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Chapter 10: Problem 8

Show directly that \(\int_{0}^{\infty} e^{-t x} d x=1 / t\) and \(\int_{0}^{\infty} x e^{-t x} d x=1 / t^{2}\) for \(t>0\), thus confirming the results in Examples \(10.4 .6(\mathrm{~d}, \mathrm{e})\) when \(f(x):=1\).

Short Answer

Expert verified
The first integral evaluates to \(\frac{1}{t}\), and the second integral evaluates to \(\frac{1}{t^{2}}\), under the condition that \(t > 0\).

Step by step solution

01

Solve the first integral

Begin with the first integral \(\int_{0}^{\infty} e^{-t x} d x\). We know that \(\int e^{-t x} dx = -\frac{e^{-t x}}{t}\). So use the fundamental theorem of calculus and evaluate the integral from 0 to infinity. That is, \(-\frac{e^{-t (\infty)}}{t} - -\frac{e^{-t (0)}}{t} = 0 - -\frac{1}{t} = \frac{1}{t}\), thus showing the first integral equals to \(\frac{1}{t}\) as stated.
02

Solve the second integral

We now approach the second integral, \(\int_{0}^{\infty} x e^{-t x} d x\). Here, we have to use integration by parts, for which a handy formula is \(\int u dv = uv - \int v du\), where one part is easily integratable (dv) and the other part is simple to differentiate (u). So if we let \(u = x\) and \(dv = e^{-t x} dx\), we can compute \(du = dx\) and \(v = -\frac{e^{-t x}}{t}\). Then the integral becomes: \(-\frac{x e^{-t x}}{t}\Big|_{0}^{\infty} - \int_{0}^{\infty} -\frac{e^{-t x}}{t} dx = 0 - -\frac{1}{t}\int_{0}^{\infty} e^{-t x} dx = \frac{1}{t^{2}}\), as we know from the first part that \(\int_{0}^{\infty} e^{-t x} dx = \frac{1}{t}\). So the second integral is proven to be \(\frac{1}{t^{2}}\).

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Most popular questions from this chapter

If \(f_{n}(x):=x^{n}\) for \(n \in \mathbb{N}\), show that \(f_{n} \in \mathcal{L}[0,1]\) and that \(\left\|f_{n}\right\| \rightarrow 0\). Thus \(\left\|f_{n}-\theta\right\| \rightarrow 0\), where \(\theta\) denotes the function identically equal to \(0 .\)

Let \(k_{n}(x):=n\) for \(x \in\left(0.1 / n^{2}\right)\) and \(k_{n}(x):=0\) elsewhere in \([0,1]\). Does there exist \(k \in \mathcal{L}[0,1]\) such that \(\left\|k_{n}-k\right\| \rightarrow 0 ?\)

Show that the integral \(\int_{0}^{\infty} \sqrt{x} \cdot \sin \left(x^{2}\right) d x\) is convergent, even though the integrand is not bounded as \(x \rightarrow \infty\). [Hint: Make a substitution.]

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