/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 . (a) Show that the integral \(\... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 12

. (a) Show that the integral \(\int_{1}^{\infty} x^{-1} \ln x d x\) does not converge. (b) Show that if \(\alpha>1\), then \(\int_{1}^{\infty} x^{-\alpha} \ln x d x=1 /(\alpha-1)^{2}\).

Short Answer

Expert verified
The first integral does not converge and the value of the second integral for \(\alpha > 1\) is \(\frac{1}{(\alpha -1)^2}\)

Step by step solution

01

Show that the first integral does not converge

Use the comparison test for improper integrals. For this, compare the integral \( \int_{1}^{\infty} x^{-1} \ln x \, dx \) with \(\int_{1}^{\infty} x^{-1} \, dx\). The integral \(\int_{1}^{\infty} x^{-1} \, dx \) is a divergent integral as it evaluates to \(\ln \infty\), which is infinite. Since for \( x>1 \) we have \(\ln x > 0\), it follows that \(x^{-1}\ln x >x^{-1} \), for \( x>1 \). Hence, the given integral is larger than a divergent integral, so it also diverges.
02

Setup for showing the second integral

For \(\alpha > 1\), we want to evaluate the integral \(\int_{1}^{\infty} x^{-\alpha} \ln x \, dx\), We choose to use integration by parts to solve this integral. Recall, the formula for integration by parts is \(\int u v \, dx = u V - \int V du \), where \(u= \ln x ,dv = x^{-\alpha} dx\), we can then find \(v = -\frac{1}{(\alpha -1)x^{\alpha -1}}\) and \(du = x^{-1} dx\).
03

Show the value of the second integral

After applying integration by parts on the integral \(\int_{1}^{\infty} x^{-\alpha} \ln x \, dx\), one ends up with the equation \(\int_{1}^{\infty} x^{-\alpha} \ln x \, dx= -\frac{1}{\alpha -1} [x^{1-\alpha} \ln x]_{1}^{\infty} - \int_{1}^{\infty} \frac{1}{\alpha - 1}x^{-\alpha} dx\). Evaluating the first term yields 0 and the definite integral can be evaluated as \(\frac{1}{(\alpha -1)^2}\), therefore, removing the minus sign, one finds that the integral evaluates to \(\frac{1}{(\alpha -1)^2}\).

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Most popular questions from this chapter

If \(f_{k}(x):=k\) for \(x \in[1 / k, 2 / k]\) and \(f_{k}(x):=0\) elsewhere on \([0,2]\), show that \(f_{k}(x) \rightarrow 0\) but that \(\int_{0}^{2} f_{k}=1\)

Show that the following functions belong to \(\mathcal{R}^{*}[0,1]\) by finding a function \(F_{k}\) that is continuous on \([0,1]\) and such that \(F_{k}^{\prime}(x)=f_{k}(x)\) for \(x \in[0,1] \backslash E_{k}\), for some finite set \(E_{k}\). (a) \(f_{1}(x):=(x+1) / \sqrt{x} \quad\) for \(x \in(0.1]\) and \(f_{1}(0):=0\). (b) \(f_{2}(x):=x / \sqrt{1-x} \quad\) for \(x \in[0,1)\) and \(f_{2}(1):=0\). (c) \(f_{3}(x):=\sqrt{x} \ln x \quad\) for \(x \in(0,1]\) and \(f_{3}(0):=0\). (d) \(f_{4}(x):=(\ln x) / \sqrt{x} \quad\) for \(x \in(0,1]\) and \(f_{4}(0):=0\). (e) \(f_{5}(x):=\sqrt{(1+x) /(1-x)}\) for \(x \in[0,1)\) and \(f_{5}(1):=0\) (f) \(f_{6}(x):=1 /(\sqrt{x} \sqrt{2-x}) \quad\) for \(x \in(0,1]\) and \(f_{6}(0):=0\).

Let \(\delta\) be a gauge on \([a, b]\) and let \(\dot{\mathcal{P}}\) be a \(\delta\) -fine partition of \([a, b]\). (a) Show that there exists a \(\delta\) -fine partition \(\mathcal{Q}_{1}\) such that (i) no tag belongs to two subintervals in \(\dot{\mathcal{Q}}_{1}\), and (ii) \(S\left(f ; \dot{\mathcal{Q}}_{1}\right)=S(f ; \dot{\mathcal{P}})\) for any function \(f\) on \([a, b]\). (b) Does there exist a \(\delta\) -fine partition \(Q_{2}\) such that (j) every tag belongs to two subintervals in \(\dot{\mathcal{Q}}_{2}\), and (jj) \(S\left(f ; \mathcal{Q}_{2}\right)=S(f ; \dot{\mathcal{P}})\) for any function \(f\) on \([a, b]\) ? (c) Show that there exists a \(\delta\) -fine partition \(\mathcal{Q}_{3}\) such that ( \(\mathrm{k}\) ) every tag is an endpoint of its subinterval, and (kk) \(S\left(f ; \dot{\mathcal{Q}}_{3}\right)=S(f ; \dot{\mathcal{P}})\) for any function \(f\) on \([a, b]\).

Discuss the following sequences of functions and their integrals on \([0,1]\). Evaluate the limit of the integrals, when possible. (a) \(e^{-k x}\). (b) \(e^{-k x} / x\) (c) \(k x e^{-k x}\), (d) \(k^{2} x e^{-k x}\), (e) \(k x e^{-k^{2} x^{2}}\) (f) \(k x e^{-k x^{2}}\).

Let \(f \in \mathcal{R}^{*}[a, \gamma]\) for all \(\gamma \geq a\). Show that \(f \in R^{*}[a, \infty)\) if and only if for every \(\varepsilon>0\) there exists \(K(\varepsilon) \geq a\) such that if \(q>p \geq K(\varepsilon)\), then \(\left|\int_{\rho}^{q} f\right|<\varepsilon\).
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