# Conditionally convergent integral

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.

• No, because Integrals from -inf to inf are defined as integral from -inf to c + integral from c to inf , where both integrals must converge. This isn't what Rearrangement is Mar 23 at 5:50

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.