Proving Definite Integrals ∫₀¹ (x-1)/log(x) Dx = Log 2 And ∫₀¹ Log(x)/(x-1) Dx = Π²/6

Jul 24, 2025 by JurnalWarga.com 86 views

Hey guys! Today, we're diving deep into the fascinating world of definite integrals. We're going to unravel the mysteries behind two intriguing integrals and learn how to prove them. These aren't just any integrals; they have elegant solutions that connect different areas of mathematics. So, buckle up and let's embark on this mathematical journey together!

Proving ∫₀¹ (x-1)/log(x) dx = log 2

Let's tackle the first integral: ∫₀¹ (x-1)/log(x) dx = log 2. This one might look a bit intimidating at first, but don't worry, we'll break it down step by step. The key here is to use a clever trick called Feynman's technique, also known as differentiation under the integral sign. This method involves introducing a parameter into the integral, differentiating with respect to that parameter, and then integrating back to get our final result. Trust me; it's as cool as it sounds!

Embracing Feynman's Technique: A Step-by-Step Approach

Introducing the Parameter: We start by defining a function I(a) as follows:

I(a) = ∫₀¹ xᵃ⁻¹ dx

Notice that when a = 1, our integral I(1) becomes the integral we want to solve. So, our goal is to find I(1). This is a crucial step, guys, because it sets the stage for the rest of the proof. By introducing this parameter a, we've essentially transformed our single integral into a family of integrals, each corresponding to a different value of a. This family of integrals is much easier to work with than the original integral.
Differentiating Under the Integral Sign: Now comes the magic! We differentiate I(a) with respect to a. This is where Feynman's technique shines. We're allowed to differentiate under the integral sign, which means we differentiate the integrand (the function inside the integral) with respect to a, treating x as a constant:

dI/da = d/da ∫₀¹ (xᵃ⁻¹ * (x-1) / log(x)) dx = ∫₀¹ ∂/∂a (xᵃ⁻¹ * (x-1) / log(x)) dx

When we perform the differentiation, we get:

dI/da = ∫₀¹ xᵃ⁻¹ dx

This simplifies to:

dI/da = [xᵃ / a] from 0 to 1

Evaluating the limits, we get:

dI/da = 1/a

Isn't that neat? By differentiating under the integral sign, we've transformed a complicated integral into a simple expression.
Integrating Back to Find I(a): We now have dI/da = 1/a. To find I(a), we need to integrate both sides with respect to a:

I(a) = ∫ (1/a) da = log(a) + C

Here, C is the constant of integration. We need to determine its value. This is another key step in the process. Constants of integration can sometimes be tricky, but we have a way to deal with them.
Determining the Constant of Integration: To find C, we need a known value of I(a) for some specific a. Let's consider the limit as a approaches infinity:

lim (a→∞) I(a) = lim (a→∞) ∫₀¹ (xᵃ⁻¹ * (x-1) / log(x)) dx = 0

This is because as a becomes very large, xᵃ⁻¹ approaches zero for 0 < x < 1, making the integral vanish. So, we have:

lim (a→∞) [log(a) + C] = 0

This implies that C = 0. Now we have the complete expression for I(a):

I(a) = log(a)
Finding I(1): Remember, we wanted to find I(1), which is the value of our original integral. Plugging a = 1 into our expression for I(a), we get:

I(1) = log(1) = 0

Wait a minute! This doesn't seem right. We made a mistake somewhere. Let's go back and check our steps. Ah, I see the error! We differentiated the wrong integral. We should have defined I(a) as:

I(a) = ∫₀¹ (xᵃ - 1) / log(x) dx

Let's redo the steps with this corrected definition:
1. Introducing the Parameter (Corrected):
  
  I(a) = ∫₀¹ (xᵃ - 1) / log(x) dx
2. Differentiating Under the Integral Sign (Corrected):
  
  dI/da = d/da ∫₀¹ (xᵃ - 1) / log(x) dx = ∫₀¹ ∂/∂a [(xᵃ - 1) / log(x)] dx
  
  dI/da = ∫₀¹ xᵃ dx
  
  dI/da = [xᵃ⁺¹ / (a+1)] from 0 to 1
  
  dI/da = 1 / (a+1)
3. Integrating Back to Find I(a) (Corrected):
  
  I(a) = ∫ [1 / (a+1)] da = log(a+1) + C
4. Determining the Constant of Integration (Corrected):
  
  When a = 0, I(0) = ∫₀¹ (x⁰ - 1) / log(x) dx = ∫₀¹ (1 - 1) / log(x) dx = 0. So,
  
  I(0) = log(0+1) + C = 0
  
  log(1) + C = 0
  
  0 + C = 0
  
  C = 0
5. Finding I(1) (Corrected):
  
  I(1) = log(1+1) = log(2)
There we have it! We've successfully proven that ∫₀¹ (x-1)/log(x) dx = log 2 using Feynman's technique. This method, while a bit involved, is a powerful tool in dealing with definite integrals.

Cracking the Code of ∫₀¹ log(x)/(x-1) dx = π²/6

Now, let's move on to the second integral: ∫₀¹ log(x)/(x-1) dx = π²/6. This integral has a beautiful result involving π², which hints at a connection to trigonometric functions or series. To tackle this one, we'll employ a different strategy: using series representation and some clever manipulation.

Unlocking the Solution with Series Representation

Expressing 1/(x-1) as a Series: Our first step is to rewrite the term 1/(x-1) as a geometric series. Notice that:

1/(x-1) = -1/(1-x)

For |x| < 1, we can express this as a geometric series:

-1/(1-x) = - (1 + x + x² + x³ + ... ) = - Σₙ₌₀[infinity] xⁿ

This series representation is valid for 0 ≤ x < 1, which is exactly the interval we're integrating over. This is a crucial step, guys, because it allows us to transform the integral into a form that we can work with more easily.
Substituting the Series into the Integral: Now, we substitute this series representation back into our integral:

∫₀¹ log(x)/(x-1) dx = ∫₀¹ log(x) * (- Σₙ₌₀[infinity] xⁿ) dx = - ∫₀¹ Σₙ₌₀[infinity] xⁿ log(x) dx
Interchanging Summation and Integration: Under suitable conditions (which hold in this case), we can interchange the summation and integration:

- ∫₀¹ Σₙ₌₀[infinity] xⁿ log(x) dx = - Σₙ₌₀[infinity] ∫₀¹ xⁿ log(x) dx

This is a powerful move! We've transformed the integral of an infinite series into an infinite series of integrals. Now, we just need to evaluate each of these simpler integrals.
Evaluating the Integral ∫₀¹ xⁿ log(x) dx: Let's focus on the integral ∫₀¹ xⁿ log(x) dx. We can solve this using integration by parts. Let u = log(x) and dv = xⁿ dx. Then, du = (1/x) dx and v = xⁿ⁺¹ / (n+1). Applying integration by parts:

∫₀¹ xⁿ log(x) dx = [log(x) * xⁿ⁺¹ / (n+1)] from 0 to 1 - ∫₀¹ [xⁿ⁺¹ / (n+1)] * (1/x) dx

The first term evaluates to 0 (using L'Hôpital's rule for the limit as x approaches 0). So, we're left with:

∫₀¹ xⁿ log(x) dx = - ∫₀¹ xⁿ / (n+1) dx = - [xⁿ⁺¹ / (n+1)²] from 0 to 1 = -1 / (n+1)²

This is a beautiful result! We've found a simple expression for the integral of each term in the series.
Substituting Back into the Series: Now, we substitute this result back into our series:

- Σₙ₌₀[infinity] ∫₀¹ xⁿ log(x) dx = - Σₙ₌₀[infinity] [-1 / (n+1)²] = Σₙ₌₀[infinity] 1 / (n+1)²

This is a well-known series! By a simple change of index (let k = n+1), we can rewrite it as:

Σₙ₌₀[infinity] 1 / (n+1)² = Σₖ₌₁[infinity] 1 / k²

This is the famous Basel problem, which Euler solved in the 18th century. The sum of this series is π²/6.
The Grand Finale: Therefore, we have:

∫₀¹ log(x)/(x-1) dx = Σₖ₌₁[infinity] 1 / k² = π²/6

And there you have it! We've successfully proven that ∫₀¹ log(x)/(x-1) dx = π²/6 using series representation and a bit of mathematical ingenuity.

The Beauty of Definite Integrals

So, guys, we've tackled two challenging definite integrals today and uncovered their elegant solutions. We've seen how powerful techniques like Feynman's method and series representation can be in solving these problems. These integrals not only showcase the beauty of mathematics but also highlight the interconnectedness of different mathematical concepts. Keep exploring, keep questioning, and keep the mathematical spirit alive!

Keywords: Definite Integrals, Feynman's Technique, Differentiation Under the Integral Sign, Series Representation, Basel Problem, Integration by Parts, Geometric Series, Logarithm, Pi, Mathematical Proof, Calculus