# Higher Level Integration Marathon & Questions (1 Viewer)

##### Active Member
$\bg_white \noindent Evaluate \int_0^1 \frac{\ln x \ln (x + 1)}{x (x + 1)} \, dx.$

#### seanieg89

##### Well-Known Member
^ 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.

##### -insert title here-
^ 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

##### Well-Known Member
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?

##### -insert title here-
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

##### Well-Known Member
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.

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

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

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

#### He-Mann

##### Vexed?
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

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

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

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#### seanieg89

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

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

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

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

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)

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#### InteGrand

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

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

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

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

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

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

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

#### seanieg89

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

#### InteGrand

##### Well-Known Member

(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

##### Well-Known Member
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

##### Well-Known Member
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

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

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

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

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

$\bg_white \noindent Making the following substitutions$

\bg_white \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*}

$\bg_white \noindent yields$

\bg_white \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*}

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

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

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

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

$\bg_white \noindent thus$

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

$\bg_white \noindent as required to show.$

#### leehuan

##### Well-Known Member
Easy difficulty

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

#### seanieg89

##### Well-Known Member
Easy difficulty

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