3
$\begingroup$

So this might be a stupid question, but I was wondering what the following integral evaluates to:

$$\int_{\infty}^{\infty} \, f(x) \, dx$$

And no, the lower bound is not a typo.

I would imagine that the integral is $0$, for the following reason: \begin{align*} \int_{\infty}^{\infty} \, f(x) \, dx = \lim_{k \to \infty} \, \int_k^k \, f(x) \, dx = \lim_{k\to\infty} \, 0 = 0 \end{align*}

Edit (This is my real question):

If that's true, then could we evaluate the integral $$\lim_{k\to\infty} \, \int_k^{\infty} \, f(x) \, dx$$ as \begin{align*} \lim_{k\to\infty} \, \int_k^{\infty} \, f(x) \, dx &= \lim_{k\to\infty} \, \int_k^k \, f(x) \, dx \\[8pt] &= \lim_{k\to\infty} \, 0 \\[8pt] &= 0 \end{align*}

I'm pretty sure the first equation is wrong, and should be something like $$\lim_{k\to\infty} \, \int_k^{\infty} \, f(x) \, dx = \lim_{k\to\infty} \, \lim_{j \to \infty} \, \int_k^j \, f(x) \, dx$$

If so, then I was thinking that, because $k$ and $j$ are growing increasingly large, there are points $\alpha$ where the values of $k$ and $j$ will begin being the same, so

$$\lim_{k\to\infty} \, \lim_{j \to \infty} \, \int_k^j \, f(x) \, dx = \lim_{\alpha \to \infty} \, \int_{\alpha}^{\alpha} \, f(x) \, dx$$

which equals $0$ from the above arguments.

Edit: I suppose a more formal way of expressing my thoughts concerning $\alpha$ would be to let $\alpha = \operatorname{max}\{j,k\}$.

Just wanted some thoughts or criticisms, in particular.

Thanks!

$\endgroup$
  • $\begingroup$ Of course you can choose your definition at your convenience, but one may interpret this in its most possible general form: $$ \int_{\infty}^{\infty} f(x) \, dx := \lim_{j,k\to\infty} \int_{j}^{k}f(x) \, dx. $$ One may show that this is true if and only if $$ \int_{j}^{\infty} f(x)\,dx \text{ exists for all large $j$ and } \lim_{j\to\infty} \int_{j}^{\infty} f(x)\,dx = 0. $$ $\endgroup$ – Sangchul Lee Apr 30 '18 at 7:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.