Question

Analysis question

Answer 1

Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a continuous function. Suppose there exists \(x_0\) such that \(\int_{x_0}^{\infty}f(x)dx\) and \(\int_{-\infty}^{x_0}f(x)dx\) both converge. Then, is it true that for any \(x_1\) both \(\int_{x_1}^{\infty}f(x)dx\) and \(\int_{-\infty}^{x_1}f(x)dx\) converge and that \[\int_{x_0}^{\infty}f(x)dx + \int_{-\infty}^{x_0}f(x)dx=\int_{x_1}^{\infty}f(x)dx+\int_{-\infty}^{x_1}f(x)dx\]

Answer 2

suppose we draw the function along the x axis as shown then the integral \[\int_{-\infty}^\infty f(x) dx= Total\area\under\ the\curve \]

Answer 3

now if i divide the total area under the curve into two parts and wrote;
total area = area 1 + area 2
it would have no effect on the actual area under the curve
soso if i break the area at some point \[x_1\] i would get\[\int_{-\infty} ^\infty f(x) =\int_{-\infty} ^{x_1} f(x) dx+\int_{x_1} ^\infty f(x) dx \]this just tells us that the total area has been broken from -infinity to x_1 and from x_1 to +infinity. Since the choice of x_1 was completely arbitrary we can call the point x_0 or x_1 or whatever we choose to. The logic being that complete area = sum of its two parts.