Higher Level Integration Marathon & Questions (2 Viewers)

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

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 · Jan 8, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
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 · Jan 8, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
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 · Jan 8, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$$\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 · Jan 8, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$\noindent Okay, since the integrand is odd$

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

$$\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.

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

The good old tedious integral of 1/(1+t^4) haha.

Paradoxica · Jan 8, 2016

Re: Extracurricular Integration Marathon

InteGrand said:
The good old tedious integral of 1/(1+t^4) haha.

$$\noindent It's not very tedious.$ \\ 2I = \int \frac{1+t^2+1-t^2}{1+t^4}$d$t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}$d$t - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}$d$t\\2I = \int \frac{\text{d}(t-\frac{1}{t})}{(t-\frac{1}{t})^2+2} + \int \frac{\text{d}(t+\frac{1}{t})}{(t+\frac{1}{t})^2-2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+\frac{1}{2\sqrt{2}}\log\left(\frac{t^2 + \sqrt{2}t+1}{t^2-\sqrt{2}t+1}\right)$

InteGrand · Jan 8, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$\noindent It's not very tedious.$ \\ 2I = \int \frac{1+t^2+1-t^2}{1+t^4}$d$t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}$d$t - \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}$d$t\\2I = \int \frac{\text{d}(t-\frac{1}{t})}{(t-\frac{1}{t})^2+2} + \int \frac{\text{d}(t+\frac{1}{t})}{(t+\frac{1}{t})^2-2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)-\frac{1}{2\sqrt{2}}\log\left(\frac{t^2 - \sqrt{2}t+1}{t^2+\sqrt{2}t+1}\right)$

Ah right, I was thinking of the one with a t² on the numerator too (the one that comes about when doing integral of √(tan x) ). That one was more tedious iirc. (Split into partial fractions etc.)

Paradoxica · Jan 8, 2016

Re: Extracurricular Integration Marathon

InteGrand said:
Ah right, I was thinking of the one with a t² on the numerator too (the one that comes about when doing integral of √(tan x) ). That one was more tedious iirc. (Split into partial fractions etc.)

No, that one is amenable to double symmetric substitution as well.

$Any integrals of the form $\frac{2nx^{n-1}}{x^{4n}+1}$ and $\frac{2nx^{3n-1}}{x^{4n}+1}$ are equally trivial.$

RealiseNothing · Jan 9, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
$4. \lim_{T\rightarrow \infty}\int_{-T}^T \sin(x^2)\, dx$

Let $B_R(t)=Re^{it}$ for $0 \leq t \leq \frac{\pi}{4}$

Then $B'_R(t)=Rie^{it}$

So we have:

$|\int_{B_R} e^{-z^2}dz| \leq \int_0^{\frac{\pi}{4}}|e^{-(Re^{it})^2}Rie^{it}|dt =\int_0^{\frac{\pi}{4}} Re^{-R^2cos(2t)} dt \to 0 $~as~$ R \to \infty$

$\int_{B_R} e^{-z^2}dz \to 0 $~as~$ R \to \infty$

To make things neater we will define from now on $f(z)=e^{-z^2}$

We then integrate counter-clockwise about a boundary formed by $B_R$ to make a circular arc and get:

$\int_0^R e^{-x^2} dx + \int_{B_R} f(z) dz - \int_0^R f(xe^{\frac{i \pi}{4}}) dx = 0$

Now take the limit as $R \to \infty$

$\int_0^{\infty} e^{-x^2} dx = e^{\frac{i \pi}{4}} \int_0^{\infty} f(xe^{\frac{i \pi}{4}}) dx = \frac{1+i}{\sqrt{2}} \int_0^{\infty} f(xe^{\frac{i \pi}{4}})$

Now from a previously answered integral we know that:

$\int_0^{\infty} e^{-x^2} dx =\frac{\sqrt{\pi}}{2}$

So we then have:

$\sqrt{\frac{\pi}{4}} = \frac{1+i}{\sqrt{2}} \int_0^{\infty} f(xe^{\frac{i \pi}{4}}) = \frac{1+i}{\sqrt{2}} \int_0^{\infty} e^{-ix^2} dx$

Using Euler's formula we obtain:

$\sqrt{\frac{\pi}{2}}=\int_0^{\infty} cos(x^2)+sin(x^2)+icos(x^2)-isin(x^2) dx$

Since the imaginary part = 0 we get:

$\int_0^{\infty} cos(x^2) dx = \int_0^{\infty} sin(x^2) dx$

Now equating the real part gives:

$\int_0^{\infty} cos(x^2) dx + \int_0^{\infty} sin(x^2) dx = \sqrt{\frac{\pi}{2}}$

$2\int_0^{\infty} sin(x^2) dx =\sqrt{\frac{\pi}{2}}$

$\int_0^{\infty} sin(x^2) dx =\sqrt{\frac{\pi}{8}}$

Then we just double the answer cos even function to get:

$\int_{-\infty}^{\infty}sin(x^2) dx = \sqrt{\frac{\pi}{2}}$

Paradoxica · Jan 9, 2016

Re: Extracurricular Integration Marathon

$$\noindent This was posed as a challenge by my friend.$\\$He hasn't solved it himself, he got the answer using Sage.$ \\ \int_{-1}^{1} \frac{\text{d}x}{x}\sqrt{\frac{1+x}{1-x}}\log{\left(\frac{1+2x+2x^2}{1-2x+2x^2}\right)}=4\pi\cot^{-1}{\sqrt{\phi}}$

KingOfActing · Jan 9, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
$$\noindent This was posed as a challenge by my friend.$\\$He hasn't solved it himself, he got the answer using Sage.$ \\ \int_{-1}^{1} \frac{\text{d}x}{x}\sqrt{\frac{1+x}{1-x}}\log{\left(\frac{1+2x+2x^2}{1-2x+2x^2}\right)}=4\pi\cot^{-1}{\sqrt{\phi}}$

I've seen the solution to that integral before, good luck everyone.

3 substitutions and residue theorem

Paradoxica · Jan 9, 2016

Re: Extracurricular Integration Marathon

KingOfActing said:
I've seen the solution to that integral before, good luck everyone.

Is that the most straightforward way to do it?

Surely there must be more straightforward ways.

KingOfActing · Jan 10, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
Is that the most straightforward way to do it?

Surely there must be more straightforward ways.

It's the way I've seen it done, there could be a simpler way

omegadot · Jan 11, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
Here's a very deceptive integral which appears simple.

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

$$\noindent We will make use of the following result$\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.$

omegadot · Jan 11, 2016

Re: Extracurricular Integration Marathon

Paradoxica said:
Here's a very deceptive integral which appears simple.

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

$$\noindent We will evaluate a more general integral first. To do this we will make use of the following result$\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.$

$$\noindent To prove this result, consider$\\\begin{align*}(1 - 2a \cos x + a^2) \times \sum^\infty_{k = 0} a^k \cos (kx) &= \sum^\infty_{k = 0} a^k \cos (kx) - 2 \sum^\infty_{k = 0} a^{k + 1} \cos (kx) \cos x + \sum^\infty_{k = 0} a^{k + 2} \cos (kx)\\&= \sum^\infty_{k = 0} a^k \cos (kx) - 2 \sum^\infty_{k = 1} a^k \cos [(k - 1)x] \cos x + \sum^\infty_{k = 2} a^k \cos [(k - 2)x]\\&= 1 + a \cos x + \sum^\infty_{k = 2} a^k \cos (kx) - 2a \cos x - \sum^\infty_{k = 2} 2a^k \cos [(k - 1)x] \cos x + \sum^\infty_{k = 2} a^k \cos [(k - 2) x]\\&= 1 - a \cos x + \sum^\infty_{k = 2} a^k \left [\cos(kx) - 2 \cos [(k - 1)x] \cos x + \cos [(k - 2)x] \right ]\end{align*}$

$$\noindent Making use of the result $2 \cos \alpha \cos \beta = \cos (\alpha - \beta) + \cos (\alpha + \beta).\\$If we set $\alpha = (k - 1) x$ and $\beta = x$ we have$\\2 \cos [(k - 1)x] \cos x = \cos[(k - 2)x] + \cos (kx).$

$$\noindent So the term within the summation is zero and we have$\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2},\\$as desired.$

$\noindent Now to make use of this is our integral, we need to make the $\cos x$ term in the numerator of the right hand term of the above expression "disappear." So$

$\begin{align*} \sum^\infty_{k = 0} a^k \cos (kx) &= \frac{1 - a \cos x}{1 - 2a \cos x + a^2}\\1 + \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{1 - a \cos x}{1 - 2a \cos x + a^2}\\2 + 2 \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{2 - 2 \cos x}{1 - 2a \cos x + a^2}\\\Rightarrow 1 + 2 \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{2 - 2a \cos x}{1 - 2a \cos x + a^2} - 1\\\Rightarrow 1 + 2 \sum^\infty_{k = 1} a^k \cos (kx) &= \frac{1 - a^2}{1 - 2a \cos x + a^2}\\\Rightarrow \frac{1}{1 - a^2} \left [1 + 2 \sum^\infty_{k = 1} a^k \cos (kx) \right ] &= \frac{1}{1 - 2a \cos x + a^2}\end{align*}$

$\noindent Multiplying by $x^2$ and integrating from 0 to $\pi$ both sides of the above expression we have$

$\begin{align*} \int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx &= \frac{1}{1 - a^2} \int^\pi_0 x^2 \, dx + \frac{2}{1 - a^2} \sum^\infty_{k = 1} a^k \int^\pi_0 x^2 \cos (kx) \, dx\end{align*}$

$\noindent Now$

$\begin{align*} \int^\pi_0 x^2 \, dx = \frac{\pi^3}{3}\end{align*}$

$\noindent and$

$\begin{align*}\int^\pi_0 x^2 \cos (kx) \, dx &= \frac{2 \pi}{k^2} (-1)^k \quad \mbox{(using IBP twice)}\end{align*}$

$$\noindent Thus$\\\begin{align*} \int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx &= \frac{1}{1 - a^2} \left [\frac{\pi^3}{3} + 4\pi \sum^\infty_{k = 1} \frac{(-a)^k}{k^2} \right ].\end{align*}$

$$\noindent Now the \textit{polylogarithm} function $\mbox{Li}_s (x)$ is defined as $\mbox{Li}_s (x) \sum^\infty_{k = 1} \frac{x^k}{k^s}.$

$$\noindent Thus $\sum^\infty_{k = 1} \frac{(-a)^k}{k^2} = \mbox{Li}_2 (-a),$ a \textit{dilogarithmic} function. So finally we have$

$\begin{align*} \int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx &= \frac{1}{1 - a^2} \left [\frac{\pi^3}{3} + 4 \pi \,\mbox{Li}_2 (-a) \right ] \quad (*)\end{align*}$

$$\noindent Returning to the original integral, writing this as $\\\begin{align*}\int^\pi_0 \frac{x^2}{\sqrt{5} - 2 \cos x} \, dx = \frac{1}{\sqrt{5}} \int^\pi_0 \frac{x^2}{1 - \frac{2}{\sqrt{5}} \cos x} \, dx \quad (**)\end{align*}$

$$\noindent Now the integral appearing in (*) can be rewritten as$\\\begin{align*}\int^\pi_0 \frac{x^2}{1 - 2a \cos x + a^2} dx = \frac{1}{1 + a^2} \int^\pi_0 \frac{x^2}{1 - \frac{2a}{1 + a^2} \cos x} dx \quad (***) \end{align*}$

$$\noindent Comparing (**) with (***), we set $\frac{a}{1 + a^2} = \frac{1}{\sqrt{5}} \,\, \Rightarrow \,\, a^2 - a\sqrt{5} + 1 = 0.$ Thus $a = \frac{\sqrt{5} - 1}{2} = \frac{1}{\varphi},$ since $|a| < 1.$ Here $\varphi$ is the \textit{golden ratio}.$

$$\noindent Thus$\\\begin{align*}\frac{1}{\sqrt{5}} \int^\pi_0 \frac{x^2}{1 - \frac{2}{\sqrt{5}} \cos x} \, dx &= \frac{1 + \varphi^{-2}}{\sqrt{5}} \int^\pi_0 \frac{x^2}{1 + 2 \varphi^{-1} \cos x + \varphi^{-2}} dx\\\Rightarrow \int^\pi_o \frac{x^2}{\sqrt{5} - 2 \cos x} \, dx &= \frac{1 + \varphi^{-2}}{\sqrt{5}} \cdot \frac{1}{1 - \varphi^{-2}} \left [\frac{\pi^3}{3} + 4\pi \, \mbox{Li}_{2} \left (-\frac{1}{\varphi} \right ) \right ].\end{align*}$

$$\noindent Now $\frac{1 + \varphi^{-2}}{1 - \varphi^{-2}} = \sqrt{5}$ and $\mbox{Li}_2 \left (-\frac{1}{\varphi} \right ) = -\frac{\pi^2}{15} + \frac{1}{2} \ln^2 \varphi.$

$$\noindent So finally we have$\\\begin{align*}\int^\pi_0 \frac{x^2}{\sqrt{5} -2 \cos x} \, dx &= \frac{\pi^3}{3} + 4\pi \left (-\frac{\pi^2}{15} + \frac{1}{2} \ln^2 \varphi \right ) = \frac{\pi^3}{15} + 2\pi \ln^2 \varphi.\end{align*}$

seanieg89 · Jan 11, 2016

Re: Extracurricular Integration Marathon

omegadot said:
$$\noindent We will make use of the following result$\\\sum^\infty_{k = 0} a^k \cos (kx) = \frac{1 - a \cos x}{1 - 2a \cos x + a^2}, \,\, |a| < 1.$

Excellent solution to the integral

.

One minor remark is that another way of proving the summation result used is taking the real part of the geometric series summation (common ratio z=a*cis(x)).

I.e. find the real part of 1/(1-a*cis(x)), which is quite quick.

omegadot · Jan 11, 2016

Re: Extracurricular Integration Marathon

seanieg89 said:
Excellent solution to the integral .

One minor remark is that another way of proving the summation result used is taking the real part of the geometric series summation (common ratio z=a*cis(x)).

I.e. find the real part of 1/(1-a*cis(x)), which is quite quick.

Yes, I know, but I was trying to keep the problem "purely real" for the benefit of any MX2 students who may care to read this thread (though those who are reading this thread probably already know or could follow the complex way anyway).

Paradoxica · Jan 11, 2016

Re: Extracurricular Integration Marathon

omegadot said:
Yes, I know, but I was trying to keep the problem "purely real" for the benefit of any MX2 students who may care to read this thread (though those who are reading this thread probably already know or could follow the complex way anyway).

Don't bother, anyone who can follow this thread on the real analysis side probably has what it takes to follow the complex analysis side.

Anyway, nice answer. Here's the one I got from the place I found the problem.

$$\noindent Define $J(a) = \int_{-\pi}^{\pi} \frac{e^{iax}}{\sqrt{5}-2\cos{x}}$d$x\\$The integral we are after is $-\frac{1}{2}J''(0)=\int_0^\pi \frac{x^2}{\sqrt{5}-2\cos{x}}$d$x\\$Let us consider the following contour integral over $\mathbb{C}.\\ \oint_{C} \frac{z^a}{z^2 -\sqrt{5}z+1}$ where $C$ is a unit keyhole circle on the negative real axis.$\\$By the residue theorem, this is equal to $-2i\pi\phi^a\\$But this integral is also equal to $-iJ(a)+2i\sin{\pi a}\int_0^1 \frac{x^a}{x^2 +\sqrt{5}x+1} \\ $The section about the center of the circle vanishes, leaving us with $\\ J(a) = 2\pi \phi^a +2\sin{\pi a}\int_0^1 \frac{x^a}{x^2+\sqrt{5}x+1}$d$x \\$With a little computation, we have $-\frac{1}{2}J''(0) = -\pi\log^2{\phi}-2\pi\int_0^1 \frac{\log{x}}{x^2+\sqrt{5}x+1}$d$x \\$Using the facts:$\\ \frac{1}{x^2 + \sqrt{5}x+1} \equiv \frac{1}{x+\phi} - \frac{1}{x+\frac{1}{\phi}}\; ; \; \int_0^1 \frac{\log{x}}{x+a}$d$x = $ Li$_2\left(-\frac{1}{a}\right)$

$$\noindent Li$_2\left(-\frac{1}{\phi}\right) = -\frac{\pi^2}{10}-\log^2 \phi \; ; \; $Li$_2(-\phi) = -\frac{\pi^2}{15}+\frac{1}{2}\log^2 \phi\\$We conclude $-\frac{1}{2}J''(0) = -\pi\log^2 \phi -2\pi\left[\left( -\frac{\pi^2}{10}-\log^2 \phi \right)-\left(-\frac{\pi^2}{15}+\frac{1}{2}\log^2 \phi \right)\right]\\$Hence $\int_0^\pi \frac{x^2}{\sqrt{5}-2\cos{x}}$d$x=2\pi\log^2 \phi+\frac{\pi^3}{15}$

seanieg89 · Jan 12, 2016

Re: Extracurricular Integration Marathon

$\int_0^\infty x^3 e^{-x^2}\log(1-e^{-x^2})\, dx$

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