integrate $F(x)$: NO complex analysis, NO multivariable calculus
$begingroup$
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.
integration improper-integrals elementary-functions
edited Dec 7 '18 at 15:39
egreg
180k1485202
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
$endgroup$
add a comment |
$begingroup$
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.
integration improper-integrals elementary-functions
edited Dec 7 '18 at 15:39
egreg
180k1485202
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
$endgroup$
1
$begingroup$
My guess: to your "Can it be proved" question the answer is NO.
$endgroup$
– GEdgar
Dec 6 '18 at 12:00
$begingroup$
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.
$endgroup$
– tomasz
Dec 7 '18 at 14:48
$begingroup$
sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'.
$endgroup$
– Chris Sanders
Dec 7 '18 at 14:59
$begingroup$
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.
$endgroup$
– Michael Seifert
Dec 7 '18 at 15:20
$begingroup$
@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 $alpha$.
$endgroup$
– tomasz
Dec 7 '18 at 21:26
add a comment |
$begingroup$
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.
integration improper-integrals elementary-functions
edited Dec 7 '18 at 15:39
egreg
180k1485202
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
$endgroup$
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.
integration improper-integrals elementary-functions
integration improper-integrals elementary-functions
edited Dec 7 '18 at 15:39
egreg
180k1485202
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
edited Dec 7 '18 at 15:39
egreg
180k1485202
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
edited Dec 7 '18 at 15:39
egreg
180k1485202
edited Dec 7 '18 at 15:39
egreg
180k1485202
edited Dec 7 '18 at 15:39
egreg
180k1485202
180k1485202
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
asked Dec 6 '18 at 3:29
Chris SandersChris Sanders
1,203217
1,203217
1
$begingroup$
My guess: to your "Can it be proved" question the answer is NO.
$endgroup$
– GEdgar
Dec 6 '18 at 12:00
$begingroup$
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.
$endgroup$
– tomasz
Dec 7 '18 at 14:48
$begingroup$
sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'.
$endgroup$
– Chris Sanders
Dec 7 '18 at 14:59
$begingroup$
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.
$endgroup$
– Michael Seifert
Dec 7 '18 at 15:20
$begingroup$
@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 $alpha$.
$endgroup$
– tomasz
Dec 7 '18 at 21:26
add a comment |
1
$begingroup$
My guess: to your "Can it be proved" question the answer is NO.
$endgroup$
– GEdgar
Dec 6 '18 at 12:00
$begingroup$
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.
$endgroup$
– tomasz
Dec 7 '18 at 14:48
$begingroup$
sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'.
$endgroup$
– Chris Sanders
Dec 7 '18 at 14:59
$begingroup$
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.
$endgroup$
– Michael Seifert
Dec 7 '18 at 15:20
$begingroup$
@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 $alpha$.
$endgroup$
– tomasz
Dec 7 '18 at 21:26
1
1
$begingroup$
My guess: to your "Can it be proved" question the answer is NO.
$endgroup$
– GEdgar
Dec 6 '18 at 12:00
$begingroup$
My guess: to your "Can it be proved" question the answer is NO.
$endgroup$
– GEdgar
Dec 6 '18 at 12:00
$begingroup$
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.
$endgroup$
– tomasz
Dec 7 '18 at 14:48
$begingroup$
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.
$endgroup$
– tomasz
Dec 7 '18 at 14:48
$begingroup$
sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'.
$endgroup$
– Chris Sanders
Dec 7 '18 at 14:59
$begingroup$
sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'.
$endgroup$
– Chris Sanders
Dec 7 '18 at 14:59
$begingroup$
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.
$endgroup$
– Michael Seifert
Dec 7 '18 at 15:20
$begingroup$
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.
$endgroup$
– Michael Seifert
Dec 7 '18 at 15:20
$begingroup$
@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 $alpha$.
$endgroup$
– tomasz
Dec 7 '18 at 21:26
$begingroup$
@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 $alpha$.
$endgroup$
– tomasz
Dec 7 '18 at 21:26
add a comment |
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$begingroup$
My guess: to your "Can it be proved" question the answer is NO.
$endgroup$
– GEdgar
Dec 6 '18 at 12:00
$begingroup$
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.
$endgroup$
– tomasz
Dec 7 '18 at 14:48
$begingroup$
sorry @tomasz I've now made some clarificatory edits. I didn't mean 'compute', but rather 'give a proof of that exact value'.
$endgroup$
– Chris Sanders
Dec 7 '18 at 14:59
$begingroup$
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.
$endgroup$
– Michael Seifert
Dec 7 '18 at 15:20
$begingroup$
@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 $alpha$.
$endgroup$
– tomasz
Dec 7 '18 at 21:26