Higher Level Integration Marathon & Questions (1 Viewer)

leehuan · Jan 8, 2016

Higher Level Integration Marathon & Questions
This is a marathon thread for integration (mainly numerical or computation of integrals including manipulation of integrals).

Please aim to pitch your questions for first-year/second-year university level maths, although not necessary. Excelling & gifted/talented secondary school students are also invited to contribute.

Note 1: Please do not post HIGH SCHOOL integration (ext 1 or ext 2) related questions. Use the respective Maths Ext 1 & Maths Ext 2 forums instead.
Note 2: Questions involving theorems such as the FTC are better suited to the Calculus & Analysis marathon not this one.

(mod edit 7/6/17 by dan964)
===============================

This thread is for people to bombard with integrals that ARE allowed to stretch beyond the MX2 syllabus and into the world of real maths. And for me to spectate and join in later in the year.

I invite everyone to participate.

I will kick off with a VERY cliche question.

$\int _{ -\infty }^{ \infty }{ { e }^{ -{ x }^{ 2 } }dx }$

omegadot · Jan 8, 2016

Re: Extracurricular Integration Marathon

leehuan said:
This thread is for people to bombard with integrals that ARE allowed to stretch beyond the MX2 syllabus and into the world of real maths. And for me to spectate and join in later in the year.

I invite everyone to participate.

I will kick off with a VERY cliche question.

$\int _{ -\infty }^{ \infty }{ { e }^{ -{ x }^{ 2 } }dx }$

$$\noindent If we are allowed to use triple integrals, here is a slightly less familiar approach to this problem (the `standard' approach is one that typically uses a double integral).$

$$\noindent Let $I = \int^\infty_{-\infty} e^{-x^2} \, dx.$ So$

$\begin{align*}I^3 &= \left (\int^\infty_{-\infty} e^{-x^2} \, dx \right ) \left (\int^\infty_{-\infty} e^{-y^2} \, dy \right ) \left (\int^\infty_{-\infty} e^{-z^2} \, dz \right )\\&= \int^\infty_{-\infty} \int^\infty_{-\infty} \int^\infty_{-\infty} e^{-x^2 -y^2 - z^2} \, dx \, dy \, dz\end{align*}$

$\noindent Converting to spherical coordinates, namely $(\rho,\theta,\phi)$ such that $(x,y,z) = (\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi)$. Observing that $\rho^2 = x^2 + y^2 + z^2$, the integral becomes$

$\begin{align*}I^3 &= \int^{2 \pi}_0 \int^\pi_0 \int^\infty_0 \sin \phi \rho^2 e^{-\rho^2} \, d\rho \, d\phi \, d\theta\\&= 4\pi \int^\infty_0 \rho^2 e^{-\rho^2} \, d\rho\end{align*}$

$\noindent IBP where $\rho$ is differentiated and $\rho e^{-\rho^2}$ is integrated we have$

$\begin{align*}I^3 &= 2\pi \int^\infty_0 e^{-\rho^2} \, d\rho =\pi \int^\infty_{-\infty} e^{-\rho^2} \, d\rho,\end{align*}$

$$\noindent as the integral is even. Thus$\\\begin{align*} I^3 &= \pi I\\\Rightarrow I(I^2 - \pi) &= 0\end{align*}$

$$\noindent Since $e^{-x^2} > 0$ for all $x$, then $\int^\infty_{-\infty} e^{-x^2} \, dx > 0$. Thus$

$I = \int^\infty_{-\infty} e^{-x^2} \, dx = \sqrt{\pi}.$

$$\noindent \textbf{Alternatively}$

$$\noindent If special functions are allowed, from properties of the error function erf we can write$

$\begin{align*} \int^\infty_{-\infty} e^{-x^2} \, dx &= 2 \int^\infty_0 e^{-x^2} \, dx\\&= \sqrt{\pi} \left (\frac{2}{\sqrt{\pi}} \int^\infty_0 e^{-x^2} \, dx \right )\\&= \sqrt{\pi} \, \mbox{erf} \, (\infty)\\&=\sqrt{\pi}\end{align*}$

$\noindent since $\mbox{erf} \, (\infty) = 1$, but will completely understand if you think I have cheated here.$

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

Alternatively, use polar coordinates (the 'standard' approach I see that omegadot was alluding to).

omegadot · Jan 8, 2016

Re: Extracurricular Integration Marathon

$$\noindent \textbf{Question No. 2}$

$$\noindent Evaluate $\int^1_0 \frac{x^a - 1}{\ln x} \, dx$ where $a > 0.$

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

ln(a + 1)

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

InteGrand said:
ln(a + 1)

Which follows from differentiating under the integral (Feynman's favourite integration trick):

$I'(a)=\int_0^1 x^a=\frac{1}{a+1}, I(0)=0\\ \\ \Rightarrow I(a)=I(0)+\int_0^a I'(s)\, ds = \int_0^a \frac{ds}{s+1}.$

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
Which follows from differentiating under the integral (Feynmann's favourite integration trick):

$I'(a)=\int_0^1 x^a=\frac{1}{a+1}, I(0)=0\\ \\ \Rightarrow I(a)=I(0)+\int_0^a I'(s)\, ds = \int_0^a \frac{ds}{s+1}.$

$\noindent Haha yeah, or write the integrand of the original integral as $\int _0 ^a x^\theta \text{ d}\theta$, and then switch order of integration.$

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

InteGrand said:
$\noindent Haha yeah, or write the integrand of the original integral as $\int _0 ^a x^\theta \text{ d}\theta$, and then switch order of integration.$

Yep

.

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

$3. \lim_{T\rightarrow \infty}\int_{-T}^T \frac{\sin(x)}{x}\, dx$

$4. \lim_{T\rightarrow \infty}\int_{-T}^T \sin(x^2)\, dx$

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

One can read about Feynman's commentary on differentiating under the integral in these links:

• https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign#Popular_culture

• http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

Apparently this trick wasn't taught too much back then haha.

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

InteGrand said:
Apparently this trick wasn't taught too much back then haha.

Still isn't taught too much really, but it's pretty useful!

That's why I wrote that question like last month in the undergrad marathon about differentiating things of the form $\int_{a(t)}^{b(t)}f(x,t)\, dx$ w.r.t. t, as that is just a strengthened form of the differentiation under the integral trick, where domains are also allowed to vary.

In fact there is also such a trick when the integral is instead over a time varying domain in R^n, and it is quite useful too.

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

One more before I wait for the ones already mentioned to be answered:

5.

i) Find an expression for the volume of an n-dimensional ball of radius r. (Hint: This is like computing volumes of 3-d solids in MX2 by slices, but the cross sections of n-balls are (n-1)-balls.)

ii) Let $P(n,\epsilon)$ be the probability that a randomly selected* point in the unit n-ball has distance less than $1-\epsilon$ from the centre. Show that $P(n,\epsilon)\rightarrow 0$ as $n\rightarrow\infty$ , regardless of what $\epsilon>0$ is.

The perhaps surprising moral: Almost all of the mass of a high dimensional ball is concentrated near the boundary of this ball!

*Randomly w.r.t n-dimensional volume.

Paradoxica · Jan 8, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
$3. \lim_{T\rightarrow \infty}\int_{-T}^T \frac{\sin(x)}{x}\, dx$

$4. \lim_{T\rightarrow \infty}\int_{-T}^T \sin(x^2)\, dx$

$$\noindent$ \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin{x}}{x} = \int_{0}^{\infty} \frac{\sin{x}}{x} = \int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-xy}\sin{x}\text{d}y\right)$d$x=\int_{0}^{\infty} \left(\int_{0}^{\infty}e^{-xy}\sin{x}\text{d}x \right)$d$y \\ I= \int_0^\infty e^{-xy} \sin x $d$x = \left[-e^{-xy}{\cos x}\right]_0^\infty - y \int_0^\infty e^{-xy} \cos x $d$x \\ I =1 - \left[y e^{-xy}\sin{x}\right]_0^\infty - y^2 \int_0^\infty e^{-xy} \sin x $d$x = 1-y^2I \\ I = \frac{1}{1+y^2}\\ \frac{1}{2} \int_{-\infty}^\infty \frac{\sin{x}}{x}$d$x = \int_0^\infty \frac{\text{d}y}{1+y^2}=\frac{\pi}{2}\\\therefore \int_{-\infty}^\infty \frac{\sin{x}}{x}$d$x = \pi \\$I don't remember how the order of integration swap was justified.$\\ \int_{-\infty}^\infty \sin(x^2) $d$x = $Im$(\int_{\infty}^\infty e^{ix^2}\text{d}x)\\ \int_{-\infty}^\infty e^{-ax^2} = \sqrt{\frac{\pi}{a}}:a=-i, \int_{-\infty}^\infty e^{ix^2}=\sqrt{\frac{\pi}{-i}}=(1+i)\sqrt{\frac{\pi}{2}}$

$\noindent$\therefore \int_{-\infty}^\infty \cos{x^2} $d$x = \int_{-\infty}^\infty \sin{x^2} $d$x = \sqrt{\frac{\pi}{2}}\\$I forgot how the entire thing was justified.$

Drsoccerball · Jan 8, 2016

Re: Extracurricular Integration Marathon

I feel like this thread should of been made about 9 months later when we (2015'ers) know what were doing.

Paradoxica · Jan 8, 2016

Re: Extracurricular Integration Marathon

I'll post my own problem...

$$\noindent Prove the area bounded by the axes and the curve $e^{-x}+e^{-y}=1$ is equal to $\zeta(2)$

leehuan · Jan 8, 2016

Re: Extracurricular Integration Marathon

Drsoccerball said:
I feel like this thread should of been made about 9 months later when we (2015'ers) know what were doing.

Who cares, I'm pretty sure some other people will take advantage of this thread.

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$\noindent$ \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin{x}}{x} = \int_{0}^{\infty} \frac{\sin{x}}{x} = \int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-xy}\sin{x}\text{d}y\right)$d$x=\int_{0}^{\infty} \left(\int_{0}^{\infty}e^{-xy}\sin{x}\text{d}x \right)$d$y \\ I= \int_0^\infty e^{-xy} \sin x $d$x = \left[-e^{-xy}{\cos x}\right]_0^\infty - y \int_0^\infty e^{-xy} \cos x $d$x \\ I =1 - \left[y e^{-xy}\sin{x}\right]_0^\infty - y^2 \int_0^\infty e^{-xy} \sin x $d$x = 1-y^2I \\ I = \frac{1}{1+y^2}\\ \frac{1}{2} \int_{-\infty}^\infty \frac{\sin{x}}{x}$d$x = \int_0^\infty \frac{\text{d}y}{1+y^2}=\frac{\pi}{2}\\\therefore \int_{-\infty}^\infty \frac{\sin{x}}{x}$d$x = \pi \\$I don't remember how the order of integration swap was justified.$\\ \int_{-\infty}^\infty \sin(x^2) $d$x = $Im$(\int_{\infty}^\infty e^{ix^2}\text{d}x)\\ \int_{-\infty}^\infty e^{-ax^2} = \sqrt{\frac{\pi}{a}}:a=-i, \int_{-\infty}^\infty e^{ix^2}=\sqrt{\frac{\pi}{-i}}=(1+i)\sqrt{\frac{\pi}{2}}$

$\noindent$\therefore \int_{-\infty}^\infty \cos{x^2} $d$x = \int_{-\infty}^\infty \sin{x^2} $d$x = \sqrt{\frac{\pi}{2}}\\$I forgot how the entire thing was justified.$

For 3, the swapping of order of integration is fine can be justified by your choice of the many variants of Fubini's/Tonelli's theorem. You don't need an especially powerful one because the function is smooth and absolutely integrable.

For 4, I think people should still attempt it / try other methods, as it is much less clear why "letting a=-i" should be valid. Letting a=-1 would give us something nonsensical for example. Definitely need to say something more to justify the formula being the same as that of the Gaussian integral (and why that particular choice of square root and not the other).

seanieg89 · Jan 8, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
I'll post my own problem...

$$\noindent Prove the area bounded by the axes and the curve $e^{-x}+e^{-y}=1$ is equal to $\zeta(2)$

$A=\int_0^\infty -\log(1-e^{-x})\, dx \\ \\ = \int_0^\infty \left(\sum_{j\geq 0} \frac{e^{-x(j+1)}}{j+1}\right)\, dx \\ \\ = \sum_{j\geq 0} \left(\frac{1}{j+1}\int_0^\infty e^{-x(j+1)}\, dx\right) \\ \\ = \sum_{j\geq 0}\frac{1}{(j+1)^2}\\ \\=\zeta(2).$

(The interchange of summation and integration being justified by the monotone convergence theorem, for example.)

Paradoxica · Jan 8, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
For 3, the swapping of order of integration is fine can be justified by your choice of the many variants of Fubini's/Tonelli's theorem. You don't need an especially powerful one because the function is smooth and absolutely integrable.

For 4, I think people should still attempt it / try other methods, as it is much less clear why "letting a=-i" should be valid. Letting a=-1 would give us something nonsensical for example. Definitely need to say something more to justify the formula being the same as that of the Gaussian integral (and why that particular choice of square root and not the other).

I think the solution had something to do with selecting the right branch of the square root function in the complex plane, which justifies the analytical continuation for Re(a)>0, and then taking the limit as a tends towards -i.

Paradoxica · Jan 8, 2016

Re: Extracurricular Integration Marathon

Here's a very deceptive integral which appears simple.

$\int_0^\pi\frac{x^2 \text{d}x}{\sqrt{5}-2\cos{x}}$

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