Question

Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\). b. Change variables by letting \(y=1 / x\). c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0\). d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\). (Source: Mathematics Magazine \(59,1,\) Feb 1986 )

Short answer

Question: Prove that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x = \pi\). Answer: The proof involves several steps, including integration by parts, changing variables, and evaluating integrals. By following these steps, we have demonstrated that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x = \pi\).

Step-by-step solution

Integrate by parts

Let's set \(u=\sqrt{x}\ln{x}\) and \(dv=\frac{1}{(1+x)^2}dx\). Then, we can calculate the differential of \(u\) and the integral of \(dv\): $$ du=\left(\frac{\ln x}{2\sqrt{x}} + \frac{1}{2\sqrt{x}}\right)dx \\ v=-\frac{1}{1+x} $$ Now, we can use integration by parts: $$ \int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} dx = \left[-\frac{\sqrt{x}\ln{x}}{1+x}\right]_{0}^{\infty} + \int_{0}^{\infty} \frac{1 + \ln x}{2(1+x)\sqrt{x}} dx $$

Change variables

Let's change variables by letting \(y=\frac{1}{x}\), where \(x=\frac{1}{y}\): $$ dx=-\frac{1}{y^2}dy $$ Then, make the substitution: $$ \int_{0}^{\infty} \frac{1 + \ln x}{2(1+x)\sqrt{x}} dx = -\int_{\infty}^{0} \frac{1+\ln\left(\frac{1}{y}\right)}{2\left(1+\frac{1}{y}\right)\sqrt{\frac{1}{y}}} \cdot \frac{1}{y^2} dy $$ Now, simplify the integral: $$ -\int_{\infty}^{0} \frac{1-\ln y}{2(1+y)y\sqrt{y}} dy = \int_{0}^{\infty} \frac{\ln y-1}{2(1+y)y\sqrt{y}} dy $$

Show equality and convergence of integrals

Let's consider the integral: $$ I(x)=\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} dx $$ Notice that: $$ -\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} dx = \int_{0}^{\infty} \frac{\ln y-1}{2(1+y)y\sqrt{y}} dy $$ Using the substitution of variables made in step 2, we find \(I(x)=-I(y)\). Therefore, both integrals converge and their combination results in \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} dx=0\).

Evaluate the remaining integral using change of variables

Now, we need to evaluate the remaining integral: $$ \int_{0}^{\infty} \frac{1}{2(1+x)\sqrt{x}} dx $$ Let's make this change of variables: \(z=\sqrt{x}\), so \(x=z^2\). Then, \(dx=2zdz\). The integral becomes: $$ \int_{0}^{\infty} \frac{1}{2(1+z^2)(2z)} 2z dz = \int_{0}^{\infty} \frac{1}{1+z^2} dz $$ This last integral is known to have its antiderivative as \(\arctan(z)\), so we have: $$ \int_{0}^{\infty} \frac{1}{1+z^2} dz = \left[\arctan(z)\right]_{0}^{\infty}=\left(\frac{\pi}{2}-0\right)=\frac{\pi}{2} $$ Finally, considering our boundary term from integration by parts and the calculated value of \(\int_{0}^{\infty} \frac{1}{2(1+x)\sqrt{x}} dx\), we obtain: $$ \int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} dx = \left[-\frac{\sqrt{x}\ln{x}}{1+x}\right]_{0}^{\infty} + \frac{\pi}{2} = \pi $$ So, we have shown that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\).

Key concepts

Integration by Parts

Integration by Parts is a technique used to transform the integrand of an integral into a more manageable form. It is derived from the product rule for differentiation. The formula is given by: \[ \int u \, dv = uv - \int v \, du \] where:

• \( u \) is a function of \( x \)
• \( dv \) is another differential quantity

For instance, in the provided exercise, setting \( u = \sqrt{x} \ln x \) and \( dv = \frac{1}{(1+x)^2}dx \) transforms the problem. We then differentiate \( u \) and integrate \( dv \) which results in a new expression that simplifies the integration process.
The choice of \( u \) and \( dv \) is crucial. A good rule of thumb is to choose \( u \) to be a function that becomes simpler when differentiated. This technique is particularly handy for integrals involving logarithmic, algebraic, or exponential functions.

Change of Variables

Changing variables, also known as substitution, is another powerful technique in calculus used to simplify integrals. By substituting part of an integral with a new variable, we can often transform a complex integral into a simpler one.
In this exercise, a change of variables was performed using \( y = \frac{1}{x} \), transforming \( dx \) into \( -\frac{1}{y^2}dy \).
This substitution effectively changes the limits of integration and can simplify the integrand by transforming the problematic parts into more standard forms. The integral's complexity often determines the choice of substitution, aiming for an expression that resembles a standard integral.

• It's important to properly adjust the limits of integration after substitution.
• Pay careful attention to how the substitution affects the differential \( dx \), which becomes \( dy \).

Convergence of Integrals

Convergence of integrals refers to whether an integral has a finite value as the bounds approach infinity or some point where the function might be undefined. When evaluating improper integrals, understanding convergence is crucial.

• An integral is said to converge if it approaches a finite number as an endpoint approaches infinity or an undefined point.
• Conversely, it diverges if it heads toward infinity or fails to settle on a single value.

In this exercise, determining the convergence of the integrals \( \int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} dx \) and \( -\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} dx \) is important because only convergent integrals are properly defined.
The key observation is checking whether their absolute values remain bounded. Pairing these evaluated integrals and their transformations can show that these specific integrals cancel each other out resulting in zero, aiding in determining the overall solution.

Trigonometric Integrals

Trigonometric integrals often involve functions such as \( \sin(x) \), \( \cos(x) \), or inverse functions like \( \arctan(x) \). These integrals can appear in various forms in calculus problems.
In the final step of the exercise, after changing variables to \( z=\sqrt{x} \), the integral simplifies to the form \( \int_{0}^{\infty} \frac{1}{1+z^2} dz \). Integrals with this specific form have solutions involving the inverse tangent function, \( \arctan(z) \).

• The integral \( \int \frac{1}{1+z^2} dz \) directly relates to the fundamental identity defined by \( \arctan(z) \).
• This relationship simplifies the evaluation of the integral result to \( \frac{\pi}{2} \), utilized because the integral ranges from \( 0 \) to \( \infty \).

By recognizing these trigonometric integral forms, calculations can be streamlined by applying known antiderivative results.