HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

Drsoccerball · Sep 18, 2015

Re: MX2 2015 Integration Marathon

nightweaver066 said:
Pretty ugly. You can use $x^2 = \sec \theta$ and $x^2 = \tan \theta$ for each integral respectively.

Simplify etc. etc. Then use $y^2 = \sin \theta$ to end up with something like $\frac{y^2+1}{2y(1-y^4)}$ which we can quickly decompose (using our best mate the Heaviside Cover-Up Method), integrate, then get rid of our substitutions.

I still think my method is somewhat cleaner

leehuan · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I still think my method is somewhat cleaner

I don't easily see substitutions such as u^4=1+1/x^4 though

Drsoccerball · Sep 18, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
I don't easily see substitutions such as u^4=1+1/x^4 though

You just take out x from the bracket its something you pick up while doing hard integrals now you'll see when to use it

Paradoxica · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
You just take out x from the bracket its something you pick up while doing hard integrals now you'll see when to use it

Your partial solution leads to the answer quite quickly, although I suggest using the integration formula for $\frac{1}{a}\tanh^{-1}{\frac{u}{a}}$ instead of $\frac{1}{2a}\ln{\frac{a+u}{a-u}}$

leehuan · Sep 18, 2015

Re: MX2 2015 Integration Marathon

It's just too bad they don't teach hyperbolic trig functions in Ext 2.

braintic · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Your partial solution leads to the answer quite quickly, although I suggest using the integration formula for $\frac{1}{a}\tanh^{-1}{\frac{u}{a}}$ instead of $\frac{1}{2a}\ln{\frac{a+u}{a-u}}$

In an exam, you can't quote integrals that are not on the standard integrals sheet.

kawaiipotato · Sep 18, 2015

Re: MX2 2015 Integration Marathon

$$ Evaluate $ \int_0^{\frac{\pi}{2}} \ln(\sin x) dx$

$\int \ln(\ln x) + \frac{1}{(\ln x)^{2}} dx$

Drsoccerball · Sep 18, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$$ Evaluate $ \int_0^{\frac{\pi}{2}} \ln(\sin x) dx$

$\int \ln(\ln x) + \frac{1}{(\ln x)^{2}} dx$

I dont think the first one can be solved with elementary methods as IBP gives you xcotx which we have established previously can't be done...The second one may also be the same

kawaiipotato · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Both are solvable

Paradoxica · Sep 18, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$$ Evaluate $ \int_0^{\frac{\pi}{2}} \ln(\sin x) dx$

$\int \ln(\ln x) + \frac{1}{(\ln x)^{2}} dx$

First question:

$\int_0^\frac{\pi}{2} {\ln(\sin{x})}dx=\int_0^\frac{\pi}{2} {\ln(\sin{\frac{\pi}{2}-x})}dx=\int_0^\frac{\pi}{2} {\ln(\cos{x})}dx$
$2I=\int_0^\frac{\pi}{2} {\ln(\sin{x}\cos{x})}dx=\int_0^\frac{\pi}{2} {\ln(\sin{2x})-\ln{2}dx$
Let $u=2x$ and hence, $\frac{du}{2}=dx$ , $u=\pi$ when $x=\frac{\pi}{2}$
$2I=\frac{1}{2}\int_0^\pi {\ln(\sin{u})-\ln{2}du$
Observe that since we are integrating the dummy variable over a half period, it is the same as integrating over twice a quarter-period since $\sin{x}$ is even at $\frac{\pi}{2}$
$2I=\int_0^\frac{\pi}{2} {\ln(\sin{u})du-\frac{1}{2}\int_0^\pi{\ln{2}}du$
The left part of the RHS is now identical to the original question, since a shift in variable doesn't change the numerical value of the desired integrand.
$\therefore 2I = I -\frac{1}{2}\int_0^\pi{\ln{2}}du$
$I={-\pi\ln{\sqrt{2}}}$
Second question:
$\int \ln(\ln x) + \frac{1}{(\ln x)^{2}} dx = \int \ln(\ln x) - \frac{1}{\ln{x}} + \frac{1}{\ln{x}} + \frac{1}{(\ln x)^{2}} dx$
$=\int \left(\ln(\ln x) - \frac{1}{\ln{x}}\right)dx + x\left(\frac{1}{x\ln{x}} - \frac{-1}{x(\ln x)^{2}}\right)dx=\int \left(\ln(\ln x) - \frac{1}{\ln{x}}\right)dx + xd\left(\ln(\ln x) - \frac{1}{\ln{x}}\right)=x\left({\ln{(\ln{x})}-\frac{1}{\ln{x}}}\right)+C$
By virtue of reverse product rule. Now please tell me there is a more obvious solution method, or I will flip.

kawaiipotato · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Second question:
$\int \ln(\ln x) + \frac{1}{(\ln x)^{2}} dx = \int \ln(\ln x) - \frac{1}{\ln{x}} + \frac{1}{\ln{x}} + \frac{1}{(\ln x)^{2}} dx$
$=\int \left(\ln(\ln x) - \frac{1}{\ln{x}}\right)dx + x\left(\frac{1}{x\ln{x}} - \frac{-1}{x(\ln x)^{2}}\right)dx=\int \left(\ln(\ln x) - \frac{1}{\ln{x}}\right)dx + xd\left(\ln(\ln x) - \frac{1}{\ln{x}}\right)=x\left({\ln{(\ln{x})}-\frac{1}{\ln{x}}}\right)+C$
By virtue of reverse product rule. Now please tell me there is a more obvious solution method, or I will flip.

Nice solution
Another way would've been to do IBP with the first ln(lnx) and then doing IBP again to the following integral which will cancel out the original 1/(lnx)^2

Drsoccerball · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Paradoxica you didn't evaluate the first integral ?

braintic · Sep 18, 2015

Re: MX2 2015 Integration Marathon

So nobody is concerned that ln[sin(0)] does not exist?

Ekman · Sep 18, 2015

Re: MX2 2015 Integration Marathon

braintic said:
So nobody is concerned that ln[sin(0)] does not exist?

Yet according to wolfram, the integral converges

braintic · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Yet according to wolfram, the integral converges

When I try it on Wolfram, it just repeats the question as the answer, and draws the region.
It is not telling me anything about convergence.

Paradoxica · Sep 18, 2015

Re: MX2 2015 Integration Marathon

braintic said:
When I try it on Wolfram, it just repeats the question as the answer, and draws the region.
It is not telling me anything about convergence.

If you look at the graph from the perspective of the y-axis, the area is bounded by a curve that approaches zero much faster than $\frac{1}{x}$
This is not a proof of the convergence, but gives some insight as to why it does converge.
Next problem: Evaluate:
$\int{\left \lfloor{x}\right \rfloor}dx$ , given that $\frac{d}{dx}\left \lfloor{x}\right \rfloor = 0$ to prevent ambiguity from arising.

glittergal96 · Sep 18, 2015

Re: MX2 2015 Integration Marathon

braintic said:
So nobody is concerned that ln[sin(0)] does not exist?

This is not a problem, because the leading order behaviour of sin(x) is that of x for small x. And the integral of log(t) from epsilon to 1 exists and converges as epsilon -> 0. (In other words, although the logarithm blows up at zero, it does not do so fast enough to make this integral diverge.)

If you wanted to be more concrete you could use the fact that sin(x) >= 2x/pi in the interval of integration.

So

$\int_0^{\pi/2} |\log(\sin(x))|\, dx\\ \\ \leq \int_0^{\pi/2} |\log(2x/\pi)|\, dx\\ \\ \leq \pi|\log(2/\pi)|/2+\int_0^{\pi/2} |\log(x)|\, dx\\ \\ <\infty.$

glittergal96 · Sep 18, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
If you look at the graph from the perspective of the y-axis, the area is bounded by a curve that approaches zero much faster than $\frac{1}{x}$
This is not a proof of the convergence, but gives some insight as to why it does converge.
Next problem: Evaluate:
$\int{\left \lfloor{x}\right \rfloor}dx$ , given that $\frac{d}{dx}\left \lfloor{x}\right \rfloor = 0$ to prevent ambiguity from arising.

To integrate from 0 to positive real x (which is the same as finding the values one particular primitive takes on the positive reals),

Just break it into the intervals where the integrand is piecewise constant and sum. We get:

$\sum_{k=1}^{\lfloor x\rfloor-1} k+(x-\lfloor x\rfloor)\lfloor x\rfloor=\lfloor x \rfloor (x-\lfloor x \rfloor/2 -1/2).$

Similarly, we can break up integrating from 0 to negative real x in terms of an almost identical sum, and get a similar expression for our primitive at negative reals. Too lazy right now but it is exactly the same length as the above.

Putting these two pieces together gives us a primitive for the original function over the reals.

Paradoxica · Sep 18, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
To integrate from 0 to positive real x (which is the same as finding the values one particular primitive takes on the positive reals),

Just break it into the intervals where the integrand is piecewise constant and sum. We get:

$\sum_{k=1}^{\lfloor x\rfloor-1} k+(x-\lfloor x\rfloor)\lfloor x\rfloor=\lfloor x \rfloor (x-\lfloor x \rfloor/2 -1/2).$

Similarly, we can break up integrating from 0 to negative real x in terms of an almost identical sum, and get a similar expression for our primitive at negative reals. Too lazy right now but it is exactly the same length as the above.

Putting these two pieces together gives us a primitive for the original function over the reals.

My answer says you are already correct. It does appear to handle the negative case, although for certainty, it should be considered rather than assuming it holds true for negative x.

Paradoxica · Sep 19, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
This is not a problem, because the leading order behaviour of sin(x) is that of x for small x. And the integral of log(t) from epsilon to 1 exists and converges as epsilon -> 0. (In other words, although the logarithm blows up at zero, it does not do so fast enough to make this integral diverge.)

If you wanted to be more concrete you could use the fact that sin(x) >= 2x/pi in the interval of integration.

So

$\int_0^{\pi/2} |\log(\sin(x))|\, dx\\ \\ \leq \int_0^{\pi/2} |\log(2x/\pi)|\, dx\\ \\ \leq \pi|\log(2/\pi)|/2+\int_0^{\pi/2} |\log(x)|\, dx\\ \\ <\infty.$

Another way to look at it is this: Invert the function, restricting the range so that $0<y\leq \frac{\pi}{2}$ where $y=\sin^{-1}{e^x}$
For small theta, $\sin{\theta} \approx \theta$ , and as a corollary, $\theta \approx \sin^{-1}{\theta}$
This means that any improper integral "all the way out over there" to negative infinity will be very close to the improper integral from "all the way out there" to negative infinity of $e^x$ . Since the improper integral of e^x in this given bound is finite, the improper integral of our case is also finite, otherwise approximation would fail to exist. Since the rest of the integral from "all the way out there" to here at x=0 is also finite, then that means the improper integral over the entire domain is finite. Ergo, our original integral must also be finite since it is identical.

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