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.

  • $\begingroup$ 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 $\endgroup$ Mar 23 at 5:50

1 Answer 1


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.


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