\mathcal{L}\{f(t)\}(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t

Problem 11

Find

0logxx2+α2dx     for α>0\int_{0}^{\infty} \dfrac{\log {x}}{x^2+\alpha^2} \mathrm{d}x \ \ \ \ \ \text{for} \ \alpha>0

Pratik Shastri - 5 years, 3 months ago

Log in to reply

Solution 9: First, consider the following Lemma.

LEMMA : 01lnx1x2dx=π28\int_{0}^{1} \frac{\ln{x}}{1-x^2} dx=-\frac{\pi^2}{8}

Proof: Since, 11x2=n=0x2n\frac{1}{1-x^2}=\sum\limits_{n=0}^{\infty} {x^{2n}}, we have,

01lnx1x2=n=001x2nlnxdx \int_0^1 \frac{\ln{x}}{1-x^2}=\sum\limits_{n=0}^{\infty} \int_0^1 x^{2n}{\ln{x}} \: dx

=n=01(2n+1)2=-\sum\limits_{n=0}^{\infty} {\frac{1}{(2n+1)^2}} (Using Integration By Parts)

=n=01(2n)2n=01(n)2=\sum\limits_{n=0}^{\infty} {\frac{1}{(2n)^2}} - \sum\limits_{n=0}^{\infty} {\frac{1}{(n)^2}}

=34ζ(2)=π28=\frac{-3}{4}\zeta(2)= -\frac{\pi^2}{8}

Now,

I=01ln(1+x1x)dxx1x2\text{I}= \int_{0}^{1} \ln({\frac{1+x}{1-x}}) \frac{dx}{x\sqrt{1-x^2}}

Substituting x=1y21+y2x=\frac{1-y^2}{1+y^2}, we have,

I=4×01lny1y2dx\text{I}=-4 \times \int_{0}^{1} \frac{\ln{y}}{1-y^2} dx

=π22=\frac{\pi^2}{2} (Using the Lemma)

Ishan Singh - 5 years, 3 months ago

Log in to reply

Nice answer, +1! But I'm really sorry, the one who deserves to propose the next problem is Ronak since according to the timeline his answer comes first than the other answers. Maybe, you can try to answer his proposed problem. ¨\ddot\smile

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

@Anastasiya Romanova I think maybe we should change the format of the contest. Maybe each problem should start in a new comment and the solutions must be replies to that comment. It would keep the contest more organised and understandable. The present format is just messing things up with the problem and its solution in separate comments.

Sudeep Salgia - 5 years, 3 months ago

Log in to reply

@Sudeep Salgia Ya, I agree with you. I have edited the rule accordingly. Well, this is the first contest here and I hope you understand with that.

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

Solutionofproblem9Solution\quad of\quad problem\quad 9

I=01ln(1+x1x)dxx1x2\displaystyle I=\int _{ 0 }^{ 1 }{ ln(\frac { 1+x }{ 1-x } )\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }

Put x=cos(θ)x=cos(\theta) to get our integral as :

0π2ln(cot2θ2)dθcosθ\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln({ cot }^{ 2 }\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }

I=(2)0π2ln(tanθ2)dθcosθ\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(tan\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }

Put tan(θ/2)=xtan(\theta/2)=x to get our integral as :

(4)01ln(t)dt1t2\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } }

I proved in my solution to problem 4 that :

01ln(t)dt1t2=π28\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } =\frac { -{ \pi }^{ 2 } }{ 8 }

Using this I get :

I=π22I=\frac{{\pi}^{2}}{2}

Problem10Problem\quad 10

Find 0sin3xx2dx\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx }

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:

PROBLEM 9

Prove that

01ln(1+x1x)dxx1x2=π22\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi^2}{2}

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

Solution -Problem - 9 We substitute, t=1+x1x    x=1t1+t t = \frac{1+x}{1-x} \implies x = \frac{1-t}{1+t} Doing the substitution and , simplifying , gives, I=01ln(t)(t)(1t)dt I = - \int_{0}^{1} \frac{\ln(t)}{(\sqrt{t})(1-t)} dt Now we substitute, t=sin2(x) t=\sin^2(x) I=40π2ln(sin(x))cos(x)dx I = -4\int_{0}^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)} dx Using, Problem -44 I=4×π28=π22 I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2}

Shivang Jindal - 5 years, 3 months ago

Log in to reply

I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

@Ronak Agarwal Lol, time difference of 5mins. I think i was typing solution when you posted the solution .

Shivang Jindal - 5 years, 3 months ago

Log in to reply

@Shivang Jindal You can try my posted question. @Shivang Jindal

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

I am able to write the integral as a series - I=πn=0(2n)!22n(n!)2(2n+1)I=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}

Pratik Shastri - 5 years, 3 months ago

Log in to reply

Solution of Problem 7:

Proposition :

0πdxp+cosxdx=πp21\begin{aligned} \int_0^{\pi} \frac{dx}{p+\cos x}\,dx=\frac{\pi}{\sqrt{p^2-1}} \end{aligned}

Proof :

It can be proven by using Weierstrass substitution: t=tan(x2)t=\tan\left(\dfrac{x}{2}\right), then

0πdxp+cosxdx=02p+1+(p1)t2dtt=p+1p1tany=2p21arctant  0=πp21Q.E.D.\begin{aligned} \\ \int_0^{\pi} \frac{dx}{p+\cos x}\,dx &=\int_0^{\infty} \frac{2}{p+1+(p-1)t^2}\,dt\qquad\Rightarrow\qquad t=\sqrt{\frac{p+1}{p-1}}\tan y\\ &=\left.\frac{2}{\sqrt{p^2-1}}\arctan t\;\right|_0^{\infty}\\ &=\frac{\pi}{\sqrt{p^2-1}}\qquad\qquad\text{Q.E.D.} \end{aligned}

Differentiating the proposition w.r.t. pp twice and setting p=10p=\sqrt{10}, we have

2p20π1p+cosxdx=2p2[πp21]0π2(p+cosx)3dx=π(2p2+1)(p21)50π1(10+cosx)3dx=7π162\begin{aligned} \\ \frac{\partial^2}{\partial p^2}\int_0^\pi\frac{1}{p+\cos x}\, dx&=\frac{\partial^2}{\partial p^2}\left[\frac{\pi}{\sqrt{p^2-1}}\right]\\ \int_0^\pi\frac{2}{\left(p+\cos x\right)^3}\, dx&=\frac{\pi\left(2p^2+1\right)}{\sqrt{\left(p^2-1\right)^5}}\\ \int_0^\pi\frac{1}{\left(\sqrt{10}+\cos x\right)^3}\, dx&=\frac{7\pi}{162} \end{aligned}

Problem 8 :

Prove

0π2dx1+8sin2(tanx)=π6(2e2+12e21) \int_0^{\Large\frac{\pi}{2}}\frac{dx}{1+8\sin^2(\tan x)}=\frac{\pi}{6}\left(\frac{2e^2+1}{2e^2-1}\right)

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

Sorry for very late response. Using elementary techniques,-

Solution - Problem 7

0πdxpcosx \displaystyle \int_{0}^{\pi} \dfrac{dx}{p - cosx}

2I=0π2p dxp2cos2x 2I = \displaystyle \int_{0}^{\pi} \dfrac{2p~dx}{p^2 - cos^2x}

I=0πp dxp2(1+tan2x)1 I = \displaystyle \int_{0}^{\pi} \dfrac{p~dx}{p^2(1 + tan^2x) - 1}

I=p0πsec2x dxp2tan2x+(p21)2 I = p \displaystyle \int_{0}^{\pi} \dfrac{sec^2x~dx}{p^2tan^2x + \sqrt{(p^2 - 1)^2}}

tanx = t

I=2p0dtp2t2+(p21)2 I = 2p \displaystyle \int_{0}^{\infty} \dfrac{dt}{p^2t^2 + \sqrt{(p^2 - 1)^2}}

I=2pp21×1p[0tan1ptp21 I = \dfrac{2p}{ \sqrt{p^2 - 1}} \times \dfrac{1}{p} \Big[_{0}^{\infty} tan^{-1} \dfrac{pt}{ \sqrt{p^2 - 1}}

I=πp21 I = \dfrac{\pi}{ \sqrt{p^2 - 1}}

U Z - 5 years, 1 month ago

Log in to reply

Solution of Problem 8:

Substitute tanxx\tan x\mapsto x, then the integral is:

0dx(1+x2)(1+8sin2x)=0dx(1+x2)(54cos(2x))\int_0^{\infty} \frac{dx}{(1+x^2)(1+8\sin^2x)}=\int_0^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}

=011+x2(13(1+2k=1cos(2kx)2k))=π6+23k=112k0cos(2kx)1+x2dx=\int_0^{\infty} \frac{1}{1+x^2}\left(\frac{1}{3}\left(1+2\sum_{k=1}^{\infty} \frac{\cos(2kx)}{2^k}\right)\right)=\frac{\pi}{6}+\frac{2}{3}\sum_{k=1}^{\infty} \frac{1}{2^k}\int_0^{\infty} \frac{\cos(2kx)}{1+x^2}\,dx

=π6+π3k=1(12e2)k=π6+π312e21=π6(2e2+12e21)=\frac{\pi}{6}+\frac{\pi}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2e^2}\right)^k=\frac{\pi}{6}+\frac{\pi}{3}\frac{1}{2e^2-1}=\boxed{\dfrac{\pi}{6}\left(\dfrac{2e^2+1}{2e^2-1}\right)}

I have used the following result:

0cos(mx)x2+a2dx=π2aeam\int_0^{\infty} \frac{\cos(mx)}{x^2+a^2}\,dx=\frac{\pi}{2a}e^{-am}

I cannot think of a challenging problem at the moment, I request somebody else to post one. Thanks!

Pranav Arora - 5 years, 3 months ago

Log in to reply

Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function cos(2kx)2k\dfrac{\cos (2kx)}{2^k} is positive on (0,)(0,\infty)? PS : you forgot to write the dx\mathrm{d}x in the third step.

Pratik Shastri - 5 years, 3 months ago

Log in to reply

@Pratik Shastri I don't know what did you mean by positive/negative on (0,)(0,\infty), but if you meant to swap between integral & summation sign, it is valid because cos(2kx)1+x2\frac{\cos(2kx)}{1+x^2} is continuous, finite & integrable, therefore swapping those two signs can be justified by Fubini's theorem.

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

@Anastasiya Romanova Right. I confused it with Tonelli's theorem.

Pratik Shastri - 5 years, 3 months ago

Log in to reply

@Anastasiya Romanova If pranav refuses to put up a question then who put will put it up @Anastasiya Romanova .

Ronak Agarwal - 5 years, 3 months ago

Log in to reply

@Ronak Agarwal According to the rule 4, the one is the last solver and it turns out that one is me. See PROBLEM 9 below, I've just proposed it.

Anastasiya Romanova - 5 years, 3 months ago

Log in to reply

Solution of Problem 5 :

I will prove the general form of

0xm11+xndx=πnsinmπn\int_0^\infty\frac{x^{m-1}}{1+x^n}\,dx=\frac{\pi}{n\sin\frac{m\pi}{n}}

It can easily be proven by taking substitution y=11+xn\displaystyle y=\frac{1}{1+x^n} and the integral becomes Beta function

1n01y1mn1 (1y)mn1dy=Γ(1mn)Γ(mn)n=πnsinmπn\frac{1}{n}\int_0^1 y^{\large 1-\frac{m}{n}-1}\ (1-y)^{\large \frac{m}{n}-1}\,dy=\frac{\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)}{n}=\frac{\pi}{n\sin\frac{m\pi}{n}}

where the last part comes from Euler's reflection formula for the gamma function. Hence, by setting m=1m=1, we have

011+xndx=πnsin(πn)=πncsc(πn)\int_0^\infty\frac{1}{1+x^n}\,dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)