4
$\begingroup$

It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.

My question is about the following integral: $\int\limits_{ - \infty }^\infty {{{\sin \left( x \right)} \over x}dx} $

It's known that this integral is conditionally convergent: it converges, but no absolutely.

Can one say that if ${a_n}\buildrel {n \to \infty } \over \longrightarrow \infty $ and ${b_n}\buildrel {n \to \infty } \over \longrightarrow - \infty $ that the value of $$\mathop {\lim }\limits_{n \to \infty } \int\limits_{{b_n}}^{{a_n}} {{{\sin \left( x \right)} \over x}dx} $$ depends on the choice of the sequences ${a_n},{b_n}$ ?

I have experimented a bit in wolfram alpha but no matter what I tried, the result was $\pi $.

I do remember hearing and in fact I've read also here:

Is there a rearrangement theorem for conditionally convergent improper integrals?

That the value of the conditionally convergent integral depends on the rearrangement.

$\endgroup$
3
$\begingroup$

The limit is $\pi$ irrespective of what those sequences are. It is a standard fact that $\lim _{t\to \infty}\int_0^{t} \frac {\sin \, x} x \, dx=\pi/2$ from which my claim follows.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.