HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

seanieg89 · Jan 28, 2014

Re: MX2 Integration Marathon

Here are some more from me as well:

$1. \int_0^1 \frac{\log(1+x)}{1+x^2}\, dx\\ \\ 2.\int_0^\infty \frac{\log(x)}{1+x^2}\, dx\\ \\ 3. \int_0^\pi \log(1-2a\cos(x)+a^2)\, dx\quad (a\in\mathbb{R})\\ \\$

The third one is a little different in flavour.

Sy123 · Jan 28, 2014

Re: MX2 Integration Marathon

seanieg89 said:
Here are some more from me as well:

$1. \int_0^1 \frac{\log(1+x)}{1+x^2}\, dx\\ \\ 2.\int_0^\infty \frac{\log(x)}{1+x^2}\, dx\\ \\ 3. \int_0^\pi \log(1-2a\cos(x)+a^2)\, dx\quad (a\in\mathbb{R})\\ \\$

The third one is a little different in flavour.

Here is my attempt for the third one:

$\\ f(a) = I = \int_0^{\pi} \ln (a^2- 2a\cos x +1) \ dx = \int_0^{\pi} \ln (a^2 - 2a\cos(\pi -x) + 1) \ dx \ \ \ \ $since$ \ \int_0^b f(x) \ dx = \int_0^b f(b-x) \ dx \\ \\ $Therefore$ \ 2f(a) = \int_0^{\pi} \ln((a^2-2a\cos x+1)(a^2+2a\cos x+1)) \ dx = \int_0^{\pi} \ln(a^4 -2a^2 \cos(2x) + 1) \\ \\ u= 2x \\ \\ \int_)^{2\pi} \frac{1}{2}\ln(a^4-2a^2\cos u +1) \ du \\ \\ $And$ \\ \\ \int_0^{\pi} \ln(b^2 - 2b\cos z +1) \ dz = \int_{\pi}^{2\pi} \ln(b^2 - 2b\cos z + 1) \ dz \\ \\ $Therefore$ \ \ \int_0^{2\pi} \ln(b^2- 2b\cos z +1) = 2\int_0^{\pi} \ln(b^2 -2b\cos z + 1) \ dx \\ \\ \therefore \ 2f(a) = \frac{1}{2} 2f(a^2) \\ \\ \therefore \ 2f(a) = f(a^2) \\ \\ $The only function that satisifes this is$ \ f(a) = \ln a \\ \\ \therefore \ \int_0^{\pi} \ln(1-2a\cos x + a^2) \ dx = \ln a$

Is that correct?

seanieg89 · Jan 28, 2014

Re: MX2 Integration Marathon

Sy123 said:
Here is my attempt for the third one:

$\\ f(a) = I = \int_0^{\pi} \ln (a^2- 2a\cos x +1) \ dx = \int_0^{\pi} \ln (a^2 - 2a\cos(\pi -x) + 1) \ dx \ \ \ \ $since$ \ \int_0^b f(x) \ dx = \int_0^b f(b-x) \ dx \\ \\ $Therefore$ \ 2f(a) = \int_0^{\pi} \ln((a^2-2a\cos x+1)(a^2+2a\cos x+1)) \ dx = \int_0^{\pi} \ln(a^4 -2a^2 \cos(2x) + 1) \\ \\ u= 2x \\ \\ \int_)^{2\pi} \frac{1}{2}\ln(a^4-2a^2\cos u +1) \ du \\ \\ $And$ \\ \\ \int_0^{\pi} \ln(b^2 - 2b\cos z +1) \ dz = \int_{\pi}^{2\pi} \ln(b^2 - 2b\cos z + 1) \ dz \\ \\ $Therefore$ \ \ \int_0^{2\pi} \ln(b^2- 2b\cos z +1) = 2\int_0^{\pi} \ln(b^2 -2b\cos z + 1) \ dx \\ \\ \therefore \ 2f(a) = \frac{1}{2} 2f(a^2) \\ \\ \therefore \ 2f(a) = f(a^2) \\ \\ $The only function that satisifes this is$ \ f(a) = \ln a \\ \\ \therefore \ \int_0^{\pi} \ln(1-2a\cos x + a^2) \ dx = \ln a$

Is that correct?

The functional equation 2f(a)=f(a^2) is right, but your subsequent deduction isn't. You aren't far off though.

Sy123 · Jan 28, 2014

Re: MX2 Integration Marathon

seanieg89 said:
The functional equation 2f(a)=f(a^2) is right, but your subsequent deduction isn't. You aren't far off though.

Taking the seperate cases

$f(a) = \begin{cases} 0 , \ \ \ a=0 \\ \ln a, \ \ \ a > 0 \end{cases}$

?

dunjaaa · Jan 28, 2014

Re: MX2 Integration Marathon

1.

Screen shot 2014-01-28 at 9.14.49 PM.png

dunjaaa · Jan 28, 2014

Re: MX2 Integration Marathon

2.

Screen shot 2014-01-28 at 9.41.23 PM.png

which makes sense cause the positive area cancels the negative area

seanieg89 · Jan 28, 2014

Re: MX2 Integration Marathon

Sy123 said:
Taking the seperate cases

$f(a) = \begin{cases} 0 , \ \ \ a=0 \\ \ln a, \ \ \ a > 0 \end{cases}$

?

This isn't the only solution to the functional equation (double it for instance and it's still a solution). More importantly though, this cannot be the correct value for the integral, because f(a) should tend to 0 as a->0.

Sy123 · Jan 28, 2014

Re: MX2 Integration Marathon

seanieg89 said:
This isn't the only solution to the functional equation (double it for instance and it's still a solution). More importantly though, this cannot be the correct value for the integral, because f(a) should tend to 0 as a->0.

Oh yes ok, how would you propose finding the constant k in the expression k ln(a)?

Are there more properties of the functions that need to be found?

seanieg89 · Jan 28, 2014

Re: MX2 Integration Marathon

Sy123 said:
Oh yes ok, how would you propose finding the constant k in the expression k ln(a)?

Are there more properties of the functions that need to be found?

I never said that the solution was a multiple of log(a) for a > 0, I just said that any constant multiple of log(a) satisfies the functional equation.

The functional equation itself has many solutions, but by looking at what happens near zero we can make certain deductions about what f does in various places without having to explicitly find f. We can then use another identity coming from the initial integral definition of f to use this partial information to tell us what f does everywhere.

seanieg89 · Jan 29, 2014

Re: MX2 Integration Marathon

This is how I finished it:

$Since $\log$ is an increasing function, we have (assuming $a>0$ and using $-1\leq\cos(\theta)\leq 1$):\\ \\ $2\pi\log(1-a)\leq f(a) \leq 2\pi\log(1+a)$.\\ \\ From this inequality, we get that:\\ \\ $\lim_{a\rightarrow 0^+} f(a)=0.$\\ \\So for $0<a<1$ by applying $f(x)=f(x^2)/2$ recursively, we have:\\ \\ $f(a)=2^{-n}f(a^{2^{n}})\rightarrow 0.$\\ \\ Hence $f(a)=0$ for $0\leq a<1.$ For $a>1$ we have (by substitution into the integrand and simplifying using log laws):\\ \\ $0=f(a^{-1})=f(a)-2\pi\log(a).$\\ \\Finally using the symmetry about 0 that you proved, we can conclude:\\ \\ $f(a)=\left\{\begin{array}{lr}0 & : |a|\leq 1\\2\pi\log |a| &:|a|>1.\end{array}\right.$$

seanieg89 · Jan 31, 2014

Re: MX2 Integration Marathon

$Let $f$ be continuous and even, and let $a>0$.\\ Prove that:\\ \\ $\int_{-a}^a \frac{f(x)}{1+e^{x}}\, dx=\int_0^a f(x)\, dx$$

darkness70 · Jan 31, 2014

Re: MX2 Integration Marathon

Prove they're equal?

seanieg89 · Jan 31, 2014

Re: MX2 Integration Marathon

darkness70 said:
Prove they're equal?

Yes, thanks

.

seanieg89 · Feb 28, 2014

Re: MX2 Integration Marathon

seanieg89 said:
$Let $f$ be continuous and even, and let $a>0$.\\ Prove that:\\ \\ $\int_{-a}^a \frac{f(x)}{1+e^{x}}\, dx=\int_0^a f(x)\, dx$$

Bump.

Also:

$\int \sqrt{\tan(x)} \, dx$

jyu · Feb 28, 2014

Re: MX2 Integration Marathon

seanieg89 said:
$Let $f$ be continuous and even, and let $a>0$.\\ Prove that:\\ \\ $\int_{-a}^a \frac{f(x)}{1+e^{x}}\, dx=\int_0^a f(x)\, dx$$

Sy123 · Mar 1, 2014

Re: MX2 Integration Marathon

seanieg89 said:
$Let $f$ be continuous and even, and let $a>0$.\\ Prove that:\\ \\ $\int_{-a}^a \frac{f(x)}{1+e^{x}}\, dx=\int_0^a f(x)\, dx$$

$u=-x \Rightarrow \int_{-a}^a \frac{f(x)}{1+e^{x}} \ dx = \int_{-a}^a \frac{f(-x)}{1+e^{-x}} \ dx$

$\frac{1}{1+e^{-x}} = \frac{1+e^{x} - 1}{1+e^{x}} = 1 - \frac{1}{1+e^{x}}$

$\therefore \int_{-a}^a \frac{f(x)}{1+e^{-x} } \ dx = \int_{-a}^a f(-x) - \frac{f(-x)}{1+e^{x}} \ dx$

$\therefore \ \int_{-a}^a \frac{f(x) + f(-x)}{1+e^{-x}} \ dx = \int_{-a}^a f(-x) \ dx$

$$Since$ \ f \ $is even$ \\ \\ f(x) + f(-x) = 2f(x) \\ \\ \int_{-a}^a f(x) = 2\int_0^a f(x)$

$\therefore \ \int_{-a}^a \frac{f(x)}{1+e^x} \ dx = \int_0^a f(x) \ dx$

-----

$\int \frac{1-x^2}{(x^2+1) \sqrt{x^4+1}} \ dx$

seanieg89 · Mar 1, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{1-x^2}{(x^2+1) \sqrt{x^4+1}} \ dx$

Just to be clear, when you ask for an indefinite integral like this, do you want people to post a function whose derivative is equal to the integrand on some part of the real line or do you want a function whose derivative is equal to your integrand everywhere on the real line?

I think a lot of the solutions to problems like this have a slightly different expression in different parts of the domain.

Sy123 · Mar 7, 2014

Re: MX2 Integration Marathon

$\\ \int \frac{1-t^2}{t^4+8t^2+1} \ dt$

hit patel · Mar 9, 2014

Re: MX2 Integration Marathon

hey sy or carrot, you were compiling this right?

seanieg89 · Mar 9, 2014

Re: MX2 Integration Marathon

$\int \sqrt[3]{\tan(x)}\, dx$

HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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