Question

Let \(\int_{-\infty}^{\infty} f(x) d x\) be convergent and let \(a\) and \(b\) be real numbers where \(a \neq b\). Show that \(\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x\)

Short answer

The statement \(\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x\) is proven to be true, given the conditions that \(\int_{-\infty}^{\infty} f(x) d x\) is convergent and \(a \neq b\), utilizing the property of convergent integrals.

Step-by-step solution

Identify the given conditions

The given conditions of this problem are $\int_{-\infty}^{\infty} f(x) d x$ is convergent and $a \neq b$ are real numbers.

Refer to the Property of Convergent Integrals

Since the integral from \(-\infty\) to \(+\infty\) is convergent, it can be split at any point \(x = c\). Therefore, the integral \(\int_{-\infty}^{\infty} f(x) d x\) can be written as \(\int_{-\infty}^{c} f(x) d x + \int_{c}^{\infty} f(x) d x\). It is essential to note that this is true for any real value \(c\).

Apply Property to Given Integrals

As per step 2, it can be shown that: \(\int_{-\infty}^{a} f(x) d x + \int_{a}^{\infty} f(x) d x = \int_{-\infty}^{\infty} f(x) d x\) and \(\int_{-\infty}^{b} f(x) d x + \int_{b}^{\infty} f(x) d x = \int_{-\infty}^{\infty} f(x) d x\)

Show Equality

From step 3, as both sides of the equation equal \(\int_{-\infty}^{\infty} f(x) d x\), it can hence be shown that they equal one another, as such: \(\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x\). This completes the proof.