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Suppose I have an elementary function $F(x)$ for which $\int_{-\infty}^\infty F(x) \, \text{d}x $ has an elementary value. Here 'elementary value' means anything generated by $0,1,+,-,\div,\times,\exp,\sin$.

Suppose, indeed, I can compute this value by means of complex contours or multivariable calculus — the latter being involved, say, in Feynman's trick or in the Fourier transform.

Can it be proved that I could have given a proof that the integral has that exact value, by 'single-variable methods' alone? That is to say, without differentiating or integrating anything with two real variables $x,y$? Will it suffice to work in the language of single-variable calculus?

I don't profess to know any model theory, so do help me to formulate the question more precisely.

I'm astounded to learn that for $F(x)=\frac{\sin(x)}{x}$, there are such methods: Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

But a general result would be all the more fascinating.

egreg
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Chris Sanders
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    My guess: to your "Can it be proved" question the answer is NO. – GEdgar Dec 06 '18 at 12:00
  • It is not clear what you mean by "compute". I don't see how the standard meanings of "computation" would apply here (unless you give a list of permissible transformations or something). It is also not clear what you mean by "elementary value". I don't think it matters at all what the value is (you can make it zero by subtracting an elementary function from $F$), no matter how you interpret the rest of the question. – tomasz Dec 07 '18 at 14:48
  • sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'. – Chris Sanders Dec 07 '18 at 14:59
  • I was going to say that I didn't think the integral of the Gaussian over $\mathbb{R}$ could be calculated this way, but no, here are eleven proofs that the result is $\sqrt{\pi}$. At least three of them only rely on single-variable calculus. – Michael Seifert Dec 07 '18 at 15:20
  • @ChrisSanders: This is better. Still, I don't think the value matters at all. Suppose the value is $\alpha$ for some $\alpha$. Then replace $F$ with $G=F-\frac{1}{x^2}\cdot (-\alpha)/ \int_{-\infty}^\infty \frac{1}{x^2}$. Then $\int_{-\infty}^\infty G=0$ and $G$ is elementary, and as soon as you prove that the integral evaluates to $0$, you know that the integral of $F$ evaluates to $\alpha$. – tomasz Dec 07 '18 at 21:26

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