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Is the improper integral $\int_{\infty}^{\infty} f(x) \,dx $ necessary equals zero? The question is simple, but if the function diverges when it goes to infinity?

In my opinion, it is not a valid improper integral when the function diverges. But if we define this object as $\lim_{a \to +\infty} \int_{a}^{a} f(x) \,dx$ it is zero.

Guilherme Namen
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    How you define it is essential. You could equally define it as $\lim_{a\to\infty\b\to\infty}\int_b^a$ which could cause issues if $f$ diverged really badly – FShrike Jun 07 '22 at 14:29
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    Some hints in Improper integrals: "the improper integral is the limit of proper integrals as one endpoint of the interval of integration approaches either a specified real number, or ∞, or −∞. In more complicated cases, limits are required at both endpoints, or at interior points." – Mauro ALLEGRANZA Jun 07 '22 at 14:29
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    There is another way to define this as $\lim_{a \to +\infty} \lim_{b \to +\infty} \int_{a}^b f(x)dx$, and emerges another question if the interchange limits are different or diverges or one converges and another not? – Guilherme Namen Jun 07 '22 at 14:33
  • Exactly yeah. Unless the function was sufficiently nice in general this is not necessarily definable in a useful way – Riemann'sPointyNose Jun 07 '22 at 14:57
  • the notation $\int_\infty^\infty f(x),dx$ is simply meaningless – David C. Ullrich Jun 07 '22 at 15:13
  • In my previous question I used this in the 6º step, and it works. At that time I don´t care about it. For me, the integration between two same values is not a valid Reimann Integral because you can not create a partition in a point. But if you define it as an improper integral like my previous comment, you can create a partition that tends to a point. – Guilherme Namen Jun 07 '22 at 15:32

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Since the improper integral $\int_{a}^{\infty}f(x)\,dx$ is defined as $$ \int_{a}^{\infty}f(x)\,dx \overset{\text{def}}{=} \lim_{b \to \infty}\int_{a}^{b}f(x)\,dx, $$ it is natural to define $\int_{\infty}^{\infty}f(x)\,dx$ as $$ \int_{\infty}^{\infty}f(x)\,dx \overset{\text{def}}{=} \lim_{a \to \infty} \int_{a}^{\infty} f(x) \,dx. $$ With this definition, if $\int_{a}^{\infty}f(x)\,dx$ exists and is finite for all sufficiently large $a$, then for all $A > 0$, $$ \int_{a}^{\infty}f(x)\,dx - \int_{a+A}^{\infty}f(x)\,dx = \int_{a}^{a+A}f(x)\,dx \to 0 \text{ as } a \to \infty. $$ Hence, if $\int_{a}^{\infty}f(x)\,dx$ exists and is finite for all sufficiently large $a$, then $$ \int_{\infty}^{\infty}f(x)\,dx = 0. $$

If, on the other hand, $\int_{a}^{\infty}f(x)\,dx = \pm\infty$ for all sufficiently large $a$, then $$ \int_{\infty}^{\infty}f(x)\,dx = \lim_{a \to \infty}\int_{a}^{\infty}f(x)\,dx = \lim_{a \to \infty} \pm\infty = \pm\infty. $$

Finally, if $\int_{a}^{\infty}f(x)\,dx$ diverges for all $a$ without going to $\pm\infty$, then $\int_{\infty}^{\infty}f(x)\,dx$ is not defined.

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When interchanging the limits in the above definition, a sign change occurs. To fix this, we could instead define $\int_{\infty}^{\infty}f(x)\,dx$ as $$ \int_{\infty}^{\infty}f(x)\,dx \overset{\text{def}}{=} \lim_{a \to \infty} \lim_{b \to \infty} (-1)^{[b < a]}\int_{a}^{b}f(x)\,dx $$ where $$ [b < a] \overset{\text{def}}{=} \begin{cases} 1 & \text{if }b < a,\\ 0 & \text{if }b \ge a. \end{cases} $$ This has the effect of undoing the sign-change since $$ (-1)^{[b<a]}\int_{a}^{b}f(x)\,dx = \begin{cases} \int_{b}^{a}f(x)\,dx & \text{ if }b < a,\\ \int_{a}^{b}f(x)\,dx & \text{ if }b \ge a, \end{cases} $$ which means $$ \begin{aligned} \lim_{b \to \infty}\lim_{a \to \infty} (-1)^{[b<a]}\int_{a}^{b}f(x)\,dx &= \lim_{b \to \infty}\lim_{a \to \infty} -\int_{a}^{b}f(x)\,dx \\&= \lim_{b \to \infty}\lim_{a \to \infty} \int_{b}^{a}f(x)\,dx \\&= \int_{\infty}^{\infty}f(x)\,dx. \end{aligned} $$

  • This is sufficient to prove that the incomplete gamma function goes to zero when x tends to infinity. Thanks a lot. But if we define the improper integral as $\lim_{a \to +\infty}\lim_{\epsilon \to 0} \int_{a-\epsilon}^{a+\epsilon} f(x)dx$ the only condition is that for all $f(x)$ such that $x > a$ the function must be differentiable. – Guilherme Namen Jun 07 '22 at 17:28
  • In my opinion, you are right, but what is missing is that we can not interchange the limit operation in this case, as I did in my previous comment. – Guilherme Namen Jun 07 '22 at 17:41
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    @GuilhermeNamen I've edited my answer giving an alternate definition which enables one to interchange the limits by removing the sign-change that occured when doing so using the old definition. –  Jun 07 '22 at 19:23
  • An excellent answer, thanks a lot. I was afraid that it could not be possible to use the improper gamma function when x goes to infinity. – Guilherme Namen Jun 07 '22 at 19:34