Higher Level Integration Marathon & Questions (1 Viewer)

Paradoxica · Mar 11, 2017

Re: HSC 2017 MX2 Integration Marathon

Mahan1 said:
$\textrm{2.If }f(x) = \frac{4^{x}}{4^{x}+2}$
$\textrm{Find} f(\frac{1}{14}) + ... + f(\frac{13}{14}) \ \textrm{without using calculator}$

It is obvious that f(x) + f(1-x) = 1

Therefore, by pairing the sum end to end, or adding it to itself in reverse order, like Little Gauss, the value of the sum is 7.

seanieg89 · Mar 12, 2017

Re: Extracurricular Integration Marathon

omegadot said:
$$\noindent Evaluate (preferably using real methods) $\int^\infty_0 \frac{\ln x}{1 + x^3} \, dx.$

Couldn't spot a real method, but when computing this in a rather unimaginative complex way I was surprised to discover that my method yielded values for

$\int_0^\infty f(x)\log(x)\, dx$

for pretty arbitrary rational functions f(x) (of course with appropriate decay to make circular contours have no contribution, and without poles on the non-negative real axis.)

I think you can also replace the log with a positive integral power of log. (Idk if the form will be closed for arbitrary k, but you should at least get a recursive expression.)

I didn't realise till now that thse kind of integrals could be so easily computed in such generality, so it is nice to learn something new!

Anyway, can hold back from posting this for a little while if you have a strong preference for real methods.

Paradoxica · Mar 12, 2017

Re: Extracurricular Integration Marathon

seanieg89 said:
Couldn't spot a real method, but when computing this in a rather unimaginative complex way I was surprised to discover that my method yielded values for

$\int_0^\infty f(x)\log(x)\, dx$

for pretty arbitrary rational functions f(x) (of course with appropriate decay to make circular contours have no contribution, and without poles on the non-negative real axis.)

I think you can also replace the log with a positive integral power of log. (Idk if the form will be closed for arbitrary k, but you should at least get a recursive expression.)

I didn't realise till now that thse kind of integrals could be so easily computed in such generality, so it is nice to learn something new!

Anyway, can hold back from posting this for a little while if you have a strong preference for real methods.

I think differentiation under the integral sign counts as a real method. Have you considered the Beta Function? It's what I thought of immediately but I'm ceebs actually doing it.

seanieg89 · Mar 12, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
I think differentiation under the integral sign counts as a real method. Have you considered the Beta Function? It's what I thought of immediately but I'm ceebs actually doing it.

Yeah, if I look at it later today (I probably will at some point, seems more fun that thesis editing / writing assignment for my course lol), differentiation under the integral / rewriting as a double integral in a clever way is absolutely the main way I would try to do it.

It's just when several such steps are involved in a problem that I either cannot do them or it takes an inordinate amount of time for me to do so.

Paradoxica · Mar 12, 2017

Re: Extracurricular Integration Marathon

seanieg89 said:
Yeah, if I look at it later today (I probably will at some point, seems more fun that thesis editing / writing assignment for my course lol), differentiation under the integral / rewriting as a double integral in a clever way is absolutely the main way I would try to do it.

It's just when several such steps are involved in a problem that I either cannot do them or it takes an inordinate amount of time for me to do so.

true, the necessary substitution for the x³ makes mental calculations a bit difficult.

seanieg89 · Mar 12, 2017

Re: Extracurricular Integration Marathon

$Actually, this works:\\ \\ $\int_0^\infty \frac{\log{x}}{x^3+1}\, dx\\ \\ = \int_0^1 \frac{1-x}{x^3+1}\log{x}\, dx.\\ \\ $(Split into integrals over $(0,1]$ and $[1,\infty)$, and substitute $y=x^{-1}$.)\\ \\$\Rightarrow I=\int_0^1 \frac{\log{x}}{x+1}\, dx+\sum_{k=1}^\infty (-1)^k \int_0^1 x^{3k-1}\log{x}\, dx$\\ \\ by expansion of $(1+x^3)^{-1}$ as a geometric series.\\ \\ $\Rightarrow I=-\frac{\pi^2}{12}+\frac{1}{9}\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}\\ \\ = -\frac{\pi^2}{12}+\frac{1}{9}\cdot\frac{\pi^2}{12}\\ \\ = -\frac{2\pi^2}{27}.$\\ \\ \\ (Assumed knowledge is the evaluations of the integral $\int_0^1\frac{\log{x}}{x+1}\, dx$, the alternating Basel sum, and the integral $\int_0^1 x^k \log(x).$)$

seanieg89 · Mar 12, 2017

Re: Extracurricular Integration Marathon

Kinda funny how often the Basel sum crops up in these things.

It's pretty fortunate, because it is not a trivial sum to evaluate, it just happens to be burned into all of our brains more than any series of comparable difficulty due to its fame.

Paradoxica · Mar 12, 2017

Extracurricular Integration Marathon

seanieg89 said:
Kinda funny how often the Basel sum crops up in these things.

It's pretty fortunate, because it is not a trivial sum to evaluate, it just happens to be burned into all of our brains more than any series of comparable difficulty due to its fame.

I think we can do a similar thing for all integrals of the form:

$\int_0^\infty \frac{x^a \log^b{x}}{1+x^n}$

for integers a,b,n

Actually I think there's a more general formula in one of my integral resources for this case...

seanieg89 · Mar 12, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
I think we can do a similar thing for all integrals of the form:

$\int_0^\infty \frac{x^a \log^b{x}}{1+x^n}$

for integers a,b,n

Actually I think there's a more general formula in one of my integral resources for this case...

Can you post such a formula if you locate it? This particular example posed by omegadot was much nicer than the general case (although as I remarked, the complex method I had in mind generalises well).

The indices worked out perfectly modulo 3 to make this computation short, in general it seems (at a glance at least) that we would have to deal with expressions like the sum of (3k+1)^(-2) which should be possible to compute but probably not famous enough to assume without proof.

Paradoxica · Mar 12, 2017

Re: Extracurricular Integration Marathon

seanieg89 said:
Can you post such a formula if you locate it? This particular example posed by omegadot was much nicer than the general case (although as I remarked, the complex method I had in mind generalises well).

The indices worked out perfectly modulo 3 to make this computation short, in general it seems (at a glance at least) that we would have to deal with expressions like the sum of (3k+1)^(-2) which should be possible to compute but probably not famous enough to assume without proof.

The denominator being linear is a sufficient condition for the integral to have a nice closed form free of the Gamma Function.

Proof: From the alternative definition of the Beta Function

$\text{B}(p,q) = \int_0^\infty \frac{x^{p-1}}{(1+x)^{p+q}} \, \text{d}x$

Note the substitution xⁿ→x also doesn't create a mess with the logarithm since all it does is give a rescaling factor, so nothing too bad there.

The substitution xⁿ→x does not fundamentally affect the denominator in the slightest, and so we have p+q=1, regardless of the value of n

But from the Gamma definition of the Beta Function, this means that:

$\text{B}(p,q) = \pi \csc{(\pi p)}$

The integer powers of the natural logarithm come from repeatedly differentiating with respect to x, but since the integral is already in elementary form, it's higher derivatives are also in elementary form.

Therefore, a "nice" closed form exists.

$\mathcal{Q.E.D.}$

addendum: "a" doesn't have to be an integer since it's not a part of any discrete operations. I suppose "n" doesn't have to be either, though that would make things a little more messy with the substitution component.

omegadot · Mar 12, 2017

Re: Extracurricular Integration Marathon

seanieg89 said:
Can you post such a formula if you locate it? This particular example posed by omegadot was much nicer than the general case (although as I remarked, the complex method I had in mind generalises well).

The indices worked out perfectly modulo 3 to make this computation short, in general it seems (at a glance at least) that we would have to deal with expressions like the sum of (3k+1)^(-2) which should be possible to compute but probably not famous enough to assume without proof.

$$\noindent I didn't realise the problem was so interesting until I started playing around with it but I should have suspected this was the case given the original question as asked was first solved by G. H. Hardy.$

$$\noindent Starting with $\int^\infty_0 \frac{x^m}{1 + x^n} \, dx$ which as we know can be reduced to beta and finally to $\frac{\pi}{n} \csc \left [\pi\left (\frac{m + 1}{n} \right ) \right ]$ through the reflection formula.$

$\noindent Differentiating under the integral sign with respect to $m$ gives us the log term with our new log integral equalling$

$\int^\infty_0 \frac{x^m \ln x}{1 + x^n} \, dx = -\frac{\pi^2}{n^2} \csc \left [\pi \left (\frac{m + 1}{n} \right ) \right ] \cot \left [\pi \left (\frac{m + 1}{n} \right ) \right ].$

$\noindent So taking up Paradoxica's idea of repeatedly differentiating under the integral sign (which I haven't thought of doing) I suppose we could write$

$\int^\infty_0 \frac{x^m \ln^k x}{1 + x^n} \, dx = \frac{d^k}{dm^k} \left \{\frac{\pi}{n} \csc \left [\pi\left (\frac{m + 1}{n} \right ) \right ] \right \}$

$$\noindent where $k = 0,1,2,\ldots$ so what is the $k$th derivative for the cosecant function (I am sure it must be in terms of some special polynomials)?$

InteGrand · Mar 12, 2017

Re: Extracurricular Integration Marathon

omegadot said:
$$\noindent I didn't realise the problem was so interesting until I started playing around with it but I should have suspected this was the case given the original question as asked was first solved by G. H. Hardy.$

$$\noindent Starting with $\int^\infty_0 \frac{x^m}{1 + x^n} \, dx$ which as we know can be reduced to beta and finally to $\frac{\pi}{n} \csc \left [\pi\left (\frac{m + 1}{n} \right ) \right ]$ through the reflection formula.$

$\noindent Differentiating under the integral sign with respect to $m$ gives us the log term with our new log integral equalling$

$\int^\infty_0 \frac{x^m \ln x}{1 + x^n} \, dx = -\frac{\pi^2}{n^2} \csc \left [\pi \left (\frac{m + 1}{n} \right ) \right ] \cot \left [\pi \left (\frac{m + 1}{n} \right ) \right ].$

$\noindent So taking up Paradoxica's idea of repeatedly differentiating under the integral sign (which I haven't thought of doing) I suppose we could write$

$\int^\infty_0 \frac{x^m \ln^k x}{1 + x^n} \, dx = \frac{d^k}{dm^k} \left \{\frac{\pi}{n} \csc \left [\pi\left (\frac{m + 1}{n} \right ) \right ] \right \}$

$$\noindent where $k = 0,1,2,\ldots$ so what is the $k$th derivative for the cosecant function (I am sure it must be in terms of some special polynomials)?$

This paper may be of interest: https://cs.uwaterloo.ca/journals/JIS/VOL17/Neto/neto4.pdf .

seanieg89 · Apr 6, 2017

Re: Extracurricular Integration Marathon

We should add infinite series to this thread!

$\sum_{n=0}^\infty\frac{1}{1+n^2}$

As usual, "lower tech" methods are preferable.

Paradoxica · Apr 8, 2017

Re: Extracurricular Integration Marathon

seanieg89 said:
We should add infinite series to this thread!

$\sum_{n=0}^\infty\frac{1}{1+n^2}$

As usual, "lower tech" methods are preferable.

This isn't a complete answer, but the furthest progress I have made so far without invoking too much prior knowledge.

Using the Inverse Laplace Transform, we find that for all n ≠ 0,

$\int_0^\infty \frac{\sin{x}}{e^{nx}} \, \text{d}x = \frac{1}{1+n^2}$

The integral is absolutely convergent for all values of n, by using sinx < x and comparison to Basel's Problem. By the dominated convergence theorem, the change in order can be made.

The simplified geometric sum of the integrand converges for all non-negative x, including x = 0, so the integral is equal to the sum (minus 1), without further trouble.

Hence:

$\sum_{n=0}^\infty \frac{1}{1+n^2} = 1 + \int_0^\infty \frac{\sin{x}}{e^x - 1} \, \text{d}x$

Paradoxica · Apr 8, 2017

Re: Extracurricular Integration Marathon

If anyone wants to try that integral, feel free to. I've given up already

This is the most "basic" proof I can come up with.

The Weierstrass Form (valid for all complex numbers) of the Gamma Function is defined as:

$\Gamma(z) \equiv \frac{1}{z e^{\gamma z}} \prod_{n=1}^\infty \frac{n e^{\frac{z}{n}}}{n+z}$

The proof of the Gamma Functional Equation using the Integral Form, the Integral form equals the Euler Form, and the Euler Form equals the Weierstrass Form, are all within real analytic realms, so this is probably the closest I can get. There's the problem with complex logarithms and complex derivatives which I still haven't addressed...

Using the Recurrence Relation of the Gamma Function:

$\Gamma(1-z) \equiv -z \Gamma(-z) \equiv -z \frac{e^{\gamma z}}{-z} \prod_{n=1}^\infty \frac{n e^{-\frac{z}{n}}}{n-z}$

Combine the two and multiply term by term:

$\frac{\pi}{\sin{\pi z}} \equiv \Gamma(z)\Gamma(1-z) \equiv \frac{1}{z} \prod_{n=1}^\infty \frac{n^2}{n^2 - z^2}$

Take the Logarithm of both sides, differentiate and plug in z = i. Simplify both sides and rearrange to acquire the desired sum.

Don't forget your Hyperbolic Trigonometry

Kingom · Apr 10, 2017

Re: Extracurricular Integration Marathon

$\int _{ 0 }^{ 1 }{ \frac { { \log^4{(x)} } }{ x+1 } dx }$

seanieg89 · Apr 12, 2017

Re: Extracurricular Integration Marathon

Kingom said:
$\int _{ 0 }^{ 1 }{ \frac { { \log^4{(x)} } }{ x+1 } dx }$

$\int_0^1 x^\alpha \, dx = \frac{1}{\alpha+1}\\ \\ \Rightarrow \int_0^1 x^\alpha \log^4{x}\, dx = \frac{d^4}{d\alpha^4}\left(\frac{1}{\alpha+1}\right)=\frac{24}{(\alpha+1)^5}\\ \\ \Rightarrow \int_0^1 \frac{\log^4{x}}{x+1}\, dx = \sum_{n=0}^\infty (-1)^n\int_0^1 x^n\log^4{x}\,dx\\ \\ = 24\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^5}=24\left(\zeta(5)-\frac{2}{2^5}\zeta(5)\right)=\frac{45}{2}\zeta(5).$

seanieg89 · Apr 12, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
If anyone wants to try that integral, feel free to. I've given up already

This is the most "basic" proof I can come up with.

The Weierstrass Form (valid for all complex numbers) of the Gamma Function is defined as:

$\Gamma(z) \equiv \frac{1}{z e^{\gamma z}} \prod_{n=1}^\infty \frac{n e^{\frac{z}{n}}}{n+z}$

The proof of the Gamma Functional Equation using the Integral Form, the Integral form equals the Euler Form, and the Euler Form equals the Weierstrass Form, are all within real analytic realms, so this is probably the closest I can get. There's the problem with complex logarithms and complex derivatives which I still haven't addressed...

Using the Recurrence Relation of the Gamma Function:

$\Gamma(1-z) \equiv -z \Gamma(-z) \equiv -z \frac{e^{\gamma z}}{-z} \prod_{n=1}^\infty \frac{n e^{-\frac{z}{n}}}{n-z}$

Combine the two and multiply term by term:

$\frac{\pi}{\sin{\pi z}} \equiv \Gamma(z)\Gamma(1-z) \equiv \frac{1}{z} \prod_{n=1}^\infty \frac{n^2}{n^2 - z^2}$

Take the Logarithm of both sides, differentiate and plug in z = i. Simplify both sides and rearrange to acquire the desired sum.

Don't forget your Hyperbolic Trigonometry

This is pretty good. From memory when I last computed this it was via a similar complex analytic factorisation method, and I believe there was also a hackier solution via ODES.

Will look for a nicer way to do it when I have a bit more free time, I just remembered enjoying it the first time around, and wondered if any of you guys could find a cute way of doing it.

omegadot · Apr 22, 2017

Re: Extracurricular Integration Marathon

Paradoxica said:
If anyone wants to try that integral, feel free to. I've given up already

$\noindent The usual way to evaluate your integral, or its cognate $\int^\infty_0 \frac{\sin x}{e^x + 1} \, dx$, is to convert it into the infinite series we were trying to find in the first place (or $ \sum^\infty_{n = 0} \frac{(-1)^n}{1 + n^2}$ is the case of its cognate).$

$\noindent As you suggest in your solution, the simplest way I have seen in finding this sum starts with the Weierstrass product for the sinh function$

$\frac{\sinh(\pi z)}{\pi z}=\prod_{n =1}^\infty \left(1+\frac{z^2}{n^2}\right)$

$\noindent and following what you said to do.$

omegadot · Apr 22, 2017

Re: Extracurricular Integration Marathon

seanieg89 said:
We should add infinite series to this thread!

$\sum_{n=0}^\infty\frac{1}{1+n^2}$

As usual, "lower tech" methods are preferable.

Here I offer a "high tech" solution which the OP clearly did not ask for.

$\noindent If $f(z)$ is a function that is analytic at $z = 0,\pm 1,\pm 2,\ldots$ and tends to zero at least as fast as $|z|^{-2}$ as $|z| \to \infty$ it can be show that$

$\sum_{n = -\infty}^{\infty} f(n) = -[\text{sum of residues of}\,\, \pi \cot (\pi z) f(z) \text{at the poles of}\,\, f(z)],$

$$\noindent and basically comes from applying the Cauchy Residue Theorem to the function $F(z) = \pi \cot (\pi z) f(z)$. Since$

$\sum^\infty_{n = 0} \frac{1}{1 + n^2} = 1 + \sum^\infty_{n = 1} \frac{1}{1 + n^2} = \frac{1}{2} + \frac{1}{2} \sum^\infty_{n = -\infty} \frac{1}{1 + n^2}.$

$\noindent we have$

$\sum^\infty_{n = -\infty} \frac{1}{1 + n^2} = -[\text{sum of residues of}\,\, \pi \cot (\pi z)/(1 + z^2)\,\, \text{at} \,\, z = \pm i].$

$\noindent Both poles at $z = \pm i$ are simple, so the residue at $z = i$ is $-\frac{\pi}{2} \coth \pi$ while the residue at $z = -i$ is $-\frac{\pi}{2} \coth \pi$ giving$

$\sum_{n = 0}^\infty \frac{1}{1 + n^2} = \frac{1}{2} + \frac{1}{2} \coth \pi.$

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