# Thread: Higher Level Integration Marathon & Questions

1. ## Re: Extracurricular Integration Marathon

The good old tedious integral of 1/(1+t^4) haha.

2. ## Re: Extracurricular Integration Marathon

Originally Posted by InteGrand
The good old tedious integral of 1/(1+t^4) haha.
$\noindent It's not very tedious. \\ 2I = \int \frac{1+t^2+1-t^2}{1+t^4}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt\\2I = \int \frac{\text{d}(t-\frac{1}{t})}{(t-\frac{1}{t})^2+2} + \int \frac{\text{d}(t+\frac{1}{t})}{(t+\frac{1}{t})^2-2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+\frac{1}{2\sqrt{2}}\log\left( \frac{t^2 + \sqrt{2}t+1}{t^2-\sqrt{2}t+1}\right)$

3. ## Re: Extracurricular Integration Marathon

$\noindent It's not very tedious. \\ 2I = \int \frac{1+t^2+1-t^2}{1+t^4}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt\\2I = \int \frac{\text{d}(t-\frac{1}{t})}{(t-\frac{1}{t})^2+2} + \int \frac{\text{d}(t+\frac{1}{t})}{(t+\frac{1}{t})^2-2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)-\frac{1}{2\sqrt{2}}\log\left(\frac{t^2 - \sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)$
Ah right, I was thinking of the one with a t2 on the numerator too (the one that comes about when doing integral of √(tan x) ). That one was more tedious iirc. (Split into partial fractions etc.)

4. ## Re: Extracurricular Integration Marathon

Originally Posted by InteGrand
Ah right, I was thinking of the one with a t2 on the numerator too (the one that comes about when doing integral of √(tan x) ). That one was more tedious iirc. (Split into partial fractions etc.)
No, that one is amenable to double symmetric substitution as well.

$Any integrals of the form \frac{2nx^{n-1}}{x^{4n}+1} and \frac{2nx^{3n-1}}{x^{4n}+1} are equally trivial.$

5. ## Re: Extracurricular Integration Marathon

Originally Posted by seanieg89
$4. \lim_{T\rightarrow \infty}\int_{-T}^T \sin(x^2)\, dx$
Let $B_R(t)=Re^{it}$ for $0 \leq t \leq \frac{\pi}{4}$

Then $B'_R(t)=Rie^{it}$

So we have:

$|\int_{B_R} e^{-z^2}dz| \leq \int_0^{\frac{\pi}{4}}|e^{-(Re^{it})^2}Rie^{it}|dt =\int_0^{\frac{\pi}{4}} Re^{-R^2cos(2t)} dt \to 0 ~as~ R \to \infty$

$\int_{B_R} e^{-z^2}dz \to 0 ~as~ R \to \infty$

To make things neater we will define from now on $f(z)=e^{-z^2}$

We then integrate counter-clockwise about a boundary formed by $B_R$ to make a circular arc and get:

$\int_0^R e^{-x^2} dx + \int_{B_R} f(z) dz - \int_0^R f(xe^{\frac{i \pi}{4}}) dx = 0$

Now take the limit as $R \to \infty$

$\int_0^{\infty} e^{-x^2} dx = e^{\frac{i \pi}{4}} \int_0^{\infty} f(xe^{\frac{i \pi}{4}}) dx = \frac{1+i}{\sqrt{2}} \int_0^{\infty} f(xe^{\frac{i \pi}{4}})$

Now from a previously answered integral we know that:

$\int_0^{\infty} e^{-x^2} dx =\frac{\sqrt{\pi}}{2}$

So we then have:

$\sqrt{\frac{\pi}{4}} = \frac{1+i}{\sqrt{2}} \int_0^{\infty} f(xe^{\frac{i \pi}{4}}) = \frac{1+i}{\sqrt{2}} \int_0^{\infty} e^{-ix^2} dx$

Using Euler's formula we obtain:

$\sqrt{\frac{\pi}{2}}=\int_0^{\infty} cos(x^2)+sin(x^2)+icos(x^2)-isin(x^2) dx$

Since the imaginary part = 0 we get:

$\int_0^{\infty} cos(x^2) dx = \int_0^{\infty} sin(x^2) dx$

Now equating the real part gives:

$\int_0^{\infty} cos(x^2) dx + \int_0^{\infty} sin(x^2) dx = \sqrt{\frac{\pi}{2}}$

$2\int_0^{\infty} sin(x^2) dx =\sqrt{\frac{\pi}{2}}$

$\int_0^{\infty} sin(x^2) dx =\sqrt{\frac{\pi}{8}}$

Then we just double the answer cos even function to get:

$\int_{-\infty}^{\infty}sin(x^2) dx = \sqrt{\frac{\pi}{2}}$

6. ## Re: Extracurricular Integration Marathon

$\noindent This was posed as a challenge by my friend.\\He hasn't solved it himself, he got the answer using Sage. \\ \int_{-1}^{1} \frac{\text{d}x}{x}\sqrt{\frac{1+x}{1-x}}\log{\left(\frac{1+2x+2x^2}{1-2x+2x^2}\right)}=4\pi\cot^{-1}{\sqrt{\phi}}$

7. ## Re: Extracurricular Integration Marathon

$\noindent This was posed as a challenge by my friend.\\He hasn't solved it himself, he got the answer using Sage. \\ \int_{-1}^{1} \frac{\text{d}x}{x}\sqrt{\frac{1+x}{1-x}}\log{\left(\frac{1+2x+2x^2}{1-2x+2x^2}\right)}=4\pi\cot^{-1}{\sqrt{\phi}}$
I've seen the solution to that integral before, good luck everyone.
 Spoiler (rollover to view): 3 substitutions and residue theorem

8. ## Re: Extracurricular Integration Marathon

Originally Posted by KingOfActing
I've seen the solution to that integral before, good luck everyone.
Is that the most straightforward way to do it?

Surely there must be more straightforward ways.

9. ## Re: Extracurricular Integration Marathon

Is that the most straightforward way to do it?

Surely there must be more straightforward ways.
It's the way I've seen it done, there could be a simpler way

10. ## Re: Extracurricular Integration Marathon

Here's a very deceptive integral which appears simple.

$\int_0^\pi\frac{x^2 \text{d}x}{\sqrt{5}-2\cos{x}}$
$\noindent We will make use of the following result\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.$

11. ## Re: Extracurricular Integration Marathon

Here's a very deceptive integral which appears simple.

$\int_0^\pi\frac{x^2 \text{d}x}{\sqrt{5}-2\cos{x}}$
$\noindent We will evaluate a more general integral first. To do this we will make use of the following result\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.$

\noindent To prove this result, consider\\\begin{align*}(1 - 2a \cos x + a^2) \times \sum^\infty_{k = 0} a^k \cos (kx) &= \sum^\infty_{k = 0} a^k \cos (kx) - 2 \sum^\infty_{k = 0} a^{k + 1} \cos (kx) \cos x + \sum^\infty_{k = 0} a^{k + 2} \cos (kx)\\&= \sum^\infty_{k = 0} a^k \cos (kx) - 2 \sum^\infty_{k = 1} a^k \cos [(k - 1)x] \cos x + \sum^\infty_{k = 2} a^k \cos [(k - 2)x]\\&= 1 + a \cos x + \sum^\infty_{k = 2} a^k \cos (kx) - 2a \cos x - \sum^\infty_{k = 2} 2a^k \cos [(k - 1)x] \cos x + \sum^\infty_{k = 2} a^k \cos [(k - 2) x]\\&= 1 - a \cos x + \sum^\infty_{k = 2} a^k \left [\cos(kx) - 2 \cos [(k - 1)x] \cos x + \cos [(k - 2)x] \right ]\end{align*}

$\noindent Making use of the result 2 \cos \alpha \cos \beta = \cos (\alpha - \beta) + \cos (\alpha + \beta).\\If we set \alpha = (k - 1) x and \beta = x we have\\2 \cos [(k - 1)x] \cos x = \cos[(k - 2)x] + \cos (kx).$

$\noindent So the term within the summation is zero and we have\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2},\\as desired.$

$\noindent Now to make use of this is our integral, we need to make the \cos x term in the numerator of the right hand term of the above expression "disappear." So$

\begin{align*} \sum^\infty_{k = 0} a^k \cos (kx) &= \frac{1 - a \cos x}{1 - 2a \cos x + a^2}\\1 + \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{1 - a \cos x}{1 - 2a \cos x + a^2}\\2 + 2 \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{2 - 2 \cos x}{1 - 2a \cos x + a^2}\\\Rightarrow 1 + 2 \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{2 - 2a \cos x}{1 - 2a \cos x + a^2} - 1\\\Rightarrow 1 + 2 \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{1 - a^2}{1 - 2a \cos x + a^2}\\\Rightarrow \frac{1}{1 - a^2} \left [1 + 2 \sum^\infty_{k = 1} a^k \cos (kx) \right ] &= \frac{1}{1 - 2a \cos x + a^2}\end{align*}

$\noindent Multiplying by x^2 and integrating from 0 to \pi both sides of the above expression we have$

\begin{align*} \int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx &= \frac{1}{1 - a^2} \int^\pi_0 x^2 \, dx + \frac{2}{1 - a^2} \sum^\infty_{k = 1} a^k \int^\pi_0 x^2 \cos (kx) \, dx\end{align*}

$\noindent Now$

\begin{align*} \int^\pi_0 x^2 \, dx = \frac{\pi^3}{3}\end{align*}

$\noindent and$

\begin{align*}\int^\pi_0 x^2 \cos (kx) \, dx &= \frac{2 \pi}{k^2} (-1)^k \quad \mbox{(using IBP twice)}\end{align*}

\noindent Thus\\\begin{align*} \int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx &= \frac{1}{1 - a^2} \left [\frac{\pi^3}{3} + 4\pi \sum^\infty_{k = 1} \frac{(-a)^k}{k^2} \right ].\end{align*}

$\noindent Now the \textit{polylogarithm} function \mbox{Li}_s (x) is defined as \mbox{Li}_s (x) \sum^\infty_{k = 1} \frac{x^k}{k^s}.$

$\noindent Thus \sum^\infty_{k = 1} \frac{(-a)^k}{k^2} = \mbox{Li}_2 (-a), a \textit{dilogarithmic} function. So finally we have$

\begin{align*} \int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx &= \frac{1}{1 - a^2} \left [\frac{\pi^3}{3} + 4 \pi \,\mbox{Li}_2 (-a) \right ] \quad (*)\end{align*}

\noindent Returning to the original integral, writing this as \\\begin{align*}\int^\pi_0 \frac{x^2}{\sqrt{5} - 2 \cos x} \, dx = \frac{1}{\sqrt{5}} \int^\pi_0 \frac{x^2}{1 - \frac{2}{\sqrt{5}} \cos x} \, dx \quad (**)\end{align*}

\noindent Now the integral appearing in (*) can be rewritten as\\\begin{align*}\int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx = \frac{1}{1 + a^2} \int^\pi_0 \frac{x^2}{1 - \frac{2a}{1 + a^2} \cos x} dx \quad (***) \end{align*}

$\noindent Comparing (**) with (***), we set \frac{a}{1 + a^2} = \frac{1}{\sqrt{5}} \,\, \Rightarrow \,\, a^2 - a\sqrt{5} + 1 = 0. Thus a = \frac{\sqrt{5} - 1}{2} = \frac{1}{\varphi}, since |a| < 1. Here \varphi is the \textit{golden ratio}.$

\noindent Thus\\\begin{align*}\frac{1}{\sqrt{5}} \int^\pi_0 \frac{x^2}{1 - \frac{2}{\sqrt{5}} \cos x} \, dx &= \frac{1 + \varphi^{-2}}{\sqrt{5}} \int^\pi_0 \frac{x^2}{1 + 2 \varphi^{-1} \cos x + \varphi^{-2}} dx\\\Rightarrow \int^\pi_o \frac{x^2}{\sqrt{5} - 2 \cos x} \, dx &= \frac{1 + \varphi^{-2}}{\sqrt{5}} \cdot \frac{1}{1 - \varphi^{-2}} \left [\frac{\pi^3}{3} + 4\pi \, \mbox{Li}_{2} \left (-\frac{1}{\varphi} \right ) \right ].\end{align*}

$\noindent Now \frac{1 + \varphi^{-2}}{1 - \varphi^{-2}} = \sqrt{5} and \mbox{Li}_2 \left (-\frac{1}{\varphi} \right ) = -\frac{\pi^2}{15} + \frac{1}{2} \ln^2 \varphi.$

\noindent So finally we have\\\begin{align*}\int^\pi_0 \frac{x^2}{\sqrt{5} -2 \cos x} \, dx &= \frac{\pi^3}{3} + 4\pi \left (-\frac{\pi^2}{15} + \frac{1}{2} \ln^2 \varphi \right ) = \frac{\pi^3}{15} + 2\pi \ln^2 \varphi.\end{align*}

12. ## Re: Extracurricular Integration Marathon

$\noindent We will make use of the following result\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.$
Excellent solution to the integral .

One minor remark is that another way of proving the summation result used is taking the real part of the geometric series summation (common ratio z=a*cis(x)).

I.e. find the real part of 1/(1-a*cis(x)), which is quite quick.

13. ## Re: Extracurricular Integration Marathon

Originally Posted by seanieg89
Excellent solution to the integral .

One minor remark is that another way of proving the summation result used is taking the real part of the geometric series summation (common ratio z=a*cis(x)).

I.e. find the real part of 1/(1-a*cis(x)), which is quite quick.
Yes, I know, but I was trying to keep the problem "purely real" for the benefit of any MX2 students who may care to read this thread (though those who are reading this thread probably already know or could follow the complex way anyway).

14. ## Re: Extracurricular Integration Marathon

Yes, I know, but I was trying to keep the problem "purely real" for the benefit of any MX2 students who may care to read this thread (though those who are reading this thread probably already know or could follow the complex way anyway).
Don't bother, anyone who can follow this thread on the real analysis side probably has what it takes to follow the complex analysis side.

Anyway, nice answer. Here's the one I got from the place I found the problem.

$\noindent Define J(a) = \int_{-\pi}^{\pi} \frac{e^{iax}}{\sqrt{5}-2\cos{x}}dx\\The integral we are after is -\frac{1}{2}J''(0)=\int_0^\pi \frac{x^2}{\sqrt{5}-2\cos{x}}dx\\Let us consider the following contour integral over \mathbb{C}.\\ \oint_{C} \frac{z^a}{z^2 -\sqrt{5}z+1} where C is a unit keyhole circle on the negative real axis.\\By the residue theorem, this is equal to -2i\pi\phi^a\\But this integral is also equal to -iJ(a)+2i\sin{\pi a}\int_0^1 \frac{x^a}{x^2 +\sqrt{5}x+1} \\ The section about the center of the circle vanishes, leaving us with \\ J(a) = 2\pi \phi^a +2\sin{\pi a}\int_0^1 \frac{x^a}{x^2+\sqrt{5}x+1}dx \\With a little computation, we have -\frac{1}{2}J''(0) = -\pi\log^2{\phi}-2\pi\int_0^1 \frac{\log{x}}{x^2+\sqrt{5}x+1}dx \\Using the facts:\\ \frac{1}{x^2 + \sqrt{5}x+1} \equiv \frac{1}{x+\phi} - \frac{1}{x+\frac{1}{\phi}}\; ; \; \int_0^1 \frac{\log{x}}{x+a}dx = Li_2\left(-\frac{1}{a}\right)$

$\noindent Li_2\left(-\frac{1}{\phi}\right) = -\frac{\pi^2}{10}-\log^2 \phi \; ; \; Li_2(-\phi) = -\frac{\pi^2}{15}+\frac{1}{2}\log^2 \phi\\We conclude -\frac{1}{2}J''(0) = -\pi\log^2 \phi -2\pi\left[\left( -\frac{\pi^2}{10}-\log^2 \phi \right)-\left(-\frac{\pi^2}{15}+\frac{1}{2}\log^2 \phi \right)\right]\\Hence \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos{x}}dx=2\pi\log^2 \phi+\frac{\pi^3}{15}$

15. ## Re: Extracurricular Integration Marathon

$\int_0^\infty x^3 e^{-x^2}\log(1-e^{-x^2})\, dx$

16. ## Re: Extracurricular Integration Marathon

Extra question:

$\int _{ 0 }^{ \infty }{ f\left( x+\frac { 1 }{ x } \right) \frac { \ln { x } }{ x } dx } \\ assuming f(x) is bound, non-negative$

17. ## Re: Extracurricular Integration Marathon

Originally Posted by seanieg89
$\int_0^\infty x^3 e^{-x^2}\log(1-e^{-x^2})\, dx$
\noindent We start first with a preliminary result. For n \geqslant 1 using IBP it can be shown that\\\begin{align*}\int^1_0 x^n \ln x \, dx &= - \frac{1}{(n + 1)^2}.\end{align*}

\noindent Now, let u = 1 - e^{-x^2}, du = 2x e^{-x^2} \, dx \Rightarrow x^2 = - \ln (1 - u) and x e^{-x^2} \, dx = \frac{du}{2}. For the limits of integration we have: x = 0, u = 0 and x \rightarrow \infty, u \rightarrow 1. So\\\begin{align*}\int^\infty_0 x^3 e^{-x^2} \ln (1 - e^{-x^2}) \, dx &= \int^\infty_0 x^2 \cdot \ln (1 - e^{-x^2}) \cdot x e^{-x^2} \, dx = -\frac{1}{2} \int^1_0 \ln u \ln (1 - u) \, du\end{align*}

$\noindent And as \\\ln (1 - u) = -\sum^\infty_{n = 1} \frac{x^n}{n} \,\,\, \mbox{for} \,\,\, 0 \leqslant x < 1,\\after interchanging the summation with integration we can rewrite the integral as$

\begin{align*}\int^\infty_0 x^3 e^{-x^2} \ln (1 - e^{-x^2}) \, dx &= \frac{1}{2} \sum^\infty_{n = 1} \frac{1}{n} \int^1_0 u^n \ln u \, du\\ &= -\frac{1}{2} \sum^\infty_{n = 1} \frac{1}{n(n + 1)^2}\\&= -\frac{1}{2} \sum^\infty_{n = 1} \left [\frac{1}{n} - \frac{1}{n + 1} - \frac{1}{(n + 1)^2} \right ] \,\,\, \mbox{(partial fraction decomposition)}\\&= -\frac{1}{2} \sum^\infty_{n = 1} \left (\frac{1}{n} - \frac{1}{n + 1} \right ) + \frac{1}{2} \sum^\infty_{n = 1} \frac{1}{(n + 1)^2}\\&= -\frac{1}{2} \sum^\infty_{n = 1} \left (\frac{1}{n} - \frac{1}{n + 1} \right ) + \frac{1}{2} \sum^\infty_{n = 2} \frac{1}{n^2} \,\,\, \mbox{(shifting the index)}\\&= -\frac{1}{2} \sum^\infty_{n = 1} \left (\frac{1}{n} - \frac{1}{n + 1} \right ) + \frac{1}{2} \left [\left (1 + \sum^\infty_{n =2} \frac{1}{n^2} \right ) - 1 \right ]\\&= -\frac{1}{2} \sum^\infty_{n = 1} \left (\frac{1}{n} - \frac{1}{n + 1} \right ) + \frac{1}{2} \sum^\infty_{n = 1} \frac{1}{n^2} - \frac{1}{2}\end{align*}.

$\noindent Now the first sum telescopes and has a sum equal to 1 while the second sum is the famous Basel problem and has a sum equal to \frac{\pi^2}{6}. So$

\begin{align*}\int^\infty_0 x^3 e^{-x^2} \ln (1 - e^{-x^2}) \, dx &= -\frac{1}{2} + \frac{1}{2} \cdot \frac{\pi^2}{6} - \frac{1}{2} = \frac{\pi^2}{12} - 1.\end{align*}

18. ## Re: Extracurricular Integration Marathon

Originally Posted by leehuan
Extra question:

$\int _{ 0 }^{ \infty }{ f\left( x+\frac { 1 }{ x } \right) \frac { \ln { x } }{ x } dx } \\ assuming f(x) is bound, non-negative$
$\noindent Let x = e^u, dx = e^u \, du. For the limits of integration we have: x = 0, u \rightarrow -\infty and x \rightarrow \infty, u \rightarrow \infty. So$

\begin{align*} \int^\infty_0 f \left (x + \frac{1}{x} \right ) \frac{\ln x}{x} \, dx &= \int^\infty_{-\infty} f (e^u + e^{-u}) \cdot \frac{\ln e^u}{e^u} \cdot e^u \, du\\&= \int^\infty_{-\infty} u f(2 \cosh u) u \, du\\&= 0,\end{align*}\\ as the integrand is an odd function between symmetric limits.

19. ## Re: Extracurricular Integration Marathon

Originally Posted by InteGrand
Here's another nice question.

$\noindent Find \lim _{n\to \infty} \int _0 ^\infty \frac{\mathrm{e}^{-t}\cos t}{\frac{1}{n} + nt^2} \text{ d}t.$
\begin{align*}\lim _{n\to \infty} \int _0 ^\infty \frac{\mathrm{e}^{-t}\cos t}{\frac{1}{n} + nt^2} \text{ d}t &= \lim_{n \rightarrow \infty} \frac{1}{n} \int^\infty_0 \frac{e^{-t} \cos t}{\frac{1}{n^2} + t^2} \, dt\\&= \lim_{n \rightarrow \infty} \frac{1}{n} \left [ \left (n \tan^{-1} (nt) e^{-t} \cos t \right )^\infty_0 - \int^\infty_0 n \tan^{-1} (n t) d(e^{-t} \cos t) \, dt \right ]\,\,\, \mbox{(using IBP)}\\&= -\lim_{n \rightarrow \infty} \int^\infty_0 \tan^{-1} (nt) d(e^{-t} \cos t) \, dt\\&= -\int^\infty_0 \left [\lim_{n \rightarrow \infty} \tan^{-1} (nt) \right ] d(e^{-t} \cos t) \, dt\\&= -\frac{\pi}{2} \int^\infty_0 d(e^{-t} \cos t) \, dt\\&= -\frac{\pi}{2} \left [e^{-t} \cos t \right ]^\infty_0\\&=\frac{\pi}{2}\end{align*}

20. ## Re: Extracurricular Integration Marathon

$\noindent \textbf{Two New Questions}$

\noindent Evalaute\\(a) \begin{align*}\int^{\frac{\pi}{2}}_0 \cot x \ln (\sec x) \, dx \,\,\,\,\,\,\,&\mbox{(b)} \,\, \int^1_0 \frac{\tan^{-1} (\sqrt{2 + x^2})}{(1 + x^2) \sqrt{2 + x^2}} \, dx\end{align*}

21. ## Re: Extracurricular Integration Marathon

$a) \\ \\ I=-\int_0^{\pi/2}\frac{\cos{x}\log(\cos{x})}{\sin{x}}\, dx\\ \\ =-\int_0^1 \frac{u\log{u}}{1-u^2}\, du\\ \\ = -\int_0^1 \sum_{k\geq 0}u^{2k+1}\log{u}\, du\\ \\ = -\sum_{k\geq 0}\int_0^1 u^{2k+1}\log{u}\, du\\ \\ = \sum_{k\geq 0}\frac{1}{(2k+2)^2}\\ \\ = \frac{\zeta(2)}{4}=\frac{\pi^2}{24}.$

22. ## Re: Extracurricular Integration Marathon

Originally Posted by seanieg89
$a) \\ \\ I=-\int_0^{\pi/2}\frac{\cos{x}\log(\cos{x})}{\sin{x}}\, dx\\ \\ =-\int_0^1 \frac{u\log{u}}{1-u^2}\, du\\ \\ = -\int_0^1 \sum_{k\geq 0}u^{2k+1}\log{u}\, du\\ \\ = -\sum_{k\geq 0}\int_0^1 u^{2k+1}\log{u}\, du\\ \\ = \sum_{k\geq 0}\frac{1}{(2k+2)^2}\\ \\ = \frac{\zeta(2)}{4}=\frac{\pi^2}{24}.$
Beat me to it by nine hours But my solution involved factorising out the 1/4 so basel's could be seen easily. But my main problem is that I can't get rid of the negative that appears in the series expansion for log(1-x). Here's my solution, which ignores the sign error...

$\noindent 2I = \int_0^{\frac{\pi}{2}} \frac{\cos{x}}{\sin{x}} \log{(1-\sin^2{x})} dx \\ 4I = \int _0^{\frac{\pi}{2}} \frac{2\sin{x}\cos{x}}{\sin^2{x}} \log{(1-\sin^2{x})} dx = \int _0^1 \frac{1}{v}\log{(1-v)}dv \\ 4I = \int_0^1 \sum_{r=1}^\infty \frac{v^{r-1}}{r} dv = \sum_{r=1}^\infty \int_0^1 \frac{v^{r-1}}{r} dv = \sum_{r=1}^\infty \frac{1}{r^2} = \zeta(2) \\ I = \frac{\pi^2}{24}$

WHERE DID THE MISSING SIGN GO?!?!?!

23. ## Re: Extracurricular Integration Marathon

$Use Inequalities to show that for all values of x in the interval -1 \leq x \leq 1 the functions:$

$P_{n}(x)= \frac{1}{\pi}\int_0^{\pi}{(x+i\sqrt{1-x^2}\cos{\theta})^n}d\theta$

$satisfies the inequality |P_{n}(x)| \leq 1$

24. ## Re: Extracurricular Integration Marathon

Beat me to it by nine hours But my solution involved factorising out the 1/4 so basel's could be seen easily. But my main problem is that I can't get rid of the negative that appears in the series expansion for log(1-x). Here's my solution, which ignores the sign error...

$\noindent 2I = \int_0^{\frac{\pi}{2}} \frac{\cos{x}}{\sin{x}} \log{(1-\sin^2{x})} dx \\ 4I = \int _0^{\frac{\pi}{2}} \frac{2\sin{x}\cos{x}}{\sin^2{x}} \log{(1-\sin^2{x})} dx = \int _0^1 \frac{1}{v}\log{(1-v)}dv \\ 4I = \int_0^1 \sum_{r=1}^\infty \frac{v^{r-1}}{r} dv = \sum_{r=1}^\infty \int_0^1 \frac{v^{r-1}}{r} dv = \sum_{r=1}^\infty \frac{1}{r^2} = \zeta(2) \\ I = \frac{\pi^2}{24}$

WHERE DID THE MISSING SIGN GO?!?!?!
$\noindent For the missing sign, your first line would have the R.H.S. have a 2\ln (\cos x) in the integrand, but the original integral had \sec, not \cos, so you need to put in a minus sign in your first line to make it equal a \ln (\sec x), as \ln (\sec x) =-\ln (\cos x).$

25. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
$Use Inequalities to show that for all values of x in the interval -1 \leq x \leq 1 the functions:$

$P_{n}(x)= \frac{1}{\pi}\int_0^{\pi}{(x+i\sqrt{1-x^2}\cos{\theta})^n}d\theta$

$satisfies the inequality |P_{n}(x)| \leq 1$
Minkowski Inequality easily destroys the odd valued cases, but is itself destroyed by the even valued cases.

I hope that's what you meant by inequality, because olympiad inequalities don't look any more viable than brute force.

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