Question

Given that \(\int_{0}^{1} \frac{\ln x}{(1+x) \sqrt{x}} d x\) is a convergent improper integral, prove that \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}=0\).

Short answer

Question: If \(\int_{0}^{1} \frac{\ln x}{(1+x) \sqrt{x}} d x\) is converging, prove that \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}=0\). Solution: After performing a change of variable and adjusting the limits of integration, we found that \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = \int_{0}^{\infty} \frac{\ln y}{1+y^2} d y\). Adding these integrals to the converging integral, we have: $$a + \int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = 2\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}$$ where \(a\) is a finite value representing the convergent given integral. The only way for this equation to hold is if \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}=0\). Therefore, we have proven that \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}=0\).

Step-by-step solution

Introduce a new variable

Let's introduce a new variable \(y\), such that \(y = \frac{1}{x}\). Now find the values of \(x\) when \(y=0\) and \(y \to \infty\), which gives us \(x=\infty\) and \(x=0\), respectively.

Calculate the differential of x

To calculate the differential \(dx\) as a function of the new variable, differentiate \(y=\frac{1}{x}\) with respect to \(y\): $$\frac{d x}{d y}=\frac{d\left(\frac{1}{y}\right)}{d y}=-\frac{1}{y^2}$$ Now find \(dx\) in terms of \(dy\): $$d x = - \frac{1}{y^2} d y$$

Change the variable in the integral

Using the relationship \(x=\frac{1}{y}\), substitute this value into the given integral, also replacing \(dx\) with the expression found in the previous step: $$\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = -\int_{\infty}^{0} \frac{\ln(\frac{1}{y})}{\left(1+\frac{1}{y}\right) \sqrt{\frac{1}{y}}}\left(-\frac{1}{y^2} d y\right) $$

Simplify the integral and change limits of integration

Simplify this integral: $$\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = \int_{0}^{\infty} \frac{\ln y}{1+y^2} d y $$ Notice that the limits of integration have been reversed. It's important to always remember to reverse limits of integration after substituting the original variable with a new one. This integral contains the same expression as in the given integral, but with different limits of integration.

Add the given integral and the modified integral

Now that the integral expressions are the same, let's add the given integral and the modified integral together: $$\int_{0}^{1} \frac{\ln x}{(1+x) \sqrt{x}} d x + \int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = \int_{0}^{\infty} \frac{\ln y}{1+y^2} d y + \int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} $$

Apply the properties of convergent integrals

Since the given integral is convergent, the sum of these two integrals converges as well. This means that the sum of these two integrals is equal to a finite value. Also, notice that these two integrals are of the same function but with different variables (\(x\) and \(y\)), which means they must be equal: $$\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = \int_{0}^{\infty} \frac{\ln y}{1+y^2} d y$$ From the previous equation, we can write: $$\int_{0}^{1} \frac{\ln x}{(1+x) \sqrt{x}} d x + \int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = 2\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}$$ The given integral converges, which means it has some finite value, let's say \(a\): $$a + \int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = 2\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}$$ Now find \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}\): $$\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} = 2\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}} - a$$ Since \(a\) is a finite number, the only way for this equation to hold is if \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}=0\). We have therefore proven that \(\int_{0}^{\infty} \frac{\ln x d x}{(1+x) \sqrt{x}}=0\).

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics focusing on the accumulation of quantities and the areas under curves. It's the inverse operation of differentiation and is essential in calculating volumes, areas, and in solving problems in physics and engineering.

When we encounter an integral, our goal is often to find the antiderivative, which is a function F(x) such that dF(x)/dx is equal to the integrand. In the case of improper integrals, we are dealing with integrals that have infinite limits of integration or integrands that approach infinity within the interval of integration. The convergence or divergence of these improper integrals is a critical aspect of their evaluation, as it determines if there is a finite area under the curve. A convergent improper integral has a finite value, while a divergent one does not.

For example, evaluating \(\int_{0}^{1} \frac{\ln x}{(1+x) \sqrt{x}} dx\) requires understanding not just the antiderivative but also the behavior of the function as x approaches 0 and 1, because of the square root and logarithmic functions involved.

Variable Substitution

Variable substitution, or u-substitution, is a technique used in integral calculus to simplify the integration process. This method involves replacing a part of the integrand with a new variable to make the integral more manageable.

In our exercise, the substitution \( y = \frac{1}{x} \) is used. By finding the differential dx in terms of dy, we transform the integrand into a new form which is often easier to integrate. After substitution, it is crucial to also change the limits of integration to match the new variable. Substitution might introduce negative signs or other factors that we need to account for to proceed correctly with the integration. This technique is extremely powerful for complex integrals or when an integral contains a composite function, and is a staple tool within integral calculus.

Convergence of Integrals

Convergence of integrals is a concept that determines whether an integral has a finite value (converges) or not (diverges). Convergence is particularly important when dealing with improper integrals, where either the limits of integration are infinite or the integrand approaches infinity or a discontinuity within the limits of integration.

In order to determine convergence, we may need to compare the integral to other known convergent or divergent integrals, or apply specific convergence tests. Convergence is essential in many fields, as it can represent physical realities such as the total mass of an object with continuously distributed density or the total charge in a given electric field. In our textbook example, the convergence of \(\int_{0}^{1} \frac{\ln x}{(1+x) \sqrt{x}} dx\) is given, and this information is used creatively along with the properties of logarithms and the relationship between the variables x and y to prove that \(\int_{0}^{ty} \frac{\ln x dx}{(1+x) \sqrt{x}}=0\).