# [SOLVED] Another improper integral

• August 13th 2008, 01:08 PM
Spec
[SOLVED] Another improper integral
$\int \int_D \frac{dxdy}{\sqrt{xy}}$ where $D = \left\{x+y\leq1, x> 0, y> 0\right\}$

$\int_b^1 \left(\int_b^{1-x} \frac{dy}{\sqrt{xy}}\right)dx = 2\int_b^1 \frac{1}{x}(\sqrt{x-x^2} -\sqrt{bx})dx$

I can't solve the last integral, and the answer I get from Mathematica does not work for these limits. So I suspect the above isn't correct.

EDIT: Here's another tricky improper integral:

$\int \int \int_D \frac{dxdydz}{\sqrt{x^2+y^2+z^2}}$ where $\sqrt{x^2+y^2} < z < 1$

EDIT: And another one:

$\int \int_{R^2} \frac{xdxdy}{(x^2+y^2)^2 +1}$
• August 13th 2008, 01:16 PM
galactus
Are you sure that isn't dydx instead of dxdy?. The way it is you have a dependent limit.
• August 14th 2008, 04:08 AM
shawsend
Quote:

Originally Posted by Spec
I can't solve the last integral, and the answer I get from Mathematica does not work for these limits.

Hey Spec. This is what I get with Mathematica 6: What's wrong with it?

$\text{If}\left[\left(\text{Re}\left[\frac{b}{1-b}\right]\geq 0\&\&\frac{b}{-1+b}\neq 0\right)\left\|\text{Re}\left[\frac{b}{1-b}\right]\leq -1\right\|\text{Im}\left[\frac{b}{1-b}\right]\neq 0]\right]$

$
\frac{-\frac{4}{\sqrt{\frac{1}{b}}}+\left(2 \left(\sqrt{-1+b} \sqrt{b} \left(-1+2 \sqrt{-\frac{b}{-1+b}}\right)+\text{Log}\left[\sqrt{-1+b}+
\sqrt{b}\right]\right)\right)}{\left(\sqrt{-1+b} \sqrt{\frac{1}{b}} \sqrt{\frac{b}{1-b}}\right)}$

Also, if you want to cut-and-paste Mathematica as latex code, select the code in Mathmatica, then choose Edit/Copy as/Latex (with some minor limitations). For example, I didn't type in any of that code above although I did have to re-format it slightly because of some limitations on latex length in here.
• August 14th 2008, 02:49 PM
Spec
The integral (and answer) looks way too complicated, and we're supposed to be able to solve all these problems without using any tools.
• August 14th 2008, 03:50 PM
NonCommAlg
Quote:

Originally Posted by Spec
$\int \int_D \frac{dxdy}{\sqrt{xy}}$ where $D = \left\{x+y\leq1, x> 0, y> 0\right\}$

$=\int_0^1 \frac{2}{\sqrt{x}} [ \sqrt{y}]_0^{1-x} \ dx=2\int_0^1 \sqrt{\frac{1-x}{x}} \ dx=4\int_0^{\frac{\pi}{2}} \cos^2 \theta \ d \theta = \pi. \ \ \ \square$

in the last integral we put $x=\sin^2 \theta.$

Quote:

$\int \int \int_D \frac{dxdydz}{\sqrt{x^2+y^2+z^2}}$ where $\sqrt{x^2+y^2} < z < 1$

$=\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^{\sec \phi} \rho \sin \phi \ d \rho \ d \phi \ d \theta=\pi \int_0^{\frac{\pi}{4}} \sec^2 \phi \ \sin \phi \ d \phi = \pi [\sec \phi]_0^{\frac{\pi}{4}}=\pi(\sqrt{2} \ - \ 1). \ \square$

note that in spherical coordinates $z=\sqrt{x^2+y^2}$ becomes $\phi = \frac{\pi}{4}$ and $z=1$ becomes $\rho=\sec \phi.$

Quote:

$\int \int_{R^2} \frac{xdxdy}{(x^2+y^2)^2 +1}$

$= \int_0^{\infty} \int_0^{2 \pi} \frac{r^2 \cos \theta}{r^4 + 1} \ d \theta \ dr = \int_0^{\infty} \frac{r^2}{r^4 + 1} [\sin \theta]_0^{2 \pi} \ dr=0. \ \ \ \square$

note that $\int_0^{\infty} \frac{r^2}{r^4 + 1} \ dr$ is convergent.
• August 15th 2008, 05:26 AM
Spec
Quote:

Originally Posted by NonCommAlg
$= \int_0^{\infty} \int_0^{2 \pi} \frac{r^2 \cos \theta}{r^4 + 1} \ d \theta \ dr = \int_0^{\infty} \frac{r^2}{r^4 + 1} [\sin \theta]_0^{2 \pi} \ dr=0. \ \ \ \square$

note that $\int_0^{\infty} \frac{r^2}{r^4 + 1} \ dr$ is convergent.

Does that matter though? Since your multiplying it with zero either way. I actually came that far myself, but struggled with the calculation of $\int_0^{\infty} \frac{r^2dr}{r^4 + 1}$ (I tried to integrate it before $d\theta$, and thus didn't notice that the $d\theta$ integral was zero).