Thread: Higher Level Integration Marathon & Questions

1. Re: Higher Level Integration Marathon & Questions

Originally Posted by seanieg89
Eh I wouldn't say that necessarily, I did the integral in a not particularly imaginative way...there could be a clever shortcut.

To me it does seem like computing this integral is roughly the same mathematical "depth" as computing some particular values of the di/trilogarithm though.

Anyway, is there a reason why you were seeking to avoid them?
because I like a nice balance of low tech and efficient solutions to integrals

2. Re: Higher Level Integration Marathon & Questions

because I like a nice balance of low tech and efficient solutions to integrals
Fair enough, seems a rather subjective matter...to me polylog and polygamma manipulations are okay, but I also have no qualms about using contour integration etc.

3. Re: Higher Level Integration Marathon & Questions

because I like a nice balance of low tech and efficient solutions to integrals
$\noindent So do I, so if you can give me a low tech approach that makes use of real methods only for the integral$

$\int_{-\infty}^{\infty}\frac{dx}{(e^{x}-x)^{2}+{\pi}^{2}}=\frac{1}{1+\Omega},$

$\noindent where \Omega is the omega constant, I'd be really impressed.$

4. Re: Higher Level Integration Marathon & Questions

I think you guys should enrol in Extreme Integration (MATH1052), or if you're feeling ambitious try Insane Integration (MATH3025)!

PM me for details.

5. Re: Higher Level Integration Marathon & Questions

$\text{Show that }\int_0^{2\pi} (\cos \theta)^{2n}\,d\theta = \frac{\pi}{2^{2n-1}}\binom{2n}{n}$

(Try ignoring the lengthy 4U way in this thread :P)

6. Re: Higher Level Integration Marathon & Questions

Originally Posted by leehuan
$\text{Show that }\int_0^{2\pi} (\cos \theta)^{2n}\,d\theta = \frac{\pi}{2^{2n-1}}\binom{2n}{n}$

(Try ignoring the lengthy 4U way in this thread :P)
$I=\int_0^{2\pi} (\frac{e^{it}+e^{-it}}{2})^{2n}\, dt \\ \\ = 2^{-2n}\binom{2n}{n}\int_0^{2\pi}\, dt\\ \\=\frac{\pi}{2^{2n-1}}\binom{2n}{n}.\\ \\ (all nonconstant trig monomials integrate to zero.)$

7. Re: Higher Level Integration Marathon & Questions

$\noindent Assuming ax^2 + 2bxy + cy^2 = z (where z is positive) \\ describes a general ellipse in the Cartesian Plane, evaluate:$

$\huge \noindent \iint_{\mathbb{R}^2} e^{-(ax^2 + 2bxy + cy^2)} \text{d}x \text{d}y$

Prove the integral diverges if the argument of the exponential does not describe an ellipse.

Challenge: Formulate a generalisation to the above integral in higher dimensions. (Experience with Linear Algebra will be very handy)

8. Re: Higher Level Integration Marathon & Questions

$\noindent Assuming ax^2 + 2bxy + cy^2 = z (where z is positive) \\ describes a general ellipse in the Cartesian Plane, evaluate:$

$\huge \noindent \iint_{\mathbb{R}^2} e^{-(ax^2 + 2bxy + cy^2)} \text{d}x \text{d}y$

Prove the integral diverges if the argument of the exponential does not describe an ellipse.

Challenge: Formulate a generalisation to the above integral in higher dimensions. (Experience with Linear Algebra will be very handy)
$\noindent A sketch of a method. Rewrite the quadratic form ax^2 + 2bxy + cy^2 \equiv \mathbf{x}^{T}A\mathbf{x} \Bigg{(}where \mathbf{x} = \begin{bmatrix}x \\ y \end{bmatrix}, A = \begin{bmatrix}a & b \\ b & c\end{bmatrix}\Bigg{)} as \lambda_{1}u^{2} + \lambda_{2}v^{2}, where \lambda_{1}, \lambda_{2} are the eigenvalues of A and \mathbf{u} := \begin{bmatrix}u \\ v\end{bmatrix} = Q^{T}\mathbf{x}, where we have orthogonally diagonalised A as QDQ^{T}. The \text{d}x \, \text{d}y becomes \text{d}u\, \text{d}v under the change of variables \mathbf{u} = Q^{T}\mathbf{x} because the Jacobian determinant is 1 (since orthogonal matrices have determinant with absolute value 1). The region of the integral remains \mathbb{R}^{2} as we are using an invertible matrix as our transform.$

$\noindent Since the curve describes an ellipse, the \lambda_{i} are positive, so the resulting integral converges and can be split up and be essentially computed as a product of Gaussian integrals. A similar method applies to when we have a quadratic form in higher dimensions.$

9. Re: Higher Level Integration Marathon & Questions

Originally Posted by InteGrand
$\noindent A sketch of a method. Rewrite the quadratic form ax^2 + 2bxy + cy^2 \equiv \mathbf{x}^{T}A\mathbf{x} \Bigg{(}where \mathbf{x} = \begin{bmatrix}x \\ y \end{bmatrix}, A = \begin{bmatrix}a & b \\ b & c\end{bmatrix}\Bigg{)} as \lambda_{1}u^{2} + \lambda_{2}v^{2}, where \lambda_{1}, \lambda_{2} are the eigenvalues of A and \mathbf{u} := \begin{bmatrix}u \\ v\end{bmatrix} = Q^{T}\mathbf{x}, where we have orthogonally diagonalised A as QDQ^{T}. The \text{d}x \, \text{d}y becomes \text{d}u\, \text{d}v under the change of variables \mathbf{u} = Q^{T}\mathbf{x} because the Jacobian determinant is 1 (since orthogonal matrices have determinant with absolute value 1). The region of the integral remains \mathbb{R}^{2} as we are using an invertible matrix as our transform.$

$\noindent Since the curve describes an ellipse, the \lambda_{i} are positive, so the resulting integral converges and can be split up and be essentially computed as a product of Gaussian integrals. A similar method applies to when we have a quadratic form in higher dimensions.$

(As in you technically didn't do the computation, not that you're being cryptic)

10. Re: Higher Level Integration Marathon & Questions

$Let Q be the symmetric matrix associated to the quadratic form defining the conic. Q yields an ellipse iff it is a positive-definite matrix. Standard spectral theory shows that Q is diagonalisable by orthogonal matrices. The resulting matrix is \textrm{diag}(\lambda_j), so after our orthogonal transformation (rotating the ellipse so its axes coincide with the coordinate axes), the integral is: \\ \\ \int_{\mathbb{R}^n}\exp(-\sum_i \lambda_i^2x_i^2)\, dx=\prod_i\lambda_i^{-1}\int_{\mathbb{R}}\exp(-x_i^2)\, dx_i=\frac{\pi^{n/2}}{\det(Q)}.$

11. Re: Higher Level Integration Marathon & Questions

(As in you technically didn't do the computation, not that you're being cryptic)
Well I did say I was sketching a method (i.e. not giving the full answer).

12. Re: Higher Level Integration Marathon & Questions

Oh didn't see you edited it, was the divergence thing always there and I just blind before?

In any case, we can still represent the quadratic form by a symmetric, hence diagonalisable matrix Q. If the quadratic form does not represent an ellipse, then at least one eigenvalue is non-positive, which implies that at least one of the integrals in the above product of single-variable integrals is infinite and we are done.

13. Re: Higher Level Integration Marathon & Questions

Originally Posted by seanieg89
Oh didn't see you edited it, was the divergence thing always there and I just blind before?

In any case, we can still represent the quadratic form by a symmetric, hence diagonalisable matrix Q. If the quadratic form does not represent an ellipse, then at least one eigenvalue is non-positive, which implies that at least one of the integrals in the above product of single-variable integrals is infinite and we are done.
I think it was there before. I remember answering it in my head but then forgetting to comment on it in my post. Haha

14. Re: Higher Level Integration Marathon & Questions

Originally Posted by leehuan
$\text{Show that }\int_0^{2\pi} (\cos \theta)^{2n}\,d\theta = \frac{\pi}{2^{2n-1}}\binom{2n}{n}$

(Try ignoring the lengthy 4U way in this thread :P)
$\noindent Here is a real method, though it is nowhere near as slick as that given by seanieg89.$

$\noindent We begin by splitting up the interval of integration as follows:$

$\int^{2\pi}_0 \cos^{2n} \theta \, d\theta = \int^{\frac{\pi}{2}}_0 \cos^{2n} \theta \, d\theta + \int^\pi_{\frac{\pi}{2}} \cos^{2n} \theta \, d\theta + \int^{\frac{3\pi}{2}}_\pi \cos^{2n} \theta \, d\theta + \int^{2\pi}_{\frac{3\pi}{2}} \cos^{2n} \theta \, d\theta$

$\noindent Making the following substitutions$

\begin{align*}\text{Second integral} \quad & \theta \mapsto \frac{\pi}{2} + \theta\\\text{Third integral} \quad & \theta \mapsto \pi + \theta\\\text{Fourth integral} \quad & \theta \mapsto \frac{3\pi}{2} + \theta\end{align*}

$\noindent yields$

\begin{align*} \int^{2\pi}_0 \cos^{2n} \theta \, d\theta &= \int^{\frac{\pi}{2}}_0 \cos^{2n} \theta \, d\theta + \int^{\frac{\pi}{2}}_0 (-1)^{2n} \sin^{2n} \theta \, d\theta + \int^{\frac{\pi}{2}}_0 (-1)^{2n} \cos^{2n} \theta \, d\theta + \int^{\frac{\pi}{2}}_0 \sin^{2n} \theta \, d\theta\\ &= 2 \int^{\frac{\pi}{2}}_0 \cos^{2n} \theta \, d\theta + 2 \int^{\frac{\pi}{2}}_0 \sin^{2n} \theta \, d\theta\\ &= 2 \int^{\frac{\pi}{2}}_0 \cos^{2(\frac{2n + 1}{2})} \theta \sin ^{2 (\frac{1}{2}) - 1} \theta \, d\theta + \int^{\frac{\pi}{2}}_0 \cos^{2(\frac{1}{2}) - 1} \theta \sin^{2(\frac{2n + 1}{2})} \theta \, d\theta\\ &= \text{B} \left (\frac{2n + 1}{2}, \frac{1}{2} \right ) + \text{B} \left (\frac{1}{2}, \frac{2n + 1}{2} \right )\\&= 2 \, \text{B} \left (\frac{1}{2}, n + \frac{1}{2} \right )\\ &= \frac{2 \Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2})}{\Gamma (n + 1)}.\end{align*}

$\noindent Since n is a positive integer we have$

$\Gamma (\frac{1}{2}) = \sqrt{\pi}, \Gamma (n + 1) = n!$

$\noindent and it is not too hard to show that$

$\Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi},$

$\noindent thus$

$\int^{2\pi}_{0} \cos^{2n} \theta \, d\theta = \frac{\pi}{2^{2n - 1}} \cdot \frac{(2n)!}{n! \, n!} = \frac{\pi}{2^{2n - 1}} \binom{2n}{n},$

$\noindent as required to show.$

15. Re: Higher Level Integration Marathon & Questions

Easy difficulty

$\int \frac{\tanh x}{\exp x}\,dx$

16. Re: Higher Level Integration Marathon & Questions

Originally Posted by leehuan
Easy difficulty

$\int \frac{\tanh x}{\exp x}\,dx$
$\int \frac{e^x-e^{-x}}{e^{2x}+1}\, dx = \int \frac{u^2-1}{u^2(u^2+1)}\, du= \int \frac{2}{u^2+1}-\frac{1}{u^2}\, du\\ \\ = 2\arctan(e^x)+e^{-x}+C\\ \\ (where u:=e^x)$

17. Re: Higher Level Integration Marathon & Questions

$\int \sqrt{\tanh{x}}\,\text{d}x$

18. Re: Higher Level Integration Marathon & Questions

Originally Posted by He-Mann
I think you guys should enrol in Extreme Integration (MATH1052), or if you're feeling ambitious try Insane Integration (MATH3025)!

PM me for details.
this feels like a meme

19. Re: Higher Level Integration Marathon & Questions

$\int \sqrt{\tanh{x}}\,\text{d}x$
At least quick partial fractions is possible this time round lol

$u^2 = \tanh x \implies 2u\,du = \text{sech}^2 x\,dx = (1-\tanh^2 x)\,dx\quad (u\ge 0)$

\begin{align*} I & = \int \sqrt{ \tanh x} \, dx\\ &= \int \frac{2u^2}{1-u^4} \, du\\ &= \int \frac{2u^2}{(1-u^2)(1+u^2)}\,du\\ &= \int\left(\frac{1}{1-u^2}-\frac{1}{1+u^2}\right)\,du\\&= \frac{1}{2}\ln \left|\frac{1+u}{1-u}\right| - \arctan u+C \\ &= \frac{1}{2} \ln \left(\frac{1+\sqrt{\tanh x}}{1-\sqrt{\tanh x}}\right) - \arctan\left(\sqrt{\tanh x}\right)\end{align*}

20. Re: Higher Level Integration Marathon & Questions

$Some high dimensional geometric weirdness that we can explore with integration:\\ \\ a) By analogy with the volumes by slices of MX2, find an expression for the volume of the n-dimensional unit ball. Express your answer in terms of factorials or the Gamma function. By using Stirling's formula, find an asymptotic for |B_n(1)| as n\rightarrow\infty. \\ \\ b) Find and prove the uniqueness of an \alpha>0 such that the ratio \frac{|B_n(1-n^{-\alpha})|}{|B_n(1)|} tends to a positive limit less than 1 as n\rightarrow\infty. (Intuitively this says that the volume of high dimensional balls concentrates near the boundary, and quantifies the exact rate of this concentration.)\\ \\$

$c) What is the limit of the above ratio for the value of \alpha you found in b)?\\ \\ d) Along the lines of b) and c), show that the volume of high dimensional balls concentrates in the equatorial slice \{|x_n|\leq n^{-\beta}\} and quantify the exact rate of this concentration, also computing the limit of \frac{|\{|x_n|\leq n^{-\beta}\}|}{|B_n(1)|} as n\rightarrow\infty in as nice a form as you can.$

21. Re: Higher Level Integration Marathon & Questions

$\int_0^\pi \ln(1+2a\cos x + a^2)\,dx\text{ splitting into appropriate cases}$

22. Re: Higher Level Integration Marathon & Questions

Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.

$I=\int_0^{\pi} \mathrm{ln}(1+2a\mathrm{cos}(x)+a^2)\mathrm{d}x$

$\frac{\mathrm{d}I}{\mathrm{d}a} =\int_0^{\pi} \frac{2\mathrm{cos}(x)+2a}{1+2a\mathrm{cos}(x)+a^2 }\mathrm{d}x = \frac{1}{a}\int_0^{\pi} (1-\frac{1-a^2}{1+2a\mathrm{cos}(x)+a^2})\mathrm{d}x = \frac{\pi}{a}-\frac{1}{a}\int_0^{\pi} \frac{1-a^2}{1+2a\mathrm{cos}(x)+a^2}$

$J = \int_0^{\pi} \frac{1-a^2}{1+2a\mathrm{cos}(x)+a^2}\mathrm{d}x$

Let $t=\mathrm{tan}\frac{x}{2}, so \mathrm{d}t = \frac{1+t^2}{2}\mathrm{d}x$

$J = \int_0^{\pi} \frac{1-a^2}{1+2a\mathrm{cos}(x)+a^2}\mathrm{d}x = \int_0^{\infty} \frac{1-a^2}{1+2a\frac{1-t^2}{1+t^2}+a^2}\frac{2}{1+t^2}\mathrm{d}t = 2\int_0^{\infty} \frac{1-a^2}{(1+2a+a^2)+(1-2a+a^2)t^2}\mathrm{d}t = 2\int_0^{\infty} \frac{(1-a)(1+a)}{(1+a)^2+(1-a)^2t^2}\mathrm{d}t = 2\int_0^{\infty} \frac{C}{C^2+t^2}\mathrm{d}t$

where $C=\frac{1+a}{1-a}$

This is easily evaluated depending on the different cases:

If $a>1$ or $a<-1$, we have $C<0$ $J = -\pi$, if $-1 then $C>0$ and $J = \pi$, and if $a=-1$ then $C=$ and $J=0$.

Case 1: $-1

$\frac{\mathrm{d}I}{\mathrm{d}a} = \frac{\pi}{a} - \frac{\pi}{a} = 0$ hence $I =C$ for some constant $C$.

Evaluating the integral for $a=0$, we obtain that:

$I=0$ when $-1.

Case $a>1$ or $a<-1$.

$I=\int_0^{\pi} \mathrm{ln}(1+2a\mathrm{cos}(x)+a^2)\mathrm{d}x = \int_0^{\pi} \mathrm{ln}(1+2\frac{1}{a}\mathrm{cos}(x)+\frac{1} {a^2})\mathrm{d}x + \int_0^{\pi} 2\mathrm{ln}a\mathrm{d}x = 0 + 2\pi \mathrm{ln}(a)$ (which follows from Case 1).

$I = 2\pi \mathrm{ln}(a)$ when $a>1$ or $a<-1$.

Case 3: $a=-1$

$I = \int_0^{\pi} \mathrm{ln}(2-2\mathrm{cos}(x))\mathrm{d}x = \pi \mathrm{ln}2 + \int_0^{\pi} \mathrm{ln}(1-\mathrm{cos}(x))\mathrm{d}x$

Let $K = \int_0^{\pi} \mathrm{ln}(1-\mathrm{cos}(x))\mathrm{d}x$

By symmetry, $K = \int_0^{\frac{\pi}{2}} \mathrm{ln}(1-\mathrm{cos}(x))\mathrm{d}x+\int_0^{\frac{\pi}{2}} \mathrm{ln}(1+\mathrm{cos}(x))\mathrm{d}x=\int_0^{ \frac{\pi}{2}} \mathrm{ln}(1-\mathrm{cos^2}(x))\mathrm{d}x=2\int_0^{\frac{\pi}{ 2}} \mathrm{ln}(\mathrm{sin}(x))\mathrm{d}x = 2\int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{cos}(x))\mathrm{d}x$

Let $L = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{cos}(x))\mathrm{d}x = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{sin}(x))\mathrm{d}x$

$2L = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{cos}(x)\mathrm{sin}(x))\mathrm {d}x = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\frac{1}{2}\mathrm{sin}(2x))\mathrm{d} x = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\frac{1}{2})\mathrm{d}x+\int_0^{\frac{ \pi}{2}} \mathrm{ln}(\mathrm{sin}(2x))\mathrm{d}x + = \int_0^{\frac{\pi}{2}} \mathrm{ln}(\frac{1}{2})\mathrm{d}x+\frac{1}{2}\in t_0^{\pi} \mathrm{ln}(\mathrm{sin}(u))\mathrm{d}u$

By symmetry, $\int_0^{\pi} \mathrm{ln}(\mathrm{sin}(u))\mathrm{d}u = 2\int_0^{\frac{\pi}{2}} \mathrm{ln}(\mathrm{sin}(u))\mathrm{d}u$

Therefore $2L = -\frac{\pi}{2}\mathrm{ln}(2) + L$, so $L = -\frac{\pi}{2}\mathrm{ln}(2)$.

Subbing the values back in, we obtain $I=0$.

Case 4: $a=1$

$I = \int_0^{\pi} \mathrm{ln}(2+2\mathrm{cos}(x))\mathrm{d}x$

This can be done by the same method as case 3, and so $I=0$.

Hence, when $-1\leq a \leq 1$, $I=0$, and $I=2\pi \mathrm{ln}(a)$ otherwise.

23. Re: Higher Level Integration Marathon & Questions

Originally Posted by calamebe
Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.
Fourier Expansion :P

24. Re: Higher Level Integration Marathon & Questions

Fourier Expansion :P
No idea what that is, I'm not all that far beyond HSC maths

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