We are tasked with proving the following:

$\int_0^1 \left( \log\frac{1}{x} \right)^{n-1} dx = \Gamma(n), \quad \text{for } n > 0.$

Proof:

Step 1: Simplify the logarithmic term

First, rewrite the logarithm: $\log\frac{1}{x} = -\log x.$ Substituting this into the integral, we get: $\int_0^1 \left( \log\frac{1}{x} \right)^{n-1} dx = \int_0^1 \left( -\log x \right)^{n-1} dx.$

Since $(-\log x)^{n-1} = (-1)^{n-1} (\log x)^{n-1}$, we can factor out $(-1)^{n-1}$: $\int_0^1 \left( \log\frac{1}{x} \right)^{n-1} dx = (-1)^{n-1} \int_0^1 (\log x)^{n-1} dx.$

Step 2: Substitution

Let $\log x = -t$, so that $x = e^{-t}$ and $dx = -e^{-t} dt$.

The limits of integration change as follows:

• When $x = 1$, $t = 0$,
• When $x = 0$, $t \to \infty$.

Thus, the integral becomes: $\int_0^1 (\log x)^{n-1} dx = \int_0^\infty t^{n-1} e^{-t} dt.$

Step 3: Recognize the Gamma function

The definition of the Gamma function is: $\Gamma(n) = \int_0^\infty t^{n-1} e^{-t} dt, \quad n > 0.$

From the above substitution, we see that: $\int_0^1 (\log x)^{n-1} dx = \Gamma(n).$

Therefore: $\int_0^1 \left( \log\frac{1}{x} \right)^{n-1} dx = \Gamma(n), \quad \text{for } n > 0.$

Conclusion:

The integral is proven to equal the Gamma function: $\boxed{\int_0^1 \left( \log\frac{1}{x} \right)^{n-1} dx = \Gamma(n), \quad n > 0.}$

Do you want a more detailed breakdown of any step? Let me know!

Follow-up Questions:

1. What is the significance of the Gamma function in mathematics?
2. Can the Gamma function be extended to complex numbers?
3. What are the practical applications of such integrals in physics or engineering?
4. How does this result relate to factorials ($n!$) for integer $n$?
5. Can you explore the Beta function and its relation to the Gamma function?

Tip: The Gamma function generalizes the factorial to non-integer values, offering immense flexibility in mathematical analysis and computation.