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Show that if \(n \in \mathbb{N}, s>0\), then \(\int_{0}^{\infty} x^{n} e^{-s x} d x=n ! / s^{n+1}\).

Short Answer

Expert verified
Proved, \(\int_{0}^{\infty} x^{n} e^{-s x} dx = \frac{n!}{s^{n+1}}\).

Step by step solution

01

Performing Integration by Parts

We start by recognizing that the integral is in a form suitable for integration by parts. The formula for integration by parts is \(\int u dv = uv - \int v du\). Let \(u = x^{n}\) and \(dv = e^{-s x} dx\). We find that \( du = n x^{n-1} dx \) and \( v = -\frac{1}{s} e^{-s x}\). Substituting these into our formula and integrating gives: \(\int_{0}^{\infty} x^{n} e^{-s x} dx = - \frac{x^n}{s} e^{-s x} |_{0}^{\infty} + \frac{1}{s} \int_{0}^{\infty} n x^{n-1} e^{-s x} dx \)
02

Evaluating the Integral

Following Step 1, we simplify the first term and it yields 0 given that \(\lim_{x\to\infty} - \frac{x^n}{s} e^{-s x} = 0\) and when \(x = 0\) the expression also equals zero. Hence, we are only left with the integral term. Now, the remaining integral is like the original integral formula. Hence, we can apply the same method on this remaining integral, by which we recursively integrate by parts until \(n=0\). Then the result becomes the factorial of \(n\), represented as \(n!\), divided by \(s^{n+1}\). Final output: \(0 + \frac{n!}{s^{n+1}} = \frac{n!}{s^{n+1}}\)
03

Interpret the Result

This result shows that the integral of a function in the form \(x^{n} e^{-s x}\) from 0 to infinity equals to \(\frac{n!}{s^{n+1}}\). This result is significant in calculations involving integrals of functions, especially those related to the Gamma function.

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Most popular questions from this chapter

Ler \(f\) and \(|f|\) belong to \(R^{*}[a, \gamma]\) for every \(\gamma \geq a\). Show that \(f \in \mathcal{L}[a, \infty)\) if and only if the set \(V:=\left\\{\int_{a}^{x}|f|: x \geq a\right\\}\) is bounded in \(\mathbb{R}\).

Determine whether the following integrals are convergent or divergent. (Define the integrands to be 0 where they are not already defined.) (a) \(\int_{0}^{1} \frac{\sin x d x}{x^{3 / 2}}\), (b) \(\int_{0}^{1} \frac{\cos x d x}{x^{3 / 2}}\). (c) \(\int_{0}^{1} \frac{\ln x d x}{x \sqrt{1-x^{2}}}\), (d) \(\int_{0}^{1} \frac{\ln x d x}{1-x}\), (c) \(\int_{0}^{1}(\ln x)(\sin (1 / x)) d x\), (f) \(\int_{0}^{1} \frac{d x}{\sqrt{x}(1-x)}\).

Let \(g_{n}(x):=-1\) for \(x \in[-1,-1 / n)\). let \(g_{n}(x):=n x\) for \(x \in[-1 / n, 1 / n]\) and let \(g_{n}(x):=1\) for \(x \in(1 / n, 1]\). Show that \(\left\|g_{m}-g_{n}\right\| \rightarrow 0\) as \(m, n \rightarrow \infty\), so that the Completeness Theorem \(10.2 .12\) implies that there exists \(g \in \mathcal{L}[-1,1]\) such that \(\left(g_{n}\right)\) converges to \(g\) in \(\mathcal{L}[-1,1]\). Find such a function \(g\).

Consider the following sequences of functions with the indicated domains. Does the sequence converge? If so, to what? Is the convergence uniform? Is it bounded? If not bounded, is it dominated? Is it monotone? Evaluate the limit of the sequence of integrals. (a) \(\frac{k x}{1+k x}\) \([0,1]\) (b) \(\frac{x^{k}}{1+x^{k}} \quad[0,2]\), (c) \(\frac{1}{1+x^{k}}\) \([0,1]\), (d) \(\frac{1}{1+x^{k}}\) \([0,2]\)

Let \(\Phi(x):=x|\cos (\pi / x)|\) for \(x \in(0,1)\) and let \(\Phi(0):=0 .\) Then \(\Phi\) is continuous on \([0,1]\) and \(\Phi^{\prime}(x)\) exists for \(x \notin E:=\\{0\\} \cup\left\\{a_{k}: k \in \mathbb{N}\right\\}\), where \(a_{k}:=2 /(2 k+1) .\) Let \(\varphi(x):=\Phi^{\prime}(x)\) for \(x \notin E\) and \(\varphi(x):=0\) for \(x \in E\). Show that \(\varphi\) is not bounded on \([0,1]\). Using the Fundamental Theorem \(10.1 .9\) with \(E\) countable, conclude that \(\varphi \in \mathcal{R}^{*}[0,1]\) and that \(\int_{a}^{b} \varphi=\Phi(b)-\Phi(a)\) for \(a, b \in[0,1] .\) As in Exercise 19, show that \(|\varphi| \notin \mathcal{R}^{*}[0,1]\).

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