Evaluate ∫ 0 4 Ln ⁡ X 4 X − X 2 D X \int_0^4 \frac{\ln X}{\sqrt{4x-x^2}} \,\mathrm Dx ∫ 0 4 ​ 4 X − X 2 ​ L N X ​ D X

Evaluating definite integrals, especially those with singularities or involving transcendental functions, can be a fascinating journey through the world of calculus. In this article, we will delve into the evaluation of the integral:

$$\displaystyle\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx$$

This particular integral presents a unique challenge due to the presence of both a logarithmic function and a square root in the denominator, along with the integration limits that include a singularity at x = 0. We will explore a step-by-step approach to tackle this integral, employing techniques such as trigonometric substitution and leveraging the properties of definite integrals. The journey will not only lead us to the solution but also provide insights into the broader realm of integral calculus.

Understanding the Challenges

Before diving into the solution, it's crucial to understand the challenges this integral poses. The integrand, ${\frac{\ln x}{\sqrt{4x-x^2}}}$, has two potential points of concern: x = 0 and the points where the denominator becomes zero. The denominator, ${\sqrt{4x-x^2}}$, is zero when 4x - x^2 = 0, which simplifies to x(4-x) = 0. This gives us x = 0 and x = 4 as the points where the denominator is zero. Additionally, the natural logarithm, ln(x), is undefined for x ≤ 0, further complicating the behavior of the integrand near x = 0. These singularities necessitate a careful approach, often involving the use of improper integrals and limit techniques. The presence of both the logarithmic function and the square root term suggests that a clever substitution might be needed to simplify the integral. The interplay of these factors makes this integral an interesting problem to solve, requiring a blend of analytical skills and a solid understanding of calculus principles.

Trigonometric Substitution

The key to evaluating this integral lies in making a clever substitution. We observe that the expression inside the square root, 4x - x^2, resembles the form that can be completed into a perfect square. By completing the square, we can rewrite the expression as 4 - (x - 2)^2. This form suggests a trigonometric substitution, specifically one involving the sine function. Let's make the substitution:

x - 2 = 2 \sin \theta

This substitution implies that x = 2 + 2 \sin \theta and dx = 2 \cos \theta d\theta. Now, we need to change the limits of integration according to this substitution. When x = 0, we have 2 \sin \theta = -2, which means \sin \theta = -1, so \theta = -\frac{\pi}{2}. When x = 4, we have 2 \sin \theta = 2, which means \sin \theta = 1, so \theta = \frac{\pi}{2}. Next, we rewrite the square root term in terms of \theta:

$$\sqrt{4x - x^2} = \sqrt{4 - (x - 2)^2} = \sqrt{4 - 4\sin^2 \theta} = 2\sqrt{1 - \sin^2 \theta} = 2 \cos \theta$$

Now, we substitute these expressions into the original integral:

$$\displaystyle\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\ln(2 + 2\sin \theta)}{2 \cos \theta} 2 \cos \theta \,\mathrm d\theta$$

Simplifying, we get:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(2 + 2\sin \theta) \,\mathrm d\theta$$

This substitution has transformed the integral into a more manageable form, allowing us to proceed with further analysis. The logarithmic term, however, still presents a challenge, and we will address it in the subsequent steps.

Simplifying the Integral

After the trigonometric substitution, our integral now looks like this:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(2 + 2\sin \theta) \,\mathrm d\theta$$

We can simplify this further by factoring out the constant 2 from the argument of the logarithm:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln[2(1 + \sin \theta)] \,\mathrm d\theta$$

Using the property of logarithms, ${\ln(ab) = \ln a + \ln b}$, we can split the logarithm:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\ln 2 + \ln(1 + \sin \theta)] \,\mathrm d\theta$$

Now, we can separate the integral into two parts:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln 2 \,\mathrm d\theta + \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta$$

The first integral is straightforward to evaluate:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln 2 \,\mathrm d\theta = \ln 2 \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm d\theta = \ln 2 [\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \ln 2 \left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right) = \pi \ln 2$$

So, we have:

$$\pi \ln 2 + \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta$$

The remaining integral, ${\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta}$, requires a bit more manipulation. We will address this in the next section.

Utilizing Symmetry

To evaluate the remaining integral, let's denote it as I:

$$I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta$$

We can use the property of definite integrals that states ${\displaystyle\int_a^b f(x) \, dx = \displaystyle\int_a^b f(a + b - x) \, dx}$. Applying this property to our integral, we get:

$$I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln\left(1 + \sin\left(-\frac{\pi}{2} + \frac{\pi}{2} - \theta\right)\right) \,\mathrm d\theta = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 - \sin \theta) \,\mathrm d\theta$$

Now, let's add the two expressions for I:

$$2I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta + \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 - \sin \theta) \,\mathrm d\theta$$

Combining the integrals, we get:

$$2I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\ln(1 + \sin \theta) + \ln(1 - \sin \theta)] \,\mathrm d\theta$$

Using the property of logarithms, ${\ln a + \ln b = \ln(ab)}$, we can simplify the integrand:

$$2I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln[(1 + \sin \theta)(1 - \sin \theta)] \,\mathrm d\theta = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 - \sin^2 \theta) \,\mathrm d\theta$$

Since ${1 - \sin^2 \theta = \cos^2 \theta}$, we have:

$$2I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos^2 \theta) \,\mathrm d\theta = 2 \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos \theta) \,\mathrm d\theta$$

Thus,

$$I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos \theta) \,\mathrm d\theta$$

This integral is a well-known result, and its value is ${-\pi \ln 2}$. We will delve into its evaluation in the next section. For now, we have expressed I in terms of this known integral, which will help us find the final solution.

Evaluating the Known Integral

The integral ${\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos \theta) \,\mathrm d\theta}$ is a classic example that can be evaluated using various techniques. One common method involves using the Fourier series expansion of ${\ln(\cos \theta)}$. However, for the sake of brevity, we will state the result and provide a brief justification. The integral evaluates to:

$$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos \theta) \,\mathrm d\theta = -\pi \ln 2$$

To see why this is the case, one could consider the integral representation of the Dirichlet eta function or use complex analysis techniques. A more elementary approach involves using the identity ${\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)}$ and then employing integration by parts or other trigonometric manipulations. The detailed derivation is beyond the scope of this article, but the result is well-established in calculus literature.

Now, we recall that we had:

$$I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta$$

And we found that:

$$2I = 2 \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos \theta) \,\mathrm d\theta$$

So,

$$I = \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos \theta) \,\mathrm d\theta = -\pi \ln 2$$

Now we can substitute this back into our original expression:

$$\pi \ln 2 + \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin \theta) \,\mathrm d\theta = \pi \ln 2 + I = \pi \ln 2 - \pi \ln 2 = 0$$

Thus, we have successfully evaluated the integral.

Final Result

After navigating through the intricacies of trigonometric substitution, symmetry arguments, and leveraging known integral results, we arrive at the final answer:

$$\displaystyle\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx = 0$$

This result showcases the power of calculus techniques in tackling seemingly complex integrals. The combination of substitution, logarithmic properties, and symmetry considerations allowed us to simplify the integral and arrive at a concise solution. The journey through this problem has not only provided us with the answer but also highlighted the elegance and interconnectedness of various concepts in integral calculus. The integral, with its blend of logarithmic and algebraic functions, serves as a valuable example in the repertoire of problems that demonstrate the beauty and utility of mathematical analysis.