Math Problem Statement

Beta functions integral from 0 to infinity x²/(1+x)^6=1/30 prove that

Solution

To prove that the integral

0x2(1+x)6dx=130\int_{0}^{\infty} \frac{x^2}{(1 + x)^6} \, dx = \frac{1}{30}

we'll show how it can be evaluated using the Beta function.

Step 1: Rewrite the Integral

We start by rewriting the integrand in a form that suggests a substitution related to the Beta function.

Consider the integral:

I=0x2(1+x)6dxI = \int_{0}^{\infty} \frac{x^2}{(1 + x)^6} \, dx

Now, let's express the integrand as:

I=0x2(1+x)6dx=0x2(1+x)61dxI = \int_{0}^{\infty} \frac{x^2}{(1 + x)^6} \, dx = \int_{0}^{\infty} \frac{x^2 (1 + x)^{-6}}{1} \, dx

Step 2: Substitution to Convert to Beta Function

Let us use the substitution:

x=t1twheret=x1+xx = \frac{t}{1 - t} \quad \text{where} \quad t = \frac{x}{1 + x}

With this substitution:

dx=dt(1t)2dx = \frac{dt}{(1 - t)^2}

So, the integral becomes:

I=01(t1t)2(1+t1t)6dt(1t)2I = \int_{0}^{1} \frac{\left(\frac{t}{1 - t}\right)^2}{(1 + \frac{t}{1 - t})^6} \cdot \frac{dt}{(1 - t)^2}

Simplify the expression:

I=01t2(1t)21(1t)6dt(1t)2=01t2(1t)622dtI = \int_{0}^{1} \frac{t^2}{(1 - t)^2} \cdot \frac{1}{(1 - t)^6} \cdot \frac{dt}{(1 - t)^2} = \int_{0}^{1} t^2 (1 - t)^{6 - 2 - 2} \, dt

I=01t2(1t)2dtI = \int_{0}^{1} t^2 (1 - t)^2 \, dt

Step 3: Recognize the Beta Function

The integral 01ta1(1t)b1dt\int_{0}^{1} t^{a-1} (1 - t)^{b-1} \, dt is the Beta function B(a,b)B(a, b). Here, we identify:

a=3andb=3a = 3 \quad \text{and} \quad b = 3

Thus, the integral II is:

I=B(3,3)I = B(3, 3)

Step 4: Use the Beta Function Property

We know that the Beta function is related to the Gamma function by the identity:

B(a,b)=Γ(a)Γ(b)Γ(a+b)B(a, b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a + b)}

For a=b=3a = b = 3, we have:

B(3,3)=Γ(3)Γ(3)Γ(6)B(3, 3) = \frac{\Gamma(3) \Gamma(3)}{\Gamma(6)}

Using the fact that Γ(n)=(n1)!\Gamma(n) = (n-1)! for a positive integer nn:

Γ(3)=2!=2,Γ(6)=5!=120\Gamma(3) = 2! = 2, \quad \Gamma(6) = 5! = 120

Therefore:

B(3,3)=2×2120=4120=130B(3, 3) = \frac{2 \times 2}{120} = \frac{4}{120} = \frac{1}{30}

Conclusion

We have shown that:

0x2(1+x)6dx=B(3,3)=130\int_{0}^{\infty} \frac{x^2}{(1 + x)^6} \, dx = B(3, 3) = \frac{1}{30}

This completes the proof.

Would you like to explore any of these concepts further? Here are 5 related questions to consider:

  1. What are the properties and applications of the Beta function in various fields?
  2. How is the substitution method useful in solving integrals related to Beta and Gamma functions?
  3. How do the Beta and Gamma functions relate to combinatorial identities?
  4. Can you derive the integral using a different substitution method?
  5. What are some other important integrals that involve the Beta function?

Tip: When evaluating integrals, recognizing forms that match Beta or Gamma functions can simplify your work significantly.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Beta Function
Gamma Function

Formulas

Beta function definition
Gamma function identity

Theorems

Beta-Gamma relationship

Suitable Grade Level

Advanced Mathematics