Question

1. if $\int_{-\infty}^{\infty} f(x) d x$ is convergent and $a$ and $b$ are real numbers, show that $\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x$

Explanation:

Step1: Recall the property of improper integrals

For a convergent improper integral \(\int_{-\infty}^{\infty}f(x)dx\), we know that \(\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{c}f(x)dx+\int_{c}^{\infty}f(x)dx\) for any real number \(c\).
Since \(\int_{-\infty}^{\infty}f(x)dx\) is convergent, we have \(\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx\) and \(\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{b}f(x)dx+\int_{b}^{\infty}f(x)dx\).

Step2: Equate the two expressions

Because \(\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx=\int_{-\infty}^{\infty}f(x)dx\) and \(\int_{-\infty}^{b}f(x)dx+\int_{b}^{\infty}f(x)dx=\int_{-\infty}^{\infty}f(x)dx\), we can set \(\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx=\int_{-\infty}^{b}f(x)dx+\int_{b}^{\infty}f(x)dx\)

Answer:

By the property of convergent improper integrals \(\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{c}f(x)dx+\int_{c}^{\infty}f(x)dx\) for any real \(c\), and since \(\int_{-\infty}^{\infty}f(x)dx\) is convergent, \(\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx=\int_{-\infty}^{b}f(x)dx+\int_{b}^{\infty}f(x)dx\) holds.