Higher Level Integration Marathon & Questions (2 Viewers)

leehuan · Jan 17, 2016

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$

omegadot · Jan 20, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
$\int_0^\infty x^3 e^{-x^2}\log(1-e^{-x^2})\, dx$

$$\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$

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

omegadot · Jan 20, 2016

Re: Extracurricular Integration Marathon

leehuan said:
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.$$

omegadot · Jan 20, 2016

Re: Extracurricular Integration Marathon

InteGrand said:
Here's another nice question.

$\noindent Find $\lim _{n\to \infty} \int _0 ^\infty \frac{\mathrm{e}^{-t}\cos t}{\frac{1}{n} + nt^2} \text{ d}t$.$

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

omegadot · Jan 20, 2016

Re: Extracurricular Integration Marathon

$\noindent \textbf{Two New Questions}$

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

seanieg89 · Jan 20, 2016

Re: Extracurricular Integration Marathon

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

Paradoxica · Jan 20, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
$a) \\ \\ I=-\int_0^{\pi/2}\frac{\cos{x}\log(\cos{x})}{\sin{x}}\, dx\\ \\ =-\int_0^1 \frac{u\log{u}}{1-u^2}\, du\\ \\ = -\int_0^1 \sum_{k\geq 0}u^{2k+1}\log{u}\, du\\ \\ = -\sum_{k\geq 0}\int_0^1 u^{2k+1}\log{u}\, du\\ \\ = \sum_{k\geq 0}\frac{1}{(2k+2)^2}\\ \\ = \frac{\zeta(2)}{4}=\frac{\pi^2}{24}.$

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

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

Drsoccerball · Jan 24, 2016

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$

InteGrand · Jan 24, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
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})} $d$x \\ 4I = \int _0^{\frac{\pi}{2}} \frac{2\sin{x}\cos{x}}{\sin^2{x}} \log{(1-\sin^2{x})} $d$x = \int _0^1 \frac{1}{v}\log{(1-v)}$d$v \\ 4I = \int_0^1 \sum_{r=1}^\infty \frac{v^{r-1}}{r} $d$v = \sum_{r=1}^\infty \int_0^1 \frac{v^{r-1}}{r} $d$v = \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 $\ln (\sec x) =-\ln (\cos x)$.$

Paradoxica · Jan 24, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
$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.

seanieg89 · Jan 25, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
$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$

The integrand has its modulus squared equal to (x^2 + (1-x^2)cos^2(theta))^n =< (x^2 + (1-x^2))^n=1.

The triangle inequality completes the proof.

Paradoxica · Jan 25, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$\noindent \textbf{Two New Questions}$

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

The Laurent series for arctan(z) looks unpromising.... We need a different approach.

Drsoccerball · Jan 25, 2016

Re: Extracurricular Integration Marathon

When we do Cauchy's Integrals what are we supposed to picture in our heads?

Paradoxica · Jan 31, 2016

Re: Extracurricular Integration Marathon

$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} $d$x$

seanieg89 · Feb 1, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} $d$x$

$I=\int_0^\infty \frac{2x^2}{x^4+2x^2+5}\, dx=\int_{\mathbb{R}}\frac{x^2}{x^4+2x^2+5}\, dx$

This integral is clearly a prime target for contour integration. (We could factorise into its real quadratic factors, use partial fractions, and integrate the simpler summands, but contour integration seems faster).

Note that since the quartic denominator is both even and real, the poles of the integrand occur at $\alpha,-\alpha,\overline{\alpha},-\overline{\alpha}$ where $\alpha$ is an arbitrarily chosen root, say the one in the first quadrant of the the complex plane.

Now if we take a semicircular contour (radius R) with diameter on the real axis centred at the origin, and semicircular arc in the upper half-plane, then the integral around this contour (with positive orientation) is just equal to I, because the integrand decays as 1/R^2, and so the contribution from the curved segment of length O(R) tends to zero.

On the other hand, for sufficiently large R this integral is just equal to

$2\pi i (\textrm{Res}(f;\alpha)+\textrm{Res}(f;-\overline{\alpha}))\\ \\ = 2\pi i\left(\frac{\alpha^2}{2\alpha(\alpha-\overline{\alpha})(\alpha+\overline{\alpha})}+\frac{\overline{\alpha}^2}{2\overline{\alpha}(\alpha-\overline{\alpha})(\alpha+\overline{\alpha})}\right)\\ \\ =\frac{i\pi}{\alpha-\overline{\alpha}}=\frac{\pi}{2\textrm{Im}(\alpha)}$

It remains to compute $\alpha$ , which is just the principal square root of $-1+2i$ .

Writing $\alpha=x+iy$ , we get $x^2-y^2=-1,xy=1.$

The resulting biquadratic yields $x=\phi^{-1/2},y=\phi^{1/2}$ , where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.

Hence

$I=\frac{\pi}{2\sqrt{\phi}}.$

Paradoxica · Feb 16, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$\noindent \textbf{Two New Questions}$

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

$$Now, this is a wild guess, but is the second one $\frac{5\pi^2}{96} ?$

$Okay, that is the answer, but proving it seems slightly difficult.$

Paradoxica · Feb 18, 2016

Re: Extracurricular Integration Marathon

$\int_0^\infty \frac{(x+1)\ln^2{(x+1)}}{(4x^2 +8x +5)^{\frac{3}{2}}}\,$d$x$

omegadot · Feb 18, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$Now, this is a wild guess, but is the second one $\frac{5\pi^2}{96} ?$

$Okay, that is the answer, but proving it seems slightly difficult.$

$$\noindent Yes, $\frac{5\pi^2}{96}$ is the correct answer but surely you didn't just guess it?$

Paradoxica · Feb 18, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$$\noindent Yes, $\frac{5\pi^2}{96}$ is the correct answer but surely you didn't just guess it?$

#ISC+MasterRace

Drsoccerball · Feb 20, 2016

Re: Extracurricular Integration Marathon

$$ Find : $ \int_0^1{x^x}dx$

Higher Level Integration Marathon & Questions (2 Viewers)

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