Higher Level Integration Marathon & Questions

**leehuan** · 8 Jan 2016, 1:46 PM

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** · 8 Jan 2016, 3:33 PM

Originally Posted by leehuan

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** · 8 Jan 2016, 3:34 PM

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

**omegadot** · 8 Jan 2016, 3:35 PM

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

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

**InteGrand** · 8 Jan 2016, 3:39 PM

ln(a + 1)

**seanieg89** · 8 Jan 2016, 3:52 PM

Originally Posted by InteGrand

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** · 8 Jan 2016, 3:55 PM

Originally Posted by seanieg89

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** · 8 Jan 2016, 3:59 PM

Originally Posted by InteGrand

$\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** · 8 Jan 2016, 4:01 PM

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

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

**InteGrand** · 8 Jan 2016, 4:02 PM

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

• https://en.wikipedia.org/wiki/Differ...opular_culture

• http://www.math.uconn.edu/~kconrad/b...ffunderint.pdf

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

**seanieg89** · 8 Jan 2016, 4:07 PM

Originally Posted by InteGrand

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** · 8 Jan 2016, 6:04 PM

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** · 8 Jan 2016, 6:47 PM

Originally Posted by seanieg89

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

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

$$\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** · 8 Jan 2016, 7:12 PM

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

**Paradoxica** · 8 Jan 2016, 7:14 PM

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** · 8 Jan 2016, 7:33 PM

Originally Posted by Drsoccerball

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** · 8 Jan 2016, 7:41 PM

Originally Posted by Paradoxica

$$\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** · 8 Jan 2016, 7:45 PM

Originally Posted by Paradoxica

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** · 8 Jan 2016, 8:13 PM

Originally Posted by seanieg89

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** · 8 Jan 2016, 8:17 PM

Here's a very deceptive integral which appears simple.

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

**InteGrand** · 8 Jan 2016, 8:23 PM

Here's another nice question.

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

**omegadot** · 8 Jan 2016, 9:13 PM

Originally Posted by seanieg89

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).

$\noindent Okay, since the integrand is odd$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= 2 \int^\infty_0 \sin (x^2) \, dx\end{align*}$

$$\noindent Let $u = x^2, dx = \frac{du}{2 \sqrt{u}},$ while the limits of integration are unchanged.$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= \int^\infty_0 \frac{\sin u}{\sqrt{u}} \, du. \quad (*)\end{align*}$

$$\noindent Now taking advantage of the Gamma function, we note that$\\\frac{1}{u^a} = \frac{1}{\Gamma (a)} \int^\infty_0 z^{a - 1} e^{-uz} \, dz.$

$$\noindent If we set $a = \frac{1}{2}$ one has$\\\frac{1}{\sqrt{u}} = \frac{1}{\Gamma (1/2)} \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \, dz = \frac{1}{\sqrt{\pi}} \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \, dz.$

$\noindent So on writing (*) as a double integral we have$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= \frac{1}{\sqrt{\pi}} \int^\infty_0 \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \sin u \, dz \, du\\&= \frac{1}{\sqrt{\pi}} \int^\infty_0 \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \sin u \, du \, dz,\end{align*}$

$\noindent after changing the order of integration has been made and is justified by Fubini's theorem. The $u$ integral can be found by applying IBP twice. The result is$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= \frac{1}{\sqrt{\pi}} \int^\infty_0 \frac{dz}{\sqrt{z} (1 + z^2)}.\end{align*}$

$$\noindent Now let $z = t^2, dz = 2t \, dt$, while the limits of integration remain unchanged to give$\\\int^\infty_{-\infty} \sin(x^2) \, dx = \frac{2}{\sqrt{\pi}} \int^\infty_0 \frac{dt}{1 + t^4}.$

$$\noindent The last integral has been done to death in the 2016 MX2 Integration Marathon. Its value is $\frac{\pi \sqrt{2}}{4}.$

$$\noindent Thus $\int^\infty_{-\infty} \sin(x^2) \, dx = \frac{2}{\sqrt{\pi}} \cdot \frac{\pi \sqrt{2}}{4} = \sqrt{\frac{\pi}{2}}.$

**omegadot** · 8 Jan 2016, 9:36 PM

Originally Posted by Paradoxica

Here's a very deceptive integral which appears simple.

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

$$\noindent Do I smell a golden ratio? The whole $\sqrt{5}$ with 2 thing seems very suspicious to me.$

**Paradoxica** · 8 Jan 2016, 9:43 PM

Originally Posted by omegadot

$$\noindent Do I smell a golden ratio? The whole $\sqrt{5}$ with 2 thing seems very suspicious to me.$

Your nose does not deceive you.

**seanieg89** · 8 Jan 2016, 9:49 PM

Originally Posted by omegadot

$\noindent Okay, since the integrand is odd$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= 2 \int^\infty_0 \sin (x^2) \, dx\end{align*}$

$$\noindent Let $u = x^2, dx = \frac{du}{2 \sqrt{u}},$ while the limits of integration are unchanged.$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= \int^\infty_0 \frac{\sin u}{\sqrt{u}} \, du. \quad (*)\end{align*}$

$$\noindent Now taking advantage of the Gamma function, we note that$\\\frac{1}{u^a} = \frac{1}{\Gamma (a)} \int^\infty_0 z^{a - 1} e^{-uz} \, dz.$

$$\noindent If we set $a = \frac{1}{2}$ one has$\\\frac{1}{\sqrt{u}} = \frac{1}{\Gamma (1/2)} \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \, dz = \frac{1}{\sqrt{\pi}} \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \, dz.$

$\noindent So on writing (*) as a double integral we have$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= \frac{1}{\sqrt{\pi}} \int^\infty_0 \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \sin u \, dz \, du\\&= \frac{1}{\sqrt{\pi}} \int^\infty_0 \int^\infty_0 \frac{e^{-uz}}{\sqrt{z}} \sin u \, du \, dz,\end{align*}$

$\noindent after changing the order of integration has been made and is justified by Fubini's theorem. The $u$ integral can be found by applying IBP twice. The result is$

$\begin{align*} \int^\infty_{-\infty} \sin(x^2) \, dx &= \frac{1}{\sqrt{\pi}} \int^\infty_0 \frac{dz}{\sqrt{z} (1 + z^2)}.\end{align*}$

$$\noindent Now let $z = t^2, dz = 2t \, dt$, while the limits of integration remain unchanged to give$\\\int^\infty_{-\infty} \sin(x^2) \, dx = \frac{2}{\sqrt{\pi}} \int^\infty_0 \frac{dt}{1 + t^4}.$

$$\noindent The last integral has been done to death in the 2016 MX2 Integration Marathon. Its value is $\frac{\pi \sqrt{2}}{4}.$

$$\noindent Thus $\int^\infty_{-\infty} \sin(x^2) \, dx = \frac{2}{\sqrt{\pi}} \cdot \frac{\pi \sqrt{2}}{4} = \sqrt{\frac{\pi}{2}}.$

Nice

, it's a bit lengthier than you can do it using complex analysis but I like the fact that a high school student could understand it.

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