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Chapter 2: Problem 2

Prove that if \(J_{m}=\int_{1}^{e} \ln ^{m} x d x\), then \(J_{m}=e-m J_{m-1^{\prime}}\) ( \(\mathrm{m}\) is a positive integer).

Short Answer

Expert verified
Short Answer: To prove the relation \(J_m = e - mJ_{m-1}\), we used integration by parts by defining \(u = \ln^m x\) and \(dv = dx\), then computed \(du\) and \(v\). After applying integration by parts and simplifying the expression, we related the result to the given \(J_m\) terms and arrived at the desired relation, \(J_m = e - mJ_{m-1}\).

Step by step solution

01

Define u and dv for integration by parts

Let's begin by defining \(u\) and \(dv\) in terms of the given integral: \(u = \ln^m x\) and \(dv = dx\).
02

Compute du and v

Now we need to calculate \(du\) and \(v\): \(du = \frac{d (\ln^m x)}{dx} dx = m\cdot\ln^{m-1} x\cdot\frac{1}{x} dx\) \(v = \int dv = \int dx = x\)
03

Apply integration by parts

Now we apply integration by parts to the given integral: \(J_m=\int_{1}^{e} \ln^m x \ dx = uv \big|_{1}^{e} - \int_{1}^{e} v\ du\) \(J_m = \left[x\ln^m x \right]_1^e - \int_{1}^{e} x\cdot m\cdot\ln^{m-1} x\cdot\frac{1}{x} \ dx\)
04

Simplify the expression

Simplifying the expression, we get: \(J_m = e\ln^m e - 1\ln^m 1 - m\int_{1}^{e} \ln^{m-1}x \ dx\) Since \(\ln e = 1\) and \(\ln 1 = 0\), the expression becomes: \(J_m = e - m \int_{1}^{e} \ln^{m-1}x \ dx\)
05

Relate to J_m terms

Recall that \(J_{m}=\int_{1}^{e} \ln ^{m} x d x\). Notice that the last term in Step 4 is just the definition of \(J_{m-1}\): \(J_m = e - mJ_{m-1}\). This completes the proof.

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