Question

Prove that $$ \begin{aligned} \frac{d}{d p} \Gamma(p) &=\int_{0}^{\infty} x^{p-1} e^{-x} \ln x d x \\ \frac{d^{n}}{d p^{n}} \Gamma(p) &=\int_{0}^{\infty} x^{p-1} e^{-x}(\ln x)^{n} d x \end{aligned} $$

Short answer

Differentiating under the integral sign using Leibniz's rule leads to \ \( \frac{d}{d p} \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} \ ln x \ dx \) and \ \( \frac{d^n}{d p^n} \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} (\ln x)^n \ dx \).

Step-by-step solution

Understand the Gamma Function

The Gamma function is defined as \ \ \[ \ \ \Gamma(p) = \int_{0}^{\textbackslash infty} x^{p-1} e^{-x} \textbackslash dx \]. It generalizes the factorial function to non-integer values.

Differentiate the Gamma Function with Respect to p

Use Leibniz's rule for differentiating under the integral sign. \ \ \[ \frac{d}{d p} \Gamma(p) = \int_{0}^{\infty} \frac{\partial}{\partial p} (x^{p-1} e^{-x}) \ dx \].

Compute the Partial Derivative inside the Integral

The partial derivative of \ \(x^{p-1} e^{-x}\) with respect to \ \(p\) is: \ \ \[ \frac{\partial}{\partial p} (x^{p-1} e^{-x}) = x^{p-1} e^{-x} \ln x \].

Combine Results for First Derivative

Substitute back into the integral to get: \ \ \[ \frac{d}{d p} \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} \ ln x \ dx \].

Extend to Higher Order Derivatives

To find the \ n-th derivative, differentiate \ \(x^{p-1} e^{-x} (ln x)^n\) under the integral sign. \ \ \[ \frac{d^n}{d p^n} \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} (\ln x)^n \ dx \].

Key concepts

Differentiation under the integral sign

In calculus, differentiation under the integral sign is a powerful technique. It allows us to differentiate an integral with respect to a parameter. In the given exercise, we have an integral:
\[ \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} \, dx\].
Here, the parameter is \(p\). To differentiate it with respect to \(p\), we use:

• Leibniz’s rule

\[\frac{d}{dp} \Gamma(p) = \int_{0}^{\infty} \frac{\partial}{\partial p} (x^{p-1} e^{-x}) \, dx\].
This means that we take the partial derivative inside the integral with respect to \(p\). After computing this partial derivative, we integrate the result as usual.

Leibniz's rule

Leibniz's rule is a formula for differentiation under the integral sign. It states:
\[ \frac{d}{d p} \int_{a(p)}^{b(p)} f(x,p) \, dx = \int_{a(p)}^{b(p)} \frac{\partial f}{\partial p}\, dx + f(b(p),p) \cdot \frac{d b}{d p} - f(a(p),p) \cdot \frac{d a}{d p}\].
However, in our case, the limits are constants (0 and \(\infty\)). Therefore, the simplified version is applied:
\[ \frac{d}{d p} \int_{a}^{b} f(x,p) \, dx = \int_{a}^{b} \frac{\partial f}{\partial p} \, dx\].
We identify our function inside the integral as \( f(x,p) = x^{p-1} e^{-x}\). To find the first derivative:
\[ \frac{\partial f}{\partial p} = x^{p-1} e^{-x} \ln x\].
Thus, we get:
\[ \frac{d}{dp} \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} \ln x \, dx\].

Higher-order derivatives

To find higher-order derivatives, we use the same principles but apply them multiple times. For example, the n-th derivative of the Gamma function:
\[ \frac{d^n}{d p^n} \Gamma(p) = \int_{0}^{\infty} x^{p-1} e^{-x} (\ln x)^n \, dx\].
This means, for the second derivative, we look at:
\[ \frac{d^2}{d p^2} \Gamma(p) = \int_{0}^{\infty} \frac{\partial}{\partial p} \left( x^{p-1} e^{-x} \ln x \right)\, dx \].
We repeat this process for each subsequent differentiation:

• Differentiate under the integral sign
• Compute the partial derivative involving \((\ln x)^n\)
• Integrate the result

This method effectively uses the power of repeated applications of Leibniz's rule.