Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that $$ \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\left(1+x^{2}+y^{2}\right)^{2}} d y d x=\frac{\pi}{4} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{4} \).

Step 1: Rewrite the Integral in Polar Coordinates

To evaluate the double integral \( \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{(1+x^2+y^2)^2} \, dy \, dx \), it's easier to convert it into polar coordinates. In polar coordinates, \( x = r \cos\theta \) and \( y = r \sin\theta \), and the Jacobian of the transformation from Cartesian to polar coordinates is \( r \, dr \, d\theta \). Thus, the integrand becomes \( \frac{1}{(1+r^2)^2} \times r \) and the limits of integration become \( 0 \leq r < \infty \) and \( 0 \leq \theta \leq \frac{\pi}{2} \). This gives us the new integral:\[\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{r}{(1+r^2)^2} \, dr \, d\theta.\]

Step 2: Evaluate the Inner Integral with Respect to \(r\)

Now we focus on the inner integral with respect to \( r \):\[\int_{0}^{\infty} \frac{r}{(1+r^2)^2} \, dr.\]To solve this, use the substitution \( u = 1 + r^2 \), which implies \( du = 2r \, dr \), or equivalently \( r \, dr = \frac{1}{2} du \). The limits of \( r \) from \( 0 \) to \( \infty \) transform to \( u \) from \( 1 \) to \( \infty \), and the integral becomes:\[\int_{1}^{\infty} \frac{1}{2u^2} \, du = \frac{1}{2} \int_{1}^{\infty} u^{-2} \, du.\]

Step 3: Solve the Integral \( \int_{1}^{\infty} u^{-2} \, du \)

Calculate the antiderivative:\[\int_{1}^{\infty} u^{-2} \, du = \left[ -u^{-1} \right]_{1}^{\infty} = \left[-\frac{1}{u}\right]_{1}^{\infty}.\]Evaluating this at the bounds gives\[\left(-\frac{1}{\infty}\right) - \left(-\frac{1}{1}\right) = 0 + 1 = 1.\]Hence, the original integral evaluates to \( \frac{1}{2} \times 1 = \frac{1}{2} \).

Step 4: Evaluate the Outer Integral with Respect to \(\theta\)

Substitute the result from the inner integral back into the outer integral:\[\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \, d\theta.\]This simplifies to:\[\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{1}{2} \cdot \left[ \theta \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.\]

Step 5: Conclusion: Final Result

After evaluating both the inner and outer integrals, we arrive at the final result:\[\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{(1+x^2+y^2)^2} \, dy \, dx = \frac{\pi}{4}.\]This confirms that the given integral equals \( \frac{\pi}{4} \).

Key Concepts

Polar Coordinates

When working with double integrals, sometimes Cartesian coordinates (\(x, y\)) become cumbersome, especially with circular symmetry. That's where polar coordinates (\(r, \theta\)) shine. By converting to polar coordinates, integration can be simplified. In the given problem, replace \(x = r \cos\theta\) and \(y = r \sin\theta\). The circle's radius \(r\) spans from 0 to infinity, and the angle \(\theta\) ranges from 0 to \(\frac{\pi}{2}\).

When changing to polar coordinates, remember the Jacobian determinant \(r \, dr \, d\theta\). It accounts for the "stretching" factor of the coordinate transformation, so it's included in the integrand. This substitution transformed our integrand to \(\frac{r}{(1+r^2)^2}\), facilitating easier evaluation of the integral.

Substitution Method

The substitution method is pivotal for simplifying integrals. In the inner integral concerning \(r\), the substitution \(u = 1 + r^2\) helps us handle the expression algebraically. This technique translates the problem into terms easier to integrate.

Upon substitution, the differential \(du = 2r \, dr\) implies \(r \, dr = \frac{1}{2} \, du\). Substitution then changes the complex integral into a simpler form \(\int_{1}^{\infty} \frac{1}{2u^2} \, du\). Key to this method is also adjusting the integration limits: when \(r = 0, u = 1\) and as \(r\) heads to infinity, \(u\) does too.

Integration Limits

Integration limits define where the integration starts and ends for a given variable. They're essential for solving definite integrals and must match the coordinate system being used.

In Cartesian coordinates, both \(x\) and \(y\) initially ranged from 0 to infinity. However, after transforming to polar coordinates, \(r\) holds from 0 to infinity, while \(\theta\) ranges from 0 to \(\frac{\pi}{2}\). This equivalence needs attention when converting variables. If improper limits are used, the result won't be valid, reaffirming the significance of exact integration ranges.

Antiderivative Calculation

Successfully computing integrals heavily relies on finding correct antiderivatives. Consider the integral \(\int_{1}^{\infty} u^{-2} \, du\); its antiderivative is given by \(-u^{-1}\). Evaluating between its limits yields \(-\frac{1}{\infty} + 1 = 1\).

Finding antiderivatives requires recognizing common integral patterns, such as power rules or trigonometric rules, and applying them efficiently. Once the antiderivative is found, assessing it at the given limits provides the definite integral’s value. In the problem, multiplying the result by \(\frac{1}{2}\) and integrating over \(\theta\) gives the final answer \(\frac{\pi}{4}\). Precision in antiderivative selection assures accuracy in double integral calculations.