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Forgot how to do these : $$\displaystyle\int_0^1\int_0^1\frac{\text{d}x\text{d}y}{1-x^2y^2}$$

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First, integrate w.r.t. $x$ then proceed to $y$. –  hjpotter92 Jan 27 '13 at 11:30

3 Answers 3

up vote 3 down vote accepted

Consider the integral over $x$:

$$\int_0^1 dx \; \frac{1}{1-x^2 y^2} $$

Make the substitution $x = \sin{\theta}/y$, $dx = \cos{\theta}/y \: d \theta$:

$$\begin{align} &= \frac{1}{y} \int_0^{\arcsin{y}} d \theta \sec{\theta} \\ &= \frac{1}{y} [\log{(\sec{\theta} + \tan{\theta})]_{0}^{\arcsin{y}}} \\ &= \frac{1}{2 y} \log{\left ( \frac{1+y}{1-y} \right )} \end{align} $$

Now you can do the integral over $y$:

$$\begin{align} \int_0^1 dy \: \int_0^1 dx \: \frac{1}{1-x^2 y^2} &= \frac{1}{2} \int_0^1 dy \: \frac{1}{y} \log{\left ( \frac{1+y}{1-y} \right )}\\ &= \sum_{n=0}^{\infty} \int_0^1 dy \: \frac{y^{2 n}}{2 n+1} \\ &= \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}\\ &= \frac{\pi^2}{8} \\\\ \end{align} $$

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1. $dx = \frac{cos\theta}{y}$ and 2. The integral is missing $y$ in denominator. –  hjpotter92 Jan 27 '13 at 11:36
    
@BackinaFlash: thanks - error did not affect the other results. –  Ron Gordon Jan 27 '13 at 11:43
    
@rlgordonma : Thx man! But I dont know why the answer says $\frac{\pi ^2}{8}$? –  Ryan Jan 27 '13 at 11:49
1  
@Ryan - I made an error, I apologize. I have fixed it. –  Ron Gordon Jan 27 '13 at 11:50
    
@rlgordonma I think you'll find that $$\sum_{n=1}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8} - 1 \, . $$ Your lower limit should be $n=0$ and not $n=1$. You still need to explain why the sum of $1+1/3^2+1/5^2+\cdots$ comes to $\pi^2/8$. If the OP can't remember how to integrate $1/(1-x^2y^2)$, I doubt they'll be comfortable manipulating Fourier series. –  Fly by Night Jan 27 '13 at 12:08
  • Note that your area $0\leq x\leq 1, 0\leq y\leq 1$ is a rectangular area so you can divide the integral respect to $x$ and $y$ as: $$\int_0^1~dx\times\int_0^1~f dy$$

  • We know that $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln|\frac{x+a}{x-a}|+C$.

  • Write $\frac{1}{1-x^2y^2}$ as $\frac{(1/y^2)}{\left(\frac{1}{y^2}-x^2\right)}$ and then assume $1/y^2=a^2$ temporarily.

So first solve $$A=\int_0^1\frac{a^2}{a^2-x^2}~dx$$ and then solve $$\int_0^1Ady$$

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+1 Nice outline without doing the problem for the OP! –  amWhy Jan 27 '13 at 15:13

Here is another way. $$\dfrac1{1-x^2y^2} = \sum_{k=0}^{\infty} x^{2k} y^{2k}$$ Hence, \begin{align} \int_0^1 \int_0^1 \dfrac{dx dy}{1-x^2y^2} & = \int_0^1 \int_0^1 \sum_{k=0}^{\infty} x^{2k} y^{2k} dxdy = \sum_{k=0}^{\infty} \int_0^1 \int_0^1 x^{2k} y^{2k} dxdy\\ & = \sum_{k=0}^{\infty} \left(\int_0^1 x^{2k} dx \right)^2 = \sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} = \dfrac34 \zeta(2) = \dfrac{\pi^2}8 \end{align}

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