Question

Sketch the region whose area is \(\int_{0}^{\infty} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\), and use your sketch to show that \(\int_{0}^{\infty} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}=\int_{0}^{1} \sqrt{\frac{1-\mathrm{y}}{\mathrm{y}}} \mathrm{dy}\)

Short answer

Question: Prove that the integral transformation \( \int_{0}^{\infty} \frac{1}{1+x^{2}} \mathrm{dx}=\int_{0}^{1}\sqrt{\frac{1-\mathrm{y}}{\mathrm{y}}} \mathrm{dy}\) is valid. Answer: The integral transformation is valid as demonstrated by performing a substitution with the transformation \(x = \frac{\sqrt{y}}{1 - \sqrt{y}}\), which converts the first integral into the second integral after simplification.

Step-by-step solution

Sketch the region

To sketch the region whose area we want to find, we need to graph the function \(f(x)=\frac{1}{1+x^2}\) for \(x\geq 0\). Notice that the function is decreasing as \(x\) increases.

Analyze the sketch

Analyze the sketch and notice that the region lies between the curve \(y = \frac{1}{1+x^2}\), the x-axis, and the y-axis. We want to find the area under the curve from \(0\) to \(\infty\). It might be useful to find a transformation that will make calculating the area easier.

Find a transformation to relate the integrals

Let's examine the given integral transformation. We want to show that $\int_{0}^{\infty} \frac{1}{1+x^{2}} \mathrm{dx}=\int_{0}^{1} \sqrt{\frac{1-\mathrm{y}}{\mathrm{y}}} \mathrm{dy}\(. This transformation can be shown using a change of variables technique. Consider the transformation \)x = \frac{\sqrt{y}}{1 - \sqrt{y}}\(, which implies that \)\mathrm{dx} = \frac{\mathrm{dy}}{2(1-\sqrt{y})\sqrt{y}}\(. We can also obtain \)y=\frac{x^2}{1+x^2}$ from the original transformation. Applying this change of variables to the first integral, we get: \(\int_{0}^{\infty} \frac{dx}{1+x^{2}} = \int_{0}^{1} \frac{1}{(1+\frac{y}{1-y})^2} \cdot \frac{dy}{2(1-\sqrt{y})\sqrt{y}}\) Simplifying, we get: \(\int_{0}^{1} \frac{1}{\frac{(1-y)^2+y^2}{(1-y)^2}} \cdot \frac{dy}{2(1-\sqrt{y})\sqrt{y}} = \int_{0}^{1} \frac{(1-y)^2}{(1-y)^2+y^2} \cdot \frac{dy}{2(1-\sqrt{y})\sqrt{y}}\) Since \((1-y)^2+y^2 = 1\), we finally get: \(\int_{0}^{1} \sqrt{\frac{1-\mathrm{y}}{\mathrm{y}}} \mathrm{dy}\) Thus, the given transformation is valid, and we have shown that $\int_{0}^{\infty} \frac{1}{1+x^{2}} \mathrm{dx}=\int_{0}^{1} \sqrt{\frac{1-\mathrm{y}}{\mathrm{y}}} \mathrm{dy}$.

Key concepts

Improper Integral

The term 'improper integral' refers to an integral where either the interval of integration is infinite or the function to be integrated has one or more points of discontinuity over the interval. In our exercise, we deal with the former case, \(\int_{0}^{\infty} \frac{\mathrm{dx}}{1+\mathrm{x}^{2}}\), where the upper limit of the integration is infinity. Solving improper integrals usually involves taking a limit; we interpret \(\int_{0}^{\infty} f(x) \, dx\) as \(\lim_{b \to \infty}\int_{0}^{b} f(x) \, dx\).

One common technique involves establishing a convergent comparison integral if the function behaves well as \(x\) approaches infinity. In other words, we determine whether the area under the curve has a finite value as we stretch the region over the entire positive x-axis. This question of convergence is crucial because it guides us in justifying the calculation of the area under the curve for such functions.

Area Under a Curve

In integral calculus, the 'area under a curve' is a fundamental concept, representing the accumulation of quantities, and it's computed by finding the integral of a function. The region whose area we're evaluating is defined by the curve \(y = \frac{1}{1+x^2}\), the x-axis, and the y-axis, extending from \(x = 0\) to infinity.

Interpreting the Area

For a positive function like our \(\frac{1}{1+x^2}\), the integral from 0 to \(\infty\) corresponds to the total area between the function's graph and the x-axis over the specified interval. Since the integrand decreases as \(x\) increases and does not have any discontinuity, the area is well-defined. It’s important to note that while the area under some curves can be infinite, for others, like in our exercise, it converges to a finite value.

Integration Techniques

To calculate integrals, especially those that aren't straightforward, we use various 'integration techniques.' These can include substitution, integration by parts, partial fraction decomposition, trigonometric integration, and more. Each technique has its own applications and is used based on the form of the integrand and the interval of integration.

Finding a Suitable Technique

In the provided exercise, the area under the curve is not easy to compute directly due to the infinite interval. However, with the appropriate technique, such as a substitution or a transformation, the complexity can often be reduced, and antiderivatives can be found. This problem also showcases how recognizing the algebraic identity \(1+y^2 = (1-y)^2 + y^2\) simplifies the integral after the change of variables, thus revealing the connection between two seemingly different integrals.

Change of Variables

The 'change of variables' technique, also known as u-substitution, is a powerful tool in integration that often simplifies the integration process. It involves replacing the original variable of integration with a new variable and its corresponding differential. This can transform a difficult integral into a simpler one that we are more equipped to solve.

Implementing the Change

The exercise presents a change of variables that transforms the improper integral into a proper one on a finite interval. Here, we use the transformation \(x = \frac{\sqrt{y}}{1 - \sqrt{y}}\), leading to a new differential \(\mathrm{dx}\). This results in the integrand and limits of integration being expressed in terms of the new variable \(y\), ultimately revealing the identical area under a different curve. The careful selection of this transformation is key to making the integration work, and ensuring that the transformed integral is equivalent to the original one.