# Higher Level Integration Marathon & Questions (1 Viewer)

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

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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

#### leehuan

##### Well-Known Member
$\bg_white \int \sqrt{\tanh{x}}\,\text{d}x$
At least quick partial fractions is possible this time round lol

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

\bg_white \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*}

#### seanieg89

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

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

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#### leehuan

##### Well-Known Member
$\bg_white \int_0^\pi \ln(1+2a\cos x + a^2)\,dx\text{ splitting into appropriate cases}$

#### calamebe

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

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

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

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

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

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

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

This is easily evaluated depending on the different cases:

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

Case 1: $\bg_white -1

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

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

$\bg_white I=0$ when $\bg_white -1.

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

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

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

Case 3: $\bg_white a=-1$

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

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

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

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

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

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

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

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

Case 4: $\bg_white a=1$

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

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

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

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##### -insert title here-
Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.
Fourier Expansion

#### calamebe

##### Active Member
Fourier Expansion
No idea what that is, I'm not all that far beyond HSC maths

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$\bg_white \noindent \int_0^1 (-\log(x))^n \log{(-\log{x})} \, \text{d}x where n is a non-negative integer \\\\ \int_a^1 \frac{\sin^{-1}{x}}{\sqrt{x^2-a^2}} \, \text{d}x \\\\ \int_a^\infty \frac{x}{(x^2+b^2)\sqrt{x^2-a^2}} \, \text{d}x \\\\ \int_a^\infty \frac{x}{(x^2+b^2)^{\tfrac{n}{2}}\sqrt{x^2-a^2}} \, \text{d}x where \mathbb{Z} \ni n>1$

Bonus Challenge: Prove the last two integrals are symmetric in (a,b) <=> (b,a) without evaluating it to the final result.

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#### leehuan

##### Well-Known Member
$\bg_white \text{The substitution }x=\exp (-u)\text{ turns the first integral into }\int_0^\infty u^n \log u \exp (-u)\,du$

\bg_white \text{Call it }I_n.\text{ From IBP,}\\ \begin{align*}I_n&= -e^{-u}u^n \log u \Big |_0^\infty + \int_0^\infty e^{-u}u^{n-1}\left(1 + n \log u\right)\,du\\ &= J_{n-1}+ n I_{n-1}\end{align*}\\ \text{where }J_n = \int_0^\infty u^{n} e^{-u} \,du

$\bg_white \text{It is trivial to prove that }J_n = n!.$
______________________________________________

\bg_white \text{So we have }I_n = (n-1)! + nI_{n-1}.\text{ Hence:}\\ \begin{align*}I_n &= (n-1)! + n\left( (n-2)! + (n-1)I_{n-1}\right)\\ &\vdots \\ &= \left[(n-1)! + n(n-2)! + n(n-1)(n-3)! + \dots + n(n-1)...(4)(3) 1! + n(n-1)...(4)(3)(2) 0! \right] +n! I_0\\ &= \sum_{k=1}^n \frac{n!}{k} + n!\int_0^\infty e^{-u}\log u \, du\end{align*}

...but unfortunately I don't know how to prove that final integral equals to negative gamma.

#### Kingom

##### Member
Show that

$\bg_white \int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}dx=\dfrac{2}{3}(2-\sqrt{3})G$

where G is Catalan's constant.

#### leehuan

##### Well-Known Member
@Paradoxica I think I finally figured what it should've been.

$\bg_white \int_1^e x \sqrt[6]{x^{-1} \sqrt[20]{x \sqrt[42]{x^{-1} \dots}}}\,dx$

Well, hopefully.

#### leehuan

##### Well-Known Member
$\bg_white \int_0^{10} x^{\lim_{n\to \infty} f^n(x)}\,dx \text{ for }f(x) = \frac12 \left( x + \frac2x\right)$

$\bg_white \text{where }f^n(x) = \underbrace{f \cdots f}_{n\text{-times}}(x)$

Shouldn't be hard assuming I didn't mess up typing the question.

##### -insert title here-
$\bg_white \int_0^{10} x^{\lim_{n\to \infty} f^n(x)}\,dx \text{ for }f(x) = \frac12 \left( x + \frac2x\right)$

$\bg_white \text{where }f^n(x) = \underbrace{f \cdots f}_{n\text{-times}}(x)$

Shouldn't be hard assuming I didn't mess up typing the question.
For all positive real numbers, the iteration derived from Newton's Method applied to the equation q(x)=x²-2 results in the f(x) described above, and converges to the positive real root of the equation.

Hence, fn(x) converges to √ ̅2 for x>0

The point x=0 can be discarded, as it is a set of zero measure.

Hence, the integral is equivalent to

$\bg_white \int_0^{10} x^{\sqrt{2}} \text{d}x \\\\ = \frac{10^{\sqrt{2}+1}}{\sqrt{2}+1} = 10^{\sqrt{2}+1}\left(\sqrt{2}-1\right)$

My only problem with this question is that students don't know how to handle problematic points on the domain of an integral.

#### leehuan

##### Well-Known Member
$\bg_white \int_0^\infty \frac{\sin x}{x^2+1}dx$

#### blyatman

##### Well-Known Member
$\bg_white \int_0^\infty \frac{\sin x}{x^2+1}dx$
Would've been so much nicer if it was a cos instead of a sin lol.

Been a while since I've done one of these, but I'm guessing: take the CCW quarter-circle contour C around the first quadrant with radius R approaching infinity, so that:
$\bg_white \oint_C\frac{e^{iz}}{z^2+1}dz=\int_{C_1}\frac{e^{iz}}{z^2+1}dz+\int_{C_2}\frac{e^{iz}}{z^2+1}dz+\int_{C_3}\frac{e^{iz}}{z^2+1}dx$
where C1 is the quarter circle arc (from z = R to z = iR), C2 goes from iR to 0, and C3 goes from 0 to R?

The contour over C can then be evaluated using Residue theorem, with with z = +i being the only singularity within C.
The contours over C1 and C2 can be evaluted by using an appropriate parameterisation (alternatively, I'm thinking that the integral over C1 just be 0 since the denominator goes as R^2?).

The original integral can then be extracted (or at least I think it can...) by taking imaginary parts of both sides, so that
$\bg_white \mathrm{Im}\int_{C_3}\frac{e^{iz}}{z^2+1}dx=\int_0^\infty\frac{\sin x}{x^2+1}dx$

Is this correct? I'm a bit hesitant on that last step, not entirely convinced myself if that last expression holds.

I don't normally spend time thinking about integration questions, so that's enough integration for me for the next year haha.

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#### leehuan

##### Well-Known Member
Lol my bad. Idk why I decided to switch the cos out for a sin when posting

#### blyatman

##### Well-Known Member
Lol my bad. Idk why I decided to switch the cos out for a sin when posting
Lol yeh would've taken 2 lines with the cos.