1. ## Tougher Integral

I found this one on a Harvard graduate level exam:

Any clues?

2. Originally Posted by RedBarchetta
I found this one on a Harvard graduate level exam:

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.

3. 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.

4. 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.
Do partial fraction decomposition as danny points out,

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

5. 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...

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

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

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

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

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

6. 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

7. Originally Posted by RedBarchetta
I found this one on a Harvard graduate level exam:

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.

8. Originally Posted by RedBarchetta
I found this one on a Harvard graduate level exam:

Any clues?
Couldnt you use Countour integration on this? . I am not sure how but I am sure TPH or someone could.

9. Originally Posted by Mathstud28
Couldnt you use Countour integration on this? . 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.

10. 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 . My question is how do you know a proper way to decompose it (as you did)?

Thank you.

11. Note that $\left( x^{2}+4 \right)-\left( x^{2}+1 \right)=3,$ hence, no partial fractions method involved.

12. 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 . 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!

13. 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 . 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,

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

14. 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