# extracting finite value from a non convergent integral, where am I wrong?

This following integral is not convergent

$$\int_0^\infty dx \, x^{ia} e^{i\omega x}$$

but I know for example that

$$\int_0^\infty dx \, x^c e^{-b x} = b^{-1-c}\, \Gamma(1+c)$$ where $$\Gamma$$ is the Euler Gamma function.

Therefore calling $$c=ia$$ and $$i\omega = -b$$ one gets

$$\int_0^\infty dx \, x^{ia} e^{i\omega x} = (-i \omega)^{1-ia} \Gamma(1+ia) = e^{-i \pi/2} \omega^{1-ia} \, \Gamma(1+ia)$$

which looks finite to me, so can you tell me where I've made a mistake or an abuse?

• When you write not convergent, do you mean not absolutely integrable or that the improper integral is not convergent? (i.e. the there is no limit to the integrals restricted on $[0,n]$ when $n\to \infty$) – LL 3.14 Jun 15 '20 at 19:14

The very short answer is "analytic continuation": we can begin with definitions that are only well-defined on some subset $$A \subseteq \mathbb{C}$$, find an analytical relation that holds within $$A$$, and then if one side of this relation is valid on a larger set $$B \supseteq A$$, use that to define an extension into $$B$$. I'll provide a Gamma function example, since it complements your question.

We can define, for $$\Re(z) > 0$$, $$\Gamma(z)$$ as $$\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} \; dt$$ Using this relation, we can show that $$\Gamma(z+1) = z \Gamma(z) \implies \Gamma(z) = \frac{\Gamma(z+1)}{z}$$ Now note that while our original definition of $$\Gamma$$ is only valid for $$\Re(z) > 0$$, the right-hand side of our new relation defines an analytical (strictly, meromorphic) function on $$\Re(z) > -1, z \neq 0$$.

This lets us analytically continue our original equation for $$\Gamma$$ into a larger subset of $$\mathbb{C}$$. Moreover, we can repeat this trick to extend it to all $$z \in \mathbb{C} \setminus \lbrace 0, -1, -2, \dots \rbrace$$. We started with a definition of an analytic function on $$\Re(z) > 0$$, and ended up defining an analytic function on almost all of $$\mathbb{C}$$.

This is more or less your question: by analytically continuing our expression, we can "sensibly" assign a value to our original non-convergent expression.

A similar example is how $$\zeta(-1) = -\frac{1}{12}$$. The starting point for our definition of $$\zeta(z)$$ does not converge for $$z = -1$$, but we can analytically continue it to a larger function which is finite at $$-1$$, where it takes the value $$-\frac{1}{12}$$. But of course, this does not mean that $$\sum_{n=1}^\infty n = -\frac{1}{12}$$

For a final example, consider how the series $$f(x) = 1 + x + x^2 + \dots$$ converges for $$\lvert x \rvert < 1$$. On this domain, we can show that $$f(x) = \frac{1}{1-x}$$ This is valid on a larger domain, and we can evaluate, say, $$f(2) = -1$$ but we can't back-substitute this into our original expression $$1 + 2 + 2^2 + \dots \neq -1$$ Instead, the (roughly) correct statement is that any analytical function that agrees with our original series on $$\lvert x \rvert < 1$$ and is valid on a larger domain (for avoidance of certain cases, let's say it's "suitably nice", say path connected) containing $$2$$, will have $$f(2) = -1$$.