Higher Level Integration Marathon & Questions (1 Viewer)

seanieg89 · Nov 12, 2017

^ That was good practice!

I need to go out for dinner v soon, so will supply gory details later. In short, the answer can be expressed solely in terms of Apery's constant zeta(3).

1. IBP to essentially reduce the integral to log^2(1+x)/x. (The bulk of the difficulty of the originally posted integral lies in this integral as I see it.)

2. Expand this out to obtain an expression in terms of polylogs of order 2 and 3, but we basically only need to know the value of these polylogs at 1/2 and 1.

3. Special functions funtimes proving some reflection formulae etc to compute these guys.

Paradoxica · Nov 12, 2017

seanieg89 said:
^ That was good practice!

I need to go out for dinner v soon, so will supply gory details later. In short, the answer can be expressed solely in terms of Apery's constant zeta(3).

1. IBP to essentially reduce the integral to log^2(1+x)/x. (The bulk of the difficulty of the originally posted integral lies in this integral as I see it.)

2. Expand this out to obtain an expression in terms of polylogs of order 2 and 3, but we basically only need to know the value of these polylogs at 1/2 and 1.

3. Special functions funtimes proving some reflection formulae etc to compute these guys.

I was looking for a polylog-free solution, but I guess this puts the nail on the coffin

seanieg89 · Nov 12, 2017

Paradoxica said:
I was looking for a polylog-free solution, but I guess this puts the nail on the coffin

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?

Paradoxica · Nov 12, 2017

seanieg89 said:
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

seanieg89 · Nov 13, 2017

Paradoxica said:
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.

omegadot · Nov 14, 2017

Paradoxica said:
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.$

He-Mann · Nov 14, 2017

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

PM me for details.

leehuan · Nov 15, 2017

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

(Try ignoring the lengthy 4U way in this thread

)

seanieg89 · Nov 15, 2017

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

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

$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.)$$

Paradoxica · Nov 15, 2017

$\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)

InteGrand · Nov 15, 2017

Paradoxica said:
$\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.$

Paradoxica · Nov 16, 2017

InteGrand said:
$\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.$

An answer that isn't an answer....

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

seanieg89 · Nov 16, 2017

InteGrand · Nov 16, 2017

Paradoxica said:
An answer that isn't an answer....

(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).

seanieg89 · Nov 16, 2017

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.

InteGrand · Nov 16, 2017

seanieg89 said:
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

omegadot · Nov 16, 2017

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

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

$\noindent Here is a real method, though it is nowhere near as slick as that given by seanieg89.$

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

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

$\noindent Making the following substitutions$

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

$\noindent yields$

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

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

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

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

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

$\noindent thus$

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

$\noindent as required to show.$

leehuan · Nov 21, 2017

Easy difficulty

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

seanieg89 · Nov 22, 2017

leehuan said:
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 $u:=e^x$)$$

Higher Level Integration Marathon & Questions (1 Viewer)

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