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Chapter 4: Problem 113

Show that the function \(f(x)=\int_{0}^{1 / x} \frac{1}{t^{2}+1} d t+\int_{0}^{x} \frac{1}{t^{2}+1} d t\) is constant for \(x>0\).

Short Answer

Expert verified
The function \(f(x)=\int_{0}^{1 / x} \frac{1}{t^{2}+1} d t+\int_{0}^{x} \frac{1}{t^{2}+1} dt\) is constant for \(x>0\).

Step by step solution

01

Identification

Identify the given function as \(f(x)=\int_{0}^{1 / x} \frac{1}{t^{2}+1} d t+\int_{0}^{x} \frac{1}{t^{2}+1} d t\). We need to prove that this function is constant for \(x>0\).
02

Let \(u=1/t\) Transformation

First, for the term \(\int_{0}^{1 / x} \frac{1}{t^{2}+1} d t\), let's do a change of variables and let \(u=1/t\), hence \(t=1/u\) and \(dt=-du/u虏\). The limits of the integral should also be changed as \(t=0\) becomes \(u=\infty\) and \(t=1/x\) becomes \(u=x\). Making these changes, we get: \(\int_{\infty}^{x} \frac{1}{1+u虏} (-du/u虏)\).
03

Simplify Integration Expression

The term involves a minus sign and the limits can be switched to handle it as \(\int_{x}^{\infty} \frac{1}{1+u虏} du/u虏\). Now, our given function can be rewritten as \(f(x) = \int_{x}^{\infty} \frac{1}{1+u^{2}} d u+\int_{0}^{x} \frac{1}{1+t^{2}} d t\).
04

Use Properties of Integrals

To bring the integrals together and still work within the limits, we can apply properties of integrals to combine our two integrals. We get: \(f(x) = \int_{0}^{x}(\frac{1}{1+t^{2}} - \frac{1}{1+t^{2}}) dt + \int_{x}^{\infty}(\frac{1}{1+u^{2}}) du\). The first integral cancels out so the function simplifies to \(f(x) = \int_{x}^{\infty}\frac{1}{1+u^{2}} du\)
05

Calculate the Integral

The integral of \(\frac{1}{1+u^{2}}\) with respect to \(u\) from \(x\) to \(\infty\) is \(\arctan(u)\) evaluated from \(x\) to \(\infty\). This is \(\arctan(\infty) - \arctan(x)\). As arctan tends to \(\pi/2\) as its argument tends to infinity, this simplifies to \(\pi/2 - \arctan(x)\).
06

Evaluate the Function

We now need to differentiate this function to show that it is constant. The derivative of \(\pi/2\) is 0, and the derivative of \(-\arctan(x)\) is \(-1/(x虏+1)\). So, the derivative of \(f(x)\) is 0, implying that \(f(x)\) is a constant function for \(x>0\).

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