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Chapter 10: Problem 64

Let \(c, d \in \mathbb{R}\) with \(0

Short Answer

Expert verified
The short answer based on the provided step-by-step solution is: To prove the given equality, we first integrate the 螕(u) function with respect to u from c to d and interchange the order of integration. Then, apply the integration by parts technique and find the integral and its derivative. Finally, split the integral on the intervals (0, 1) and (1, 鈭�) and compare the result with the given integrals. This confirms that the equality holds true.

Step by step solution

01

Analyze the 螕(u) function definition

Since the 螕(u) function is defined by the integral representation: \[ \Gamma(u) = \int_{0}^{\infty} t^{u-1} e^{-t} dt \] We will integrate the 螕(u) function with respect to u, from c to d. This will help us in comparing the result with the other two given integrals.
02

Integrate 螕(u) function with respect to u

To find the desired equality, let's integrate 螕(u) function with respect to u from c to d: \begin{align*} \int_{c}^{d}\Gamma(u)du &=\int_{c}^{d}\left(\int_{0}^{\infty} t^{u-1} e^{-t} dt\right) du \\ &= \int_{0}^{\infty}\left(\int_{c}^{d} t^{u-1} e^{-t} du\right) dt \quad \text{(Interchanging the order of integration)} \end{align*}
03

Apply the integration by parts technique

Applying integration by parts, we let: \[ I(u) = \int t^{u-1} e^{-t} du \] Now, we will find the integral and its derivative: \begin{align*} I(u) &= -t^{u-1}e^{-t} \\ I'(u) &= t^{u-2}e^{-t} \end{align*} Now, applying the integration by parts technique, we have: \begin{align*} \int_{c}^{d}\Gamma(u)du &= \int_{0}^{\infty} I'(u) du dt \\ &= \int_{0}^{\infty}\left[-t^{u-1}e^{-t}\bigg|_{u=c}^{u=d}\right] dt \\ &= \int_{0}^{\infty}\left[\frac{t^{d-1}-t^{c-1}}{e^t}\right] dt \end{align*}
04

Split the integral and compare

Finally, let's split the integral on the interval (0, 1) and (1, 鈭�) and compare the result with the given integrals: \begin{align*} \int_{c}^{d}\Gamma(u)du &= \int_{0^{+}}^{1^{-}}\frac{t^{d-1}-t^{c-1}}{e^{t}(\ln t)} dt + \int_{1^{+}}^{\infty}\frac{t^{d-1}-t^{c-1}}{e^{t}(\ln t)} dt \end{align*} This result exactly matches the given integrals; hence, the proof is complete.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by Parts is a technique that can simplify the process of integrating the product of two functions. It is a valuable tool, especially when dealing with products where simple integration methods don't apply effectively. This rule is derived from the product rule of differentiation and can be expressed as:
  • \( \int u \, dv = uv - \int v \, du \)
In practical terms, if you have two functions, say \(u(x)\) and \(v'(x)\), you let \(u = g(x)\) and \(dv = h'(x) \, dx\) to apply the technique.
In our solution, integration by parts is utilized to simplify the process of integrating \(\Gamma(u)\). By integrating \(t^{u-1} e^{-t}\), we expressed it in terms of derivative and integral, eventually leading to an easier form to evaluate the definite integrals.
Integral Representation
Integral Representation is the expression of a function as an integral. In this context, we use the integral to represent the Gamma Function \(\Gamma(u)\). The Gamma Function is crucial in mathematics due to its generalizations of factorial to real and complex numbers except for negative integers.The integral representation of the Gamma Function is given by:
  • \(\Gamma(u) = \int_{0}^{\infty} t^{u-1} e^{-t} \, dt \)
This expression allows the function to be analyzed and manipulated over the range of integration. Thus, in the exercise, integrating \(\Gamma(u)\) between 0 and \(\infty\) provides a comprehensive representation linking it with gamma distribution and other probability functions.
Order of Integration
The Order of Integration refers to the sequence in which multiple integrals are evaluated. Adjusting the order can simplify the integration process, especially for double integrals, by rearranging which integral is evaluated first.In our exercise, the ability to interchange the order of integration between \(u\) and \(t\) leads to a more straightforward computation:
  • First integrate \(u\) from \(c\) to \(d\)
  • Then integrate \(t\) from \(0\) to \(\infty\)
This was a clever move, simplifying the complex integral of the Gamma Function and making it more manageable to solve using integration techniques like integration by parts. Changing the order doesn't affect the integral's value due to the Fubini's theorem, as long as the integral is absolutely convergent in the entire range.
Splitting Integrals
Splitting Integrals involves breaking down an integral into smaller parts that are easier to handle. It simplifies complex problems, especially when different behaviors are expected across different parts of the integral's domain.In the solution, the integral of the Gamma Function was split into two separate parts:
  • From \(0^{+}\) to \(1^{-}\)
  • From \(1^{+}\) to \(\infty\)
This approach is beneficial to handle discontinuities or singularities within the interval, making integrals like the given representation more feasible to evaluate. The method ensures the contribution from each interval is considered distinctly, leading to an accurate computation overall.

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Most popular questions from this chapter

Show that the series \(\sum_{k \geq 1}(-1)^{k} /\left(x^{2}+k\right)\) converges uniformly for \(x \in \mathbb{R}\), and \(\left|\sum_{k=n}^{\infty}(-1)^{k} /\left(x^{2}+\bar{k}\right)\right| \leq 1 / n\). (Hint: Corollary 10.21)

If \(E:=(0, \infty) \times(0, \infty)\) and \(f:(0,1) \times E \rightarrow \mathbb{R}\) is defined by \(f(t, p, q):=t^{p-1}(1-t)^{q-1}\), then show that for each \(c \in(0, \infty)\), the improper integral \(\int_{0

Let \(f(t, u):=\left(t^{2}-u^{2}\right) /\left(t^{2}+u^{2}\right)^{2}\) for \((t, u) \in[1, \infty) \times[1, \infty)\). Show that $$ \int_{1}^{\infty}\left(\int_{1}^{\infty} f(t, u) d t\right) d u=\pi / 4=-\int_{1}^{\infty}\left(\int_{1}^{\infty} f(t, u) d u\right) d t . $$ (Hint: Use the substitution \(t=u \tan \theta\).)

Consider \(f:[0,1] \times[0,1] \rightarrow \mathbb{R}\) defined by \(f(t, u):=1 / t\) if \(0

Show that $$ \sum_{k=1}^{\infty} x^{k} / k=-\ln (1-x) \quad \text { and } \quad \sum_{k=1}^{\infty} k x^{k}=x /(1-x)^{2} $$ for \(x \in(-1,1)\). (Hint: Parts (i) and (ii) of Proposition \(10.29\) )
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