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An elementary proof of $int_0^1fracarctan xsqrtx(1-x^2),dx = frac132sqrt2pi,Gammaleft(tfrac14right)^2$
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When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $sum_ngeq 0binom2nn^2frac116^n(4n+1)=frac116pi^2,Gammaleft(frac14right)^4 $ is:
$$int_0^1fracarctan xsqrtx(1-x^2),dx = tfrac132sqrt2pi,Gammaleft(tfrac14right)^2tagA$$
which might be regarded as a sort of Ahmed's integral under steroids.
I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $int_0^1K(x),x^mpm 1/4,dx$, which are associated with peculiar hypergeometric functions.
definite-integrals special-functions alternative-proof hypergeometric-function legendre-polynomials
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
 |Â
show 2 more comments
up vote
17
down vote
favorite
When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $sum_ngeq 0binom2nn^2frac116^n(4n+1)=frac116pi^2,Gammaleft(frac14right)^4 $ is:
$$int_0^1fracarctan xsqrtx(1-x^2),dx = tfrac132sqrt2pi,Gammaleft(tfrac14right)^2tagA$$
which might be regarded as a sort of Ahmed's integral under steroids.
I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $int_0^1K(x),x^mpm 1/4,dx$, which are associated with peculiar hypergeometric functions.
definite-integrals special-functions alternative-proof hypergeometric-function legendre-polynomials
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
1
I guess I found it: the trick is just to enforce the substitution $$ x mapsto frac1-t1+t.$$
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
1
The LHS turns out to be a multiple of a Beta function and we are done.
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
2
...you know, you should let some of us have a chance to answer your questions before you do... =P
– Simply Beautiful Art
Sep 27 '17 at 23:12
2
@SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :)
– Jack D'Aurizio♦
Sep 27 '17 at 23:13
3
:'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag)
– Simply Beautiful Art
Sep 27 '17 at 23:16
 |Â
show 2 more comments
up vote
17
down vote
favorite
up vote
17
down vote
favorite
When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $sum_ngeq 0binom2nn^2frac116^n(4n+1)=frac116pi^2,Gammaleft(frac14right)^4 $ is:
$$int_0^1fracarctan xsqrtx(1-x^2),dx = tfrac132sqrt2pi,Gammaleft(tfrac14right)^2tagA$$
which might be regarded as a sort of Ahmed's integral under steroids.
I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $int_0^1K(x),x^mpm 1/4,dx$, which are associated with peculiar hypergeometric functions.
definite-integrals special-functions alternative-proof hypergeometric-function legendre-polynomials
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
When playing with the complete elliptic integral of the first kind and its Fourier-Legendre expansion, I discovered that a consequence of $sum_ngeq 0binom2nn^2frac116^n(4n+1)=frac116pi^2,Gammaleft(frac14right)^4 $ is:
$$int_0^1fracarctan xsqrtx(1-x^2),dx = tfrac132sqrt2pi,Gammaleft(tfrac14right)^2tagA$$
which might be regarded as a sort of Ahmed's integral under steroids.
I already have a proof of this statement (through Fourier-Legendre expansions), but I would be happy to see a more direct and elementary proof of it, also because it might have some consequences about the moments of $K(x)$ of the form $int_0^1K(x),x^mpm 1/4,dx$, which are associated with peculiar hypergeometric functions.
definite-integrals special-functions alternative-proof hypergeometric-function legendre-polynomials
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
edited Sep 27 '17 at 23:50
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
asked Sep 27 '17 at 22:42
Jack D'Aurizio♦
270k31265629
270k31265629
1
I guess I found it: the trick is just to enforce the substitution $$ x mapsto frac1-t1+t.$$
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
1
The LHS turns out to be a multiple of a Beta function and we are done.
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
2
...you know, you should let some of us have a chance to answer your questions before you do... =P
– Simply Beautiful Art
Sep 27 '17 at 23:12
2
@SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :)
– Jack D'Aurizio♦
Sep 27 '17 at 23:13
3
:'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag)
– Simply Beautiful Art
Sep 27 '17 at 23:16
 |Â
show 2 more comments
1
I guess I found it: the trick is just to enforce the substitution $$ x mapsto frac1-t1+t.$$
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
1
The LHS turns out to be a multiple of a Beta function and we are done.
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
2
...you know, you should let some of us have a chance to answer your questions before you do... =P
– Simply Beautiful Art
Sep 27 '17 at 23:12
2
@SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :)
– Jack D'Aurizio♦
Sep 27 '17 at 23:13
3
:'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag)
– Simply Beautiful Art
Sep 27 '17 at 23:16
1
1
I guess I found it: the trick is just to enforce the substitution $$ x mapsto frac1-t1+t.$$
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
I guess I found it: the trick is just to enforce the substitution $$ x mapsto frac1-t1+t.$$
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
1
1
The LHS turns out to be a multiple of a Beta function and we are done.
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
The LHS turns out to be a multiple of a Beta function and we are done.
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
2
2
...you know, you should let some of us have a chance to answer your questions before you do... =P
– Simply Beautiful Art
Sep 27 '17 at 23:12
...you know, you should let some of us have a chance to answer your questions before you do... =P
– Simply Beautiful Art
Sep 27 '17 at 23:12
2
2
@SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :)
– Jack D'Aurizio♦
Sep 27 '17 at 23:13
@SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :)
– Jack D'Aurizio♦
Sep 27 '17 at 23:13
3
3
:'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag)
– Simply Beautiful Art
Sep 27 '17 at 23:16
:'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag)
– Simply Beautiful Art
Sep 27 '17 at 23:16
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
14
down vote
accepted
A possible way is to enforce the substitution $xmapstofrac1-t1+t$, giving:
$$ mathfrakI=int_0^1fracarctan(x)sqrtx(1-x^2),dx = int_0^1fractfracpi4-arctan tsqrtt(1-t^2),dt $$
and
$$ 2mathfrakI = fracpi4int_0^1 x^-1/2(1-x^2)^-1/2,dx =tfracpi8,Bleft(tfrac14,tfrac12right).$$
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
A possible way is to enforce the substitution $xmapstofrac1-t1+t$, giving:
$$ mathfrakI=int_0^1fracarctan(x)sqrtx(1-x^2),dx = int_0^1fractfracpi4-arctan tsqrtt(1-t^2),dt $$
and
$$ 2mathfrakI = fracpi4int_0^1 x^-1/2(1-x^2)^-1/2,dx =tfracpi8,Bleft(tfrac14,tfrac12right).$$
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
add a comment |Â
up vote
14
down vote
accepted
A possible way is to enforce the substitution $xmapstofrac1-t1+t$, giving:
$$ mathfrakI=int_0^1fracarctan(x)sqrtx(1-x^2),dx = int_0^1fractfracpi4-arctan tsqrtt(1-t^2),dt $$
and
$$ 2mathfrakI = fracpi4int_0^1 x^-1/2(1-x^2)^-1/2,dx =tfracpi8,Bleft(tfrac14,tfrac12right).$$
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
add a comment |Â
up vote
14
down vote
accepted
up vote
14
down vote
accepted
A possible way is to enforce the substitution $xmapstofrac1-t1+t$, giving:
$$ mathfrakI=int_0^1fracarctan(x)sqrtx(1-x^2),dx = int_0^1fractfracpi4-arctan tsqrtt(1-t^2),dt $$
and
$$ 2mathfrakI = fracpi4int_0^1 x^-1/2(1-x^2)^-1/2,dx =tfracpi8,Bleft(tfrac14,tfrac12right).$$
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
A possible way is to enforce the substitution $xmapstofrac1-t1+t$, giving:
$$ mathfrakI=int_0^1fracarctan(x)sqrtx(1-x^2),dx = int_0^1fractfracpi4-arctan tsqrtt(1-t^2),dt $$
and
$$ 2mathfrakI = fracpi4int_0^1 x^-1/2(1-x^2)^-1/2,dx =tfracpi8,Bleft(tfrac14,tfrac12right).$$
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
answered Sep 27 '17 at 23:19
Jack D'Aurizio♦
270k31265629
270k31265629
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
add a comment |Â
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
If I may, can I ask what was your line of thinking that made you realize, "You know what, substituting $x=(1-t)/(1+t)$ is the perfect way to evaluate this problem!" I fail to see how someone even gets there in the first place.
– Frank W.
May 16 at 23:49
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
@FrankW.: the geometry of the arctangent function made me realize it. $arctanleft(frac1-t1+tright)$ is a nice object; indeed the substitution $x=frac1-t1+t$ removes the arctangent from the integrand function. Given the relation between the arctangent and the logarithm, this is more or less the same thing as $$int_0^+inftyfraclog(x)p(x),dx=0$$ for any quadratic and palindromic polynomial $p(x)$, non-vanishing over $mathbbR^+$.
– Jack D'Aurizio♦
May 17 at 0:02
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
Okay... but how did you know that the denominator would stay the same? I can see how you would arrive at the substitution for the arctan function, but it seems kind of coincidental that the denominator was unchanged.
– Frank W.
Jun 3 at 23:40
add a comment |Â
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1
I guess I found it: the trick is just to enforce the substitution $$ x mapsto frac1-t1+t.$$
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
1
The LHS turns out to be a multiple of a Beta function and we are done.
– Jack D'Aurizio♦
Sep 27 '17 at 23:08
2
...you know, you should let some of us have a chance to answer your questions before you do... =P
– Simply Beautiful Art
Sep 27 '17 at 23:12
2
@SimplyBeautifulArt: sorry, I didn't do it on purpose, I just realized it a few minutes after writing the question. I guess that happens, quite often :)
– Jack D'Aurizio♦
Sep 27 '17 at 23:13
3
:'( welp... guess we shall await for your self-answer and hopefully some nice alternative proofs (which may be a suitable tag)
– Simply Beautiful Art
Sep 27 '17 at 23:16