# Tougher Integral

• Dec 27th 2008, 11:37 AM
RedBarchetta
Tougher 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 AM
Jester
Quote:

Originally Posted by RedBarchetta
I found this one on a Harvard graduate level exam:
http://img206.imageshack.us/img206/9027/toughqj9.png

Any clues?

Yes, $\frac{x^2}{x^4+5x^2+4}$ decomposes into

$\frac{4}{3} \frac{1}{x^2+4} - \frac{1}{3} \frac{1}{x^2+1}$

then integrate each separately.
• Dec 28th 2008, 01:05 AM
pberardi
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 AM
Isomorphism
Quote:

Originally Posted by pberardi
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)

Do partial fraction decomposition as danny points out,

Quote:

Originally Posted by danny arrigo
Yes, $\frac{x^2}{x^4+5x^2+4}$ decomposes into

$\frac{4}{3} \frac{1}{x^2+4} - \frac{1}{3} \frac{1}{x^2+1}$

then integrate each separately.

$\int_0^{\infty} \frac{4}{3} \frac{1}{x^2+4} \, dx - \int_0^{\infty}\frac{1}{3} \frac{1}{x^2+1}\, dx$ $= \left. \frac{2}{3} \tan^{-1}\frac{x}{2} - \frac{1}{3}\tan^{-1} x \right|_0^{\infty} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
• Dec 28th 2008, 03:05 AM
Prove It
Quote:

Originally Posted by danny arrigo
Yes, $\frac{x^2}{x^4+5x^2+4}$ decomposes into

$\frac{4}{3} \frac{1}{x^2+4} - \frac{1}{3} \frac{1}{x^2+1}$

then integrate each separately.

I hope this is right...

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

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

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

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

$= \frac{\pi}{6}$.
• Dec 28th 2008, 07:13 AM
Jester
Quote:

Originally Posted by Prove It
I hope this is right...

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

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

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

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

$= \frac{\pi}{6}$.

Looks good to me but on a personal note, I would use something other than $\varepsilon$. For me, $\varepsilon$ is something very small not large :)
• Dec 28th 2008, 10:24 AM
pberardi
Quote:

Originally Posted by RedBarchetta
I found this one on a Harvard graduate level exam:
http://img206.imageshack.us/img206/9027/toughqj9.png

Any clues?

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 PM
Mathstud28
Quote:

Originally Posted by RedBarchetta
I found this one on a Harvard graduate level exam:
http://img206.imageshack.us/img206/9027/toughqj9.png

Any clues?

Couldnt you use Countour integration on this? (Nerd). I am not sure how but I am sure TPH or someone could.
• Dec 28th 2008, 02:45 PM
ThePerfectHacker
Quote:

Originally Posted by Mathstud28
Couldnt you use Countour integration on this? (Nerd). I am not sure how but I am sure TPH or someone could.

Yes you can but I think the first approach is the best one.

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

Now define $f(z) = \frac{z^2}{z^4+5z^2+1}$ 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 PM
duydaniel
Quote:

Originally Posted by Isomorphism
Do partial fraction decomposition as danny points out,

$\int_0^{\infty} \frac{4}{3} \frac{1}{x^2+4} \, dx - \int_0^{\infty}\frac{1}{3} \frac{1}{x^2+1}\, dx$ $= \left. \frac{2}{3} \tan^{-1}\frac{x}{2} - \frac{1}{3}\tan^{-1} x \right|_0^{\infty} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$

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 PM
Krizalid
Note that $\left( x^{2}+4 \right)-\left( x^{2}+1 \right)=3,$ hence, no partial fractions method involved.
• Dec 29th 2008, 04:04 AM
HallsofIvy
Quote:

Originally Posted by duydaniel
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.

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 AM
Jester
Quote:

Originally Posted by duydaniel
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.

$A \;\text{and}\; B$ are numbers, not variables! To decompose, say

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

notice that

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

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

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

cross multipying gives

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

Expanding and re-grouping gives

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

comparing gives the equations

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

from which we solve giving

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

Thus,

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

Originally Posted by danny arrigo
$A \;\text{and}\; B$ are numbers, not variables! To decompose, say

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

notice that

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

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

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

cross multipying gives

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

Expanding and re-grouping gives

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

comparing gives the equations

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

from which we solve giving

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

Thus,

$\frac{- \frac{1}{3}}{x^2+1} + \frac{\frac{4}{3}}{x^2+4} = \frac{x^2}{(x^2+1)(x^2+4)}$

That's great!!!
I thought that the cross multiplying would ended up x^3 as in $(Ax+B)(x^2+4) + (Cx+D)(x^2+1)$ that I try to find A and B instead... but it was probably a long way to go (Headbang)