I found this one on a Harvard graduate level exam:

http://img206.imageshack.us/img206/9027/toughqj9.png

Any clues?

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- Dec 27th 2008, 11:37 AMRedBarchettaTougher Integral
I found this one on a Harvard graduate level exam:

http://img206.imageshack.us/img206/9027/toughqj9.png

Any clues? - Dec 27th 2008, 11:48 AMJester
- Dec 28th 2008, 01:05 AMpberardi
Could someone kindly do this improper integral? It has been a while since I have done it, I am very interested in seeing the solution, and would appreciate a proper refresher. Pun intended. (Nerd)

- Dec 28th 2008, 02:44 AMIsomorphism
Do partial fraction decomposition as danny points out,

$\displaystyle \int_0^{\infty} \frac{4}{3} \frac{1}{x^2+4} \, dx - \int_0^{\infty}\frac{1}{3} \frac{1}{x^2+1}\, dx $$\displaystyle = \left. \frac{2}{3} \tan^{-1}\frac{x}{2} - \frac{1}{3}\tan^{-1} x \right|_0^{\infty} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} $ - Dec 28th 2008, 03:05 AMProve It
I hope this is right...

$\displaystyle \frac{4}{3}\int_0^\infty{\frac{1}{x^2 + 4}\,dx} - \frac{1}{3}\int_0^\infty{\frac{1}{x^2 + 1}\,dx} = \frac{4}{3}\left[\frac{1}{2}\arctan{\frac{x}{2}}\right]_0^\varepsilon - \frac{1}{3}\left[\arctan{x}\right]_0^\varepsilon$

$\displaystyle = \frac{2}{3}[\lim_{\varepsilon \to \infty}(\arctan{\frac{\varepsilon}{2}}) - \arctan{0}] - \frac{1}{3}[\lim_{\varepsilon \to \infty}(\arctan{\varepsilon}) - \arctan{0}]$

$\displaystyle = \frac{2}{3}\left[\frac{\pi}{2} - 0 \right] - \frac{1}{3}\left[\frac{\pi}{2} - 0 \right]$

$\displaystyle = \frac{\pi}{3} - \frac{\pi}{6}$

$\displaystyle = \frac{\pi}{6}$. - Dec 28th 2008, 07:13 AMJester
- Dec 28th 2008, 10:24 AMpberardi
Mr. Red,

Would you mind revealing in what class you saw this problem? I am just curious which classes at the graduate level test students on their calculus abilities. Also, was this already set up or was it part of a word problem? Any more information on this would be very interesting.

Thanks. - Dec 28th 2008, 02:27 PMMathstud28
- Dec 28th 2008, 02:45 PMThePerfectHacker
Yes you can but I think the first approach is the best one.

Notice that, $\displaystyle \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{x^4+5x^2+1} dx = \int_0^{\infty} \frac{x^2}{x^4+5x^2+1}dx$

Now define $\displaystyle f(z) = \frac{z^2}{z^4+5z^2+1} $ and find its residues in the upper half plane.

Read my tutorial on it, it should go smoothly for thee. - Dec 28th 2008, 03:48 PMduydaniel
Sorry, I don't know how to write it in proper format.

Here is what I did:

The function should have the following form:

A/(x^2+1) + B/(X^2+4)

then let A = 1 => solve for B:

=> B = -4/(x^2+1)

substitute B back into the function

=> we have something like:

1/(x^2+1) - 4/[(x^2+4)*(x^2+1)]

The second integral need to split up again (Lipssealed). My question is how do you know a proper way to decompose it (as you did)?

Thank you. - Dec 28th 2008, 04:01 PMKrizalid
Note that $\displaystyle \left( x^{2}+4 \right)-\left( x^{2}+1 \right)=3,$ hence, no partial fractions method involved.

- Dec 29th 2008, 04:04 AMHallsofIvy
That's "partial fractions" and you typically learn it in "Calculus II", a Freshman or Sophomore class (unless you learned calculus in secondary school). I see now that Krizalid did not need partial fractions because he was able to do the fractions easily- he's sharper than I am!

- Dec 29th 2008, 08:15 AMJester
$\displaystyle A \;\text{and}\; B$ are numbers, not variables! To decompose, say

$\displaystyle \frac{x^2}{x^4+5x^2+4}$

notice that

$\displaystyle \frac{x^2}{(x^2+1)(x^2+4)}$

so we seek numbers A, B, C and D such that

$\displaystyle \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}$

cross multipying gives

$\displaystyle (Ax+B)(x^2+4) + (Cx+D)(x^2+1) = x^2$.

Expanding and re-grouping gives

$\displaystyle (A+C)x^3 + (B+D)x^2 +(4A+C)x + 4B+D = x^2$

comparing gives the equations

$\displaystyle A+C=0,\;\;\;B+D = 1,\;\;\;4A+C=0,\;\;\;4B+D = 0$

from which we solve giving

$\displaystyle A = 0,\;\;\;B = - \frac{1}{3},\;\;\;C=0,\;\;\;D=\frac{4}{3}$

Thus,

$\displaystyle \frac{- \frac{1}{3}}{x^2+1} + \frac{\frac{4}{3}}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}$ - Dec 29th 2008, 08:36 AMduydaniel