HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

leehuan · Jan 11, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
This is just a standard log solution. We can simplify the sum of k's using an A.P. sum formula. Did you possibly mean product instead of sum on the denominator (makes it more interesting)?

Paradoxica said:
$$\noindent$I = \int \frac{\text{d}x}{(n+1)x + \frac{n(n+1)}{2}} =\frac{1}{n+1}\int \frac{\text{d}x}{x + \frac{n}{2}} = \frac{1}{n+1}\log(x+\frac{n}{2})+C\\I = \frac{1}{n+1}\log(2x+n)+C_0$

$$\noindent Assuming Leehuan meant $\prod$ instead of $\sum$, gives us an integral that is a bit more difficult to handle$

I did. Let me fix it lol.

$\int { \frac { dx }{ \prod _{ k=0 }^{ n }{ \left( x+k \right) } } }$

seanieg89 · Jan 11, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I did. Let me fix it lol.

$\int { \frac { dx }{ \prod _{ k=0 }^{ n }{ \left( x+k \right) } } }$

Are you sure there is a "nice" closed form? (It is easy to express it as a sum of logs, with nice coefficients.)

InteGrand · Jan 11, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
Are you sure there is a "nice" closed form? (It is easy to express it as a sum of logs, with nice coefficients.)

$\noindent I think so, just a bit tedious to get the expression to look nice.$

$\noindent Let $f(x)=\frac{1}{x(x+1)(x+2)\ldots (x+n)}$. We let $\frac{1}{x(x+1)(x+2)\ldots (x+n)}\equiv \sum_{k=0}^{n}\frac{A_k}{x+k}$ (partial fractions). Our goal is essentially just to find an expression for $A_k$ in terms of $k$. The rest is trivial (sum of logs with $A_k$'s as coefficients).$

$\noindent Using the Heaviside Cover-Up method, we find $A_k$ by substituting in $x=-k$ into the the expression for $f(x)$ except for the bracket $(x+k)$ ($k=0,1,2,\ldots,n$). I.e.$

$\begin{align*}A_k &= \frac{1}{(-k)(-k+1)(-k+2)\ldots (-k+(k-1)) (-k+(k+1))\ldots (-k+(n-1))(-k+n)}.\end{align*}$

$\noindent As can be seen, we have `skipped' the $(k-k)$.$

$\noindent Now, we have$

$\begin{align*} (-k)(-k+1)(-k+2)\ldots (-k+(k-1))&= -(k) \left[-(k-1)\right]\left[-(k-2)\right]\ldots \left[-(\underbrace{k-(k-1)}_{1})\right] \\ &= (-1)^k k! \quad (\text{as there are }k \text{ factors of } (-1) \text{ and we are multiplying all the positive integers up to }k \text{ (in the case of }k=0,\text{ we just have nothing and the conventions of }0! =1 \text{ and } (-1)^0 \text{ hold})).\end{align*}$

$\noindent Also,$

$\begin{align*}(-k+(k+1))\ldots (-k+(n-1))(-k+n) &= (n-k)!.\end{align*}$

$\noindent So $A_k = \frac{1}{(-1)^k \cdot k!\cdot (n-k)!}$, $k=0,1,2,\ldots,n$ (I think this is as `nice' as we can get). The rest (finishing off the integral as a sum of logs) is left as an exercise to the interested student.$

Edit: Oh right, misread, thought you said "is it", rather than "it is" in the brackets.

Paradoxica · Jan 11, 2016

Re: MX2 2016 Integration Marathon

We can express the integrand prior to integrating as a quotient of Gammas.

seanieg89 · Jan 11, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Edit: Oh right, misread, thought you said "is it", rather than "it is" in the brackets.

. Thanks for texing it though! I did get the same.

leehuan · Jan 11, 2016

Re: MX2 2016 Integration Marathon

You did the question last year as well InteGrand

InteGrand · Jan 11, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
You did the question last year as well InteGrand

Did I? Must have forgotten I did it!

seanieg89 · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Rings a bell with me too lol.

Paradoxica · Jan 12, 2016

Re: MX2 2016 Integration Marathon

$\noindent a) Differentiate $e^{f(x)}g(x) \\ $b) $\int e^{\cos{x}}(x\sin{x}+\cos{x}\csc^2{x})$d$x $(Hint: Add and subtract unity)$

seanieg89 · Jan 12, 2016

Re: MX2 2016 Integration Marathon

On the easy side, but:

$\lim_{R\rightarrow\infty}\int_{-R}^R \frac{\sin(ax)\sin(bx)}{x^2}\, dx \qquad (a,b\in\mathbb{R}).$

Paradoxica · Jan 12, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
On the easy side, but:

$\lim_{R\rightarrow\infty}\int_{-R}^R \frac{\sin(ax)\sin(bx)}{x^2}\, dx \qquad (a,b\in\mathbb{R}).$

Didn't I post this on last year's marathon? Drsoccerball solved it.

seanieg89 · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Didn't I post this on last year's marathon? Drsoccerball solved it.

Probably. I can't remember / didn't see it. Anyway, will leave it up as I am sure some students haven't done it.

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Didn't I post this on last year's marathon? Drsoccerball solved it.

LOL don't remember? And also integrand did solve that question because Lee and I got slightly different answers.

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent a) Differentiate $e^{f(x)}g(x) \\ $b) $\int \frac{e^{\cos{x}}(x\sin^{3}{x}+\cos{x})}{\sin^2{x}}$d$x $(Hint: Add and subtract unity)$

$a) \frac{d}{dx}{e^{f(x)}g(x)} = e^{f(x)}(g(x)f'(x)+g'(x))$

$b) $ Note that also $ \frac{d}{dx}{e^{f(x)}g(x)D(x)} = e^{f(x)}(g'(x)+f'(x)g(x)+g(x)D'(x))$

$$ Equate common terms $ (g'(x)+f'(x)g(x)+g(x)D'(x)) = x \sin{x} +\frac{\cos{x}}{\sin ^2x}$

$$ By inspection $ f(x)= \cos{x} , g(x)= \frac{-1}{\sin{x}} , D'(x)= \sin{x} -x \sin ^2x$

$\therefore I= \int \frac{e^{\cos{x}}(x\sin^{3}{x}+\cos{x})}{\sin^2{x}}$d$x = - \frac{e^{\cos{x}}}{\sin{x}} \cdot \int{(sinx-x \sin ^2x)}dx$

$I=\frac{e^{\cos{x}}}{\sin{x}}\cdot (-\frac{x^2}{4}+\frac{x \sin{2x}}{4}+\frac{\cos{2x}}{8} )+\epsilon$

EDIT: Integrated the last part wrong.

Paradoxica · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$a) \frac{d}{dx}{e^{f(x)}g(x)} = e^{f(x)}(g(x)f'(x)+g'(x))$

$b) $ Note that also $ \frac{d}{dx}{e^{f(x)}g(x)D(x)} = e^{f(x)}(g'(x)+f'(x)g(x)+g(x)D'(x))$

$$ Equate common terms $ (g'(x)+f'(x)g(x)+g(x)D'(x)) = x \sin{x} +\frac{\cos{x}}{\sin ^2x}$

$$ By inspection $ f(x)= \cos{x} , g(x)= \frac{-1}{\sin{x}} , D'(x)= \sin{x} -x \sin ^2x$

$\therefore I= \int \frac{e^{\cos{x}}(x\sin^{3}{x}+\cos{x})}{\sin^2{x}}$d$x = - \frac{e^{\cos{x}}}{\sin{x}} \cdot \int{(sinx-x \sin ^2x)}dx$

$I=\frac{e^{\cos{x}}}{\sin{x}}\cdot (\cos{x} + \frac{\cos{2x}}{8}) +\epsilon$

That's not right. The answer contains an x as a linear term.

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

My bad I differentiated part b wrong.

omegadot · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\noindent a) Differentiate $e^{f(x)}g(x) \\ $b) $\int e^{\cos{x}}(x\sin{x}+\cos{x}\csc^2{x})$d$x $(Hint: Add and subtract unity)$

$$\noindent (a) $\frac{d}{dx}(e^{f(x)} g(x) = e^{f(x)}[f'(x) g(x) + g'(x)].$

$$\noindent (b) To use the result of (a) in (b) we note that $f(x) = \cos x, f'(x) = -\sin x.$

$$\noindent So for the term in square brackets we have$\\\begin{align*} f'(x) g(x) + g'(x) &= x \sin x + \frac{\cos x}{\sin^2 x}\\\Rightarrow - \sin x g(x) + g'(x) &= x \sin x + 1 + \frac{\cos x}{\sin^2 x} - 1\end{align*}$

$$\noindent By inspection, $g(x) = - x - \frac{1}{\sin x}, g'(x) = - 1 + \frac{\cos x}{\sin^2 x}.$

$$\noindent Now finding the integral we have$\\\begin{align*} \int e^{\cos x} \left (x \sin x + \frac{\cos x}{\sin^2 x} \right ) \, dx &= \int \frac{d}{dx} \left [e^{\cos x} \left (-x - \frac{1}{\sin x} \right ) \right ] \, dx = -e^{\cos x} \left (x + \frac{1}{\sin x} \right ) + \cal{C}.\end{align*}$

leehuan · Jan 12, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent (a) $\frac{d}{dx}(e^{f(x)} g(x) = e^{f(x)}[f'(x) g(x) + g'(x)].$

$$\noindent (b) To use the result of (a) in (b) we note that $f(x) = \cos x, f'(x) = -\sin x.$

$$\noindent So for the term in square brackets we have$\\\begin{align*} f'(x) g(x) + g'(x) &= x \sin x + \frac{\cos x}{\sin^2 x}\\\Rightarrow - \sin x g(x) + g'(x) &= x \sin x + 1 + \frac{\cos x}{\sin^2 x} - 1\end{align*}$

$$\noindent By inspection, $g(x) = - x - \frac{1}{\sin x}, g'(x) = - 1 + \frac{\cos x}{\sin^2 x}.$

$$\noindent Now finding the integral we have$\\\begin{align*} \int e^{\cos x} \left (x \sin x + \frac{\cos x}{\sin^2 x} \right ) \, dx &= \int \frac{d}{dx} \left [e^{\cos x} \left (-1 - \frac{1}{\sin x} \right ) \right ] \, dx = -e^{\cos x} \left (x + \frac{1}{\sin x} \right ) + \cal{C}.\end{align*}$

Is that supposed to be x+cscx and not 1+cscx?

Drsoccerball · Jan 12, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Is that supposed to be x+cscx and not 1+cscx?

Remember the integral is in the form of $e^{f(x)} g(x), g(x) = -x- \frac{1}{\sin{x}}$

leehuan · Jan 12, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Remember the integral is in the form of $e^{f(x)} g(x), g(x) = -x- \frac{1}{\sin{x}}$

But I thought he did that already and then used the fundamental theorem of calculus to revert the derivative?

$\int \frac{d}{dx} \left [e^{\cos x} \left (-1 - \frac{1}{\sin x} \right ) \right ] \, dx = -e^{\cos x} \left (x + \frac{1}{\sin x} \right ) + \cal{C}$

Or was the wrong part of the working what was being integrated. I'm not fully processing maths today to figure it out myself

Edit: Lazied my way onto Wolfram Alpha instead. Yeah I think the integrand got written wrongly, that was all.

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