Mixing
Integral arising from QED vacuum polarization
$begingroup$
I would like to integrate
$$int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
for $a>0$.
One obvious substitution, $y=(1-x^2)^{-1/2}$, produces
$$int_1^infty y^{-4}(y^2-1)^{-1/2}e^{-a y}dy$$
but neither I nor Mathematica can do this integral either. Any ideas?
integration
asked Feb 3 at 6:35
G. SmithG. Smith
2664
$endgroup$
add a comment |
$begingroup$
I would like to integrate
$$int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
for $a>0$.
One obvious substitution, $y=(1-x^2)^{-1/2}$, produces
$$int_1^infty y^{-4}(y^2-1)^{-1/2}e^{-a y}dy$$
but neither I nor Mathematica can do this integral either. Any ideas?
integration
asked Feb 3 at 6:35
G. SmithG. Smith
2664
$endgroup$
add a comment |
$begingroup$
I would like to integrate
$$int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
for $a>0$.
One obvious substitution, $y=(1-x^2)^{-1/2}$, produces
$$int_1^infty y^{-4}(y^2-1)^{-1/2}e^{-a y}dy$$
but neither I nor Mathematica can do this integral either. Any ideas?
integration
asked Feb 3 at 6:35
G. SmithG. Smith
2664
$endgroup$
I would like to integrate
$$int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
for $a>0$.
One obvious substitution, $y=(1-x^2)^{-1/2}$, produces
$$int_1^infty y^{-4}(y^2-1)^{-1/2}e^{-a y}dy$$
but neither I nor Mathematica can do this integral either. Any ideas?
integration
integration
asked Feb 3 at 6:35
G. SmithG. Smith
2664
asked Feb 3 at 6:35
G. SmithG. Smith
2664
asked Feb 3 at 6:35
G. SmithG. Smith
2664
asked Feb 3 at 6:35
G. SmithG. Smith
2664
asked Feb 3 at 6:35
G. SmithG. Smith
2664
2664
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Excluding numerical integration, I should do it using series expansion built at $x=0$. This would give
$$left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}}=e^{-a}left(1+sum_{k=1}^n b_k , x^{2k}+O(x^{2k+2}) right)$$
$$left(
begin{array}{cc}
k & b_k \
1 & -frac{a}{2}-1 \
2 & frac{a^2}{8}+frac{a}{8} \
3 & -frac{a^3}{48}+frac{a^2}{16}+frac{a}{16} \
4 & frac{a^4}{384}-frac{5 a^3}{192}+frac{5 a^2}{128}+frac{5 a}{128} \
5 & -frac{a^5}{3840}+frac{a^4}{192}-frac{7 a^3}{256}+frac{7 a^2}{256}+frac{7
a}{256} \
6 & frac{a^6}{46080}-frac{11 a^5}{15360}+frac{23 a^4}{3072}-frac{7
a^3}{256}+frac{21 a^2}{1024}+frac{21 a}{1024} \
7 & -frac{a^7}{645120}+frac{7 a^6}{92160}-frac{a^5}{768}+frac{29
a^4}{3072}-frac{55 a^3}{2048}+frac{33 a^2}{2048}+frac{33 a}{2048} \
8 & frac{a^8}{10321920}-frac{17 a^7}{2580480}+frac{41 a^6}{245760}-frac{97
a^5}{49152}+frac{1093 a^4}{98304}-frac{429 a^3}{16384}+frac{429
a^2}{32768}+frac{429 a}{32768} \
9 & -frac{a^9}{185794560}+frac{a^8}{2064384}-frac{5 a^7}{294912}+frac{29
a^6}{98304}-frac{177 a^5}{65536}+frac{103 a^4}{8192}-frac{5005
a^3}{196608}+frac{715 a^2}{65536}+frac{715 a}{65536}
end{array}
right)$$ and integrate termwise to get
$$int_0^1left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}},dx=e^{-a}left(1+sum_{k=1}^n frac {b_k}{2k+1} right)$$
Limited to $n=9$, this would give the following results
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0 & 0.6666666667 & 0.6666666667 \
1 & 0.2123631665 & 0.2125607905 \
2 & 0.07059996854 & 0.07043172674 \
3 & 0.02383555255 & 0.02381813925 \
4 & 0.008134307224 & 0.008161937294 \
5 & 0.002812734609 & 0.002823407002 \
6 & 0.0009851151098 & 0.0009836552422 \
7 & 0.0003474991861 & 0.0003446169740 \
8 & 0.0001224830345 & 0.0001212798570 \
9 & 0.00004289847190 & 0.00004284093275 \
10 & 0.00001493941267 & 0.00001518064585
end{array}
right)$$
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
With a subsitution $x=sin{t}$, $dx=cos{t},dt$:
$$I(a)=int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
$$I(a)=int_0^{pi/2}(cos^3{t})e^{-a(sec{t})}dt$$
With the Feyman trick:
$$I'(a)=-int_0^{pi/2}(cos^2{t})e^{-a(sec{t})}dt$$
$$I''(a)=int_0^{pi/2}(cos{t})e^{-a(sec{t})}dt$$
$$I'''(a)=-int_0^{pi/2}e^{-a(sec{t})}dt$$
$$I^{(4)}(a)=int_0^{pi/2}sec{t}e^{-a(sec{t})}dt$$
$$I^{(5)}(a)=-int_0^{pi/2}e^{-a(sec{t})}sec^2{t}dt$$
$$I^{(6)}(a)=int_0^{pi/2}sec^3{t}e^{-a(sec{t})}dt$$
Obvioulsy $I''(0)=1$ and $I'''(0)=-pi/2$. Integrating by parts $I(a)$, with $u=cos^2(t)e^{-asec(t)},dt$ and $dv=cos(t)dt$, then $v=sin(t)$ and $du=Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt$ :
begin{align}
I(a) &= lim_{ tto pi/2^{-} } sin(t) cos^2(t)e^{-asec(t)} - int_{0}^{pi/2} sin(t) Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt \
I(a) &= 2int_{0}^{pi/2} sin^2(t)cos(t) e^{-asec(t)} dt + aint_{0}^{pi/2} sin^2(t) e^{-asec(t)} dt \
end{align}
Using the identity $sin^2(t)= 1- cos^2(t)$:
begin{align}
I(a) &= 2int_{0}^{pi/2} cos(t) e^{-asec(t)} dt - 2 int_{0}^{pi/2} cos^3(t) e^{-asec(t)} dt + a int_{0}^{pi/2} e^{-asec(t)} dt - a int_{0}^{pi/2} cos^2(t) e^{-asec(t)} dt \
I(a) &= 2 I''(a) - 2I(a) - aI''(a) + aI'(a) \
0 &= aI'''(a) - 2I''(a) - aI'(a) + 3I(a) label{1}tag{1}
end{align}
Remember that $I''(0)=1$, then if we evaluate $a=0$ in eqref{1}, we can demostrate that $I(0)=2/3$ . If you derivate eqref{1} :
$$
0 = aI^{(4)}(a) - I'''(a) -aI''(a) + 2I'(a) label{2}tag{2}
$$
If $lim_{ato 0^{+}}$ in eqref{2}, $I'(0)=-pi/4$ . Derivating eqref{2} :
$$
aI^{(5)}(a) - aI'''(a) + I''(a) = 0 label{3}tag{3}
$$
Derivating eqref{3} :
$$
aI^{(6)} + I^{(5)}(a) - a I^{(4)} = 0
$$
If $y(a)=I^{(4)}(a)$:
$$ a^2y''(a) + ay'(a) - a^2 y(a) = 0 $$
This is a modified Bessel equation and the solution is:
$$ y(a)= c_1 I_0(a) + c_2 K_0(a);quad text{with }c_1,c_2in Bbb{R} $$
The $I_0,K_0$ are the modified Bessel functions of the first and second kind with $alpha=0$, respectively. Then:
$$ I^{(5)}(a)= c_1 sum_{m=0}^{infty}frac{1}{(m!)^2}Big(frac{a}{2}Big)^{2m} + c_2 int_0^{infty}e^{-acosh{theta}}dtheta $$
If I integrate this with $a_0>0$ :
begin{align}
& I^{(4)}(a) - I^{(4)}(a_0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + c_2 int_{a_0}^{a}int_0^{infty}e^{-tcosh{theta}}dtheta dt \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty}int_{a_0}^{a} e^{-tcosh{theta}} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-acosh(theta)} }{cosh(theta)} dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I'''(a) - I'''(0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^a int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dtheta dt - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a \
end{align}
And $I'''(0)=-pi/2$, then:
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^{infty}int_0^a frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a - pi/2 \
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} -1 }{cosh^2(theta)} Bigg) dtheta
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
Then $displaystyle int_0^{infty}frac{1}{cosh^2(theta)}dtheta =bigg[tanh(theta)bigg]_0^{infty}=1$ :
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} }{cosh^2(theta)} Bigg) dtheta - c_2
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
In construction
answered Feb 3 at 8:54
El boritoEl borito
664216
$endgroup$
1
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098264%2fintegral-arising-from-qed-vacuum-polarization%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Excluding numerical integration, I should do it using series expansion built at $x=0$. This would give
$$left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}}=e^{-a}left(1+sum_{k=1}^n b_k , x^{2k}+O(x^{2k+2}) right)$$
$$left(
begin{array}{cc}
k & b_k \
1 & -frac{a}{2}-1 \
2 & frac{a^2}{8}+frac{a}{8} \
3 & -frac{a^3}{48}+frac{a^2}{16}+frac{a}{16} \
4 & frac{a^4}{384}-frac{5 a^3}{192}+frac{5 a^2}{128}+frac{5 a}{128} \
5 & -frac{a^5}{3840}+frac{a^4}{192}-frac{7 a^3}{256}+frac{7 a^2}{256}+frac{7
a}{256} \
6 & frac{a^6}{46080}-frac{11 a^5}{15360}+frac{23 a^4}{3072}-frac{7
a^3}{256}+frac{21 a^2}{1024}+frac{21 a}{1024} \
7 & -frac{a^7}{645120}+frac{7 a^6}{92160}-frac{a^5}{768}+frac{29
a^4}{3072}-frac{55 a^3}{2048}+frac{33 a^2}{2048}+frac{33 a}{2048} \
8 & frac{a^8}{10321920}-frac{17 a^7}{2580480}+frac{41 a^6}{245760}-frac{97
a^5}{49152}+frac{1093 a^4}{98304}-frac{429 a^3}{16384}+frac{429
a^2}{32768}+frac{429 a}{32768} \
9 & -frac{a^9}{185794560}+frac{a^8}{2064384}-frac{5 a^7}{294912}+frac{29
a^6}{98304}-frac{177 a^5}{65536}+frac{103 a^4}{8192}-frac{5005
a^3}{196608}+frac{715 a^2}{65536}+frac{715 a}{65536}
end{array}
right)$$ and integrate termwise to get
$$int_0^1left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}},dx=e^{-a}left(1+sum_{k=1}^n frac {b_k}{2k+1} right)$$
Limited to $n=9$, this would give the following results
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0 & 0.6666666667 & 0.6666666667 \
1 & 0.2123631665 & 0.2125607905 \
2 & 0.07059996854 & 0.07043172674 \
3 & 0.02383555255 & 0.02381813925 \
4 & 0.008134307224 & 0.008161937294 \
5 & 0.002812734609 & 0.002823407002 \
6 & 0.0009851151098 & 0.0009836552422 \
7 & 0.0003474991861 & 0.0003446169740 \
8 & 0.0001224830345 & 0.0001212798570 \
9 & 0.00004289847190 & 0.00004284093275 \
10 & 0.00001493941267 & 0.00001518064585
end{array}
right)$$
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Excluding numerical integration, I should do it using series expansion built at $x=0$. This would give
$$left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}}=e^{-a}left(1+sum_{k=1}^n b_k , x^{2k}+O(x^{2k+2}) right)$$
$$left(
begin{array}{cc}
k & b_k \
1 & -frac{a}{2}-1 \
2 & frac{a^2}{8}+frac{a}{8} \
3 & -frac{a^3}{48}+frac{a^2}{16}+frac{a}{16} \
4 & frac{a^4}{384}-frac{5 a^3}{192}+frac{5 a^2}{128}+frac{5 a}{128} \
5 & -frac{a^5}{3840}+frac{a^4}{192}-frac{7 a^3}{256}+frac{7 a^2}{256}+frac{7
a}{256} \
6 & frac{a^6}{46080}-frac{11 a^5}{15360}+frac{23 a^4}{3072}-frac{7
a^3}{256}+frac{21 a^2}{1024}+frac{21 a}{1024} \
7 & -frac{a^7}{645120}+frac{7 a^6}{92160}-frac{a^5}{768}+frac{29
a^4}{3072}-frac{55 a^3}{2048}+frac{33 a^2}{2048}+frac{33 a}{2048} \
8 & frac{a^8}{10321920}-frac{17 a^7}{2580480}+frac{41 a^6}{245760}-frac{97
a^5}{49152}+frac{1093 a^4}{98304}-frac{429 a^3}{16384}+frac{429
a^2}{32768}+frac{429 a}{32768} \
9 & -frac{a^9}{185794560}+frac{a^8}{2064384}-frac{5 a^7}{294912}+frac{29
a^6}{98304}-frac{177 a^5}{65536}+frac{103 a^4}{8192}-frac{5005
a^3}{196608}+frac{715 a^2}{65536}+frac{715 a}{65536}
end{array}
right)$$ and integrate termwise to get
$$int_0^1left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}},dx=e^{-a}left(1+sum_{k=1}^n frac {b_k}{2k+1} right)$$
Limited to $n=9$, this would give the following results
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0 & 0.6666666667 & 0.6666666667 \
1 & 0.2123631665 & 0.2125607905 \
2 & 0.07059996854 & 0.07043172674 \
3 & 0.02383555255 & 0.02381813925 \
4 & 0.008134307224 & 0.008161937294 \
5 & 0.002812734609 & 0.002823407002 \
6 & 0.0009851151098 & 0.0009836552422 \
7 & 0.0003474991861 & 0.0003446169740 \
8 & 0.0001224830345 & 0.0001212798570 \
9 & 0.00004289847190 & 0.00004284093275 \
10 & 0.00001493941267 & 0.00001518064585
end{array}
right)$$
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
add a comment |
$begingroup$
Excluding numerical integration, I should do it using series expansion built at $x=0$. This would give
$$left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}}=e^{-a}left(1+sum_{k=1}^n b_k , x^{2k}+O(x^{2k+2}) right)$$
$$left(
begin{array}{cc}
k & b_k \
1 & -frac{a}{2}-1 \
2 & frac{a^2}{8}+frac{a}{8} \
3 & -frac{a^3}{48}+frac{a^2}{16}+frac{a}{16} \
4 & frac{a^4}{384}-frac{5 a^3}{192}+frac{5 a^2}{128}+frac{5 a}{128} \
5 & -frac{a^5}{3840}+frac{a^4}{192}-frac{7 a^3}{256}+frac{7 a^2}{256}+frac{7
a}{256} \
6 & frac{a^6}{46080}-frac{11 a^5}{15360}+frac{23 a^4}{3072}-frac{7
a^3}{256}+frac{21 a^2}{1024}+frac{21 a}{1024} \
7 & -frac{a^7}{645120}+frac{7 a^6}{92160}-frac{a^5}{768}+frac{29
a^4}{3072}-frac{55 a^3}{2048}+frac{33 a^2}{2048}+frac{33 a}{2048} \
8 & frac{a^8}{10321920}-frac{17 a^7}{2580480}+frac{41 a^6}{245760}-frac{97
a^5}{49152}+frac{1093 a^4}{98304}-frac{429 a^3}{16384}+frac{429
a^2}{32768}+frac{429 a}{32768} \
9 & -frac{a^9}{185794560}+frac{a^8}{2064384}-frac{5 a^7}{294912}+frac{29
a^6}{98304}-frac{177 a^5}{65536}+frac{103 a^4}{8192}-frac{5005
a^3}{196608}+frac{715 a^2}{65536}+frac{715 a}{65536}
end{array}
right)$$ and integrate termwise to get
$$int_0^1left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}},dx=e^{-a}left(1+sum_{k=1}^n frac {b_k}{2k+1} right)$$
Limited to $n=9$, this would give the following results
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0 & 0.6666666667 & 0.6666666667 \
1 & 0.2123631665 & 0.2125607905 \
2 & 0.07059996854 & 0.07043172674 \
3 & 0.02383555255 & 0.02381813925 \
4 & 0.008134307224 & 0.008161937294 \
5 & 0.002812734609 & 0.002823407002 \
6 & 0.0009851151098 & 0.0009836552422 \
7 & 0.0003474991861 & 0.0003446169740 \
8 & 0.0001224830345 & 0.0001212798570 \
9 & 0.00004289847190 & 0.00004284093275 \
10 & 0.00001493941267 & 0.00001518064585
end{array}
right)$$
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
$endgroup$
Excluding numerical integration, I should do it using series expansion built at $x=0$. This would give
$$left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}}=e^{-a}left(1+sum_{k=1}^n b_k , x^{2k}+O(x^{2k+2}) right)$$
$$left(
begin{array}{cc}
k & b_k \
1 & -frac{a}{2}-1 \
2 & frac{a^2}{8}+frac{a}{8} \
3 & -frac{a^3}{48}+frac{a^2}{16}+frac{a}{16} \
4 & frac{a^4}{384}-frac{5 a^3}{192}+frac{5 a^2}{128}+frac{5 a}{128} \
5 & -frac{a^5}{3840}+frac{a^4}{192}-frac{7 a^3}{256}+frac{7 a^2}{256}+frac{7
a}{256} \
6 & frac{a^6}{46080}-frac{11 a^5}{15360}+frac{23 a^4}{3072}-frac{7
a^3}{256}+frac{21 a^2}{1024}+frac{21 a}{1024} \
7 & -frac{a^7}{645120}+frac{7 a^6}{92160}-frac{a^5}{768}+frac{29
a^4}{3072}-frac{55 a^3}{2048}+frac{33 a^2}{2048}+frac{33 a}{2048} \
8 & frac{a^8}{10321920}-frac{17 a^7}{2580480}+frac{41 a^6}{245760}-frac{97
a^5}{49152}+frac{1093 a^4}{98304}-frac{429 a^3}{16384}+frac{429
a^2}{32768}+frac{429 a}{32768} \
9 & -frac{a^9}{185794560}+frac{a^8}{2064384}-frac{5 a^7}{294912}+frac{29
a^6}{98304}-frac{177 a^5}{65536}+frac{103 a^4}{8192}-frac{5005
a^3}{196608}+frac{715 a^2}{65536}+frac{715 a}{65536}
end{array}
right)$$ and integrate termwise to get
$$int_0^1left(1-x^2right) e^{-frac{a}{sqrt{1-x^2}}},dx=e^{-a}left(1+sum_{k=1}^n frac {b_k}{2k+1} right)$$
Limited to $n=9$, this would give the following results
$$left(
begin{array}{ccc}
a & text{approximation} & text{exact} \
0 & 0.6666666667 & 0.6666666667 \
1 & 0.2123631665 & 0.2125607905 \
2 & 0.07059996854 & 0.07043172674 \
3 & 0.02383555255 & 0.02381813925 \
4 & 0.008134307224 & 0.008161937294 \
5 & 0.002812734609 & 0.002823407002 \
6 & 0.0009851151098 & 0.0009836552422 \
7 & 0.0003474991861 & 0.0003446169740 \
8 & 0.0001224830345 & 0.0001212798570 \
9 & 0.00004289847190 & 0.00004284093275 \
10 & 0.00001493941267 & 0.00001518064585
end{array}
right)$$
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
answered Feb 3 at 8:28
Claude LeiboviciClaude Leibovici
126k1158135
126k1158135
add a comment |
add a comment |
$begingroup$
With a subsitution $x=sin{t}$, $dx=cos{t},dt$:
$$I(a)=int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
$$I(a)=int_0^{pi/2}(cos^3{t})e^{-a(sec{t})}dt$$
With the Feyman trick:
$$I'(a)=-int_0^{pi/2}(cos^2{t})e^{-a(sec{t})}dt$$
$$I''(a)=int_0^{pi/2}(cos{t})e^{-a(sec{t})}dt$$
$$I'''(a)=-int_0^{pi/2}e^{-a(sec{t})}dt$$
$$I^{(4)}(a)=int_0^{pi/2}sec{t}e^{-a(sec{t})}dt$$
$$I^{(5)}(a)=-int_0^{pi/2}e^{-a(sec{t})}sec^2{t}dt$$
$$I^{(6)}(a)=int_0^{pi/2}sec^3{t}e^{-a(sec{t})}dt$$
Obvioulsy $I''(0)=1$ and $I'''(0)=-pi/2$. Integrating by parts $I(a)$, with $u=cos^2(t)e^{-asec(t)},dt$ and $dv=cos(t)dt$, then $v=sin(t)$ and $du=Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt$ :
begin{align}
I(a) &= lim_{ tto pi/2^{-} } sin(t) cos^2(t)e^{-asec(t)} - int_{0}^{pi/2} sin(t) Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt \
I(a) &= 2int_{0}^{pi/2} sin^2(t)cos(t) e^{-asec(t)} dt + aint_{0}^{pi/2} sin^2(t) e^{-asec(t)} dt \
end{align}
Using the identity $sin^2(t)= 1- cos^2(t)$:
begin{align}
I(a) &= 2int_{0}^{pi/2} cos(t) e^{-asec(t)} dt - 2 int_{0}^{pi/2} cos^3(t) e^{-asec(t)} dt + a int_{0}^{pi/2} e^{-asec(t)} dt - a int_{0}^{pi/2} cos^2(t) e^{-asec(t)} dt \
I(a) &= 2 I''(a) - 2I(a) - aI''(a) + aI'(a) \
0 &= aI'''(a) - 2I''(a) - aI'(a) + 3I(a) label{1}tag{1}
end{align}
Remember that $I''(0)=1$, then if we evaluate $a=0$ in eqref{1}, we can demostrate that $I(0)=2/3$ . If you derivate eqref{1} :
$$
0 = aI^{(4)}(a) - I'''(a) -aI''(a) + 2I'(a) label{2}tag{2}
$$
If $lim_{ato 0^{+}}$ in eqref{2}, $I'(0)=-pi/4$ . Derivating eqref{2} :
$$
aI^{(5)}(a) - aI'''(a) + I''(a) = 0 label{3}tag{3}
$$
Derivating eqref{3} :
$$
aI^{(6)} + I^{(5)}(a) - a I^{(4)} = 0
$$
If $y(a)=I^{(4)}(a)$:
$$ a^2y''(a) + ay'(a) - a^2 y(a) = 0 $$
This is a modified Bessel equation and the solution is:
$$ y(a)= c_1 I_0(a) + c_2 K_0(a);quad text{with }c_1,c_2in Bbb{R} $$
The $I_0,K_0$ are the modified Bessel functions of the first and second kind with $alpha=0$, respectively. Then:
$$ I^{(5)}(a)= c_1 sum_{m=0}^{infty}frac{1}{(m!)^2}Big(frac{a}{2}Big)^{2m} + c_2 int_0^{infty}e^{-acosh{theta}}dtheta $$
If I integrate this with $a_0>0$ :
begin{align}
& I^{(4)}(a) - I^{(4)}(a_0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + c_2 int_{a_0}^{a}int_0^{infty}e^{-tcosh{theta}}dtheta dt \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty}int_{a_0}^{a} e^{-tcosh{theta}} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-acosh(theta)} }{cosh(theta)} dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I'''(a) - I'''(0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^a int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dtheta dt - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a \
end{align}
And $I'''(0)=-pi/2$, then:
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^{infty}int_0^a frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a - pi/2 \
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} -1 }{cosh^2(theta)} Bigg) dtheta
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
Then $displaystyle int_0^{infty}frac{1}{cosh^2(theta)}dtheta =bigg[tanh(theta)bigg]_0^{infty}=1$ :
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} }{cosh^2(theta)} Bigg) dtheta - c_2
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
In construction
answered Feb 3 at 8:54
El boritoEl borito
664216
$endgroup$
1
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
add a comment |
$begingroup$
With a subsitution $x=sin{t}$, $dx=cos{t},dt$:
$$I(a)=int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
$$I(a)=int_0^{pi/2}(cos^3{t})e^{-a(sec{t})}dt$$
With the Feyman trick:
$$I'(a)=-int_0^{pi/2}(cos^2{t})e^{-a(sec{t})}dt$$
$$I''(a)=int_0^{pi/2}(cos{t})e^{-a(sec{t})}dt$$
$$I'''(a)=-int_0^{pi/2}e^{-a(sec{t})}dt$$
$$I^{(4)}(a)=int_0^{pi/2}sec{t}e^{-a(sec{t})}dt$$
$$I^{(5)}(a)=-int_0^{pi/2}e^{-a(sec{t})}sec^2{t}dt$$
$$I^{(6)}(a)=int_0^{pi/2}sec^3{t}e^{-a(sec{t})}dt$$
Obvioulsy $I''(0)=1$ and $I'''(0)=-pi/2$. Integrating by parts $I(a)$, with $u=cos^2(t)e^{-asec(t)},dt$ and $dv=cos(t)dt$, then $v=sin(t)$ and $du=Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt$ :
begin{align}
I(a) &= lim_{ tto pi/2^{-} } sin(t) cos^2(t)e^{-asec(t)} - int_{0}^{pi/2} sin(t) Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt \
I(a) &= 2int_{0}^{pi/2} sin^2(t)cos(t) e^{-asec(t)} dt + aint_{0}^{pi/2} sin^2(t) e^{-asec(t)} dt \
end{align}
Using the identity $sin^2(t)= 1- cos^2(t)$:
begin{align}
I(a) &= 2int_{0}^{pi/2} cos(t) e^{-asec(t)} dt - 2 int_{0}^{pi/2} cos^3(t) e^{-asec(t)} dt + a int_{0}^{pi/2} e^{-asec(t)} dt - a int_{0}^{pi/2} cos^2(t) e^{-asec(t)} dt \
I(a) &= 2 I''(a) - 2I(a) - aI''(a) + aI'(a) \
0 &= aI'''(a) - 2I''(a) - aI'(a) + 3I(a) label{1}tag{1}
end{align}
Remember that $I''(0)=1$, then if we evaluate $a=0$ in eqref{1}, we can demostrate that $I(0)=2/3$ . If you derivate eqref{1} :
$$
0 = aI^{(4)}(a) - I'''(a) -aI''(a) + 2I'(a) label{2}tag{2}
$$
If $lim_{ato 0^{+}}$ in eqref{2}, $I'(0)=-pi/4$ . Derivating eqref{2} :
$$
aI^{(5)}(a) - aI'''(a) + I''(a) = 0 label{3}tag{3}
$$
Derivating eqref{3} :
$$
aI^{(6)} + I^{(5)}(a) - a I^{(4)} = 0
$$
If $y(a)=I^{(4)}(a)$:
$$ a^2y''(a) + ay'(a) - a^2 y(a) = 0 $$
This is a modified Bessel equation and the solution is:
$$ y(a)= c_1 I_0(a) + c_2 K_0(a);quad text{with }c_1,c_2in Bbb{R} $$
The $I_0,K_0$ are the modified Bessel functions of the first and second kind with $alpha=0$, respectively. Then:
$$ I^{(5)}(a)= c_1 sum_{m=0}^{infty}frac{1}{(m!)^2}Big(frac{a}{2}Big)^{2m} + c_2 int_0^{infty}e^{-acosh{theta}}dtheta $$
If I integrate this with $a_0>0$ :
begin{align}
& I^{(4)}(a) - I^{(4)}(a_0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + c_2 int_{a_0}^{a}int_0^{infty}e^{-tcosh{theta}}dtheta dt \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty}int_{a_0}^{a} e^{-tcosh{theta}} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-acosh(theta)} }{cosh(theta)} dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I'''(a) - I'''(0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^a int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dtheta dt - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a \
end{align}
And $I'''(0)=-pi/2$, then:
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^{infty}int_0^a frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a - pi/2 \
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} -1 }{cosh^2(theta)} Bigg) dtheta
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
Then $displaystyle int_0^{infty}frac{1}{cosh^2(theta)}dtheta =bigg[tanh(theta)bigg]_0^{infty}=1$ :
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} }{cosh^2(theta)} Bigg) dtheta - c_2
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
In construction
answered Feb 3 at 8:54
El boritoEl borito
664216
$endgroup$
1
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
add a comment |
$begingroup$
With a subsitution $x=sin{t}$, $dx=cos{t},dt$:
$$I(a)=int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
$$I(a)=int_0^{pi/2}(cos^3{t})e^{-a(sec{t})}dt$$
With the Feyman trick:
$$I'(a)=-int_0^{pi/2}(cos^2{t})e^{-a(sec{t})}dt$$
$$I''(a)=int_0^{pi/2}(cos{t})e^{-a(sec{t})}dt$$
$$I'''(a)=-int_0^{pi/2}e^{-a(sec{t})}dt$$
$$I^{(4)}(a)=int_0^{pi/2}sec{t}e^{-a(sec{t})}dt$$
$$I^{(5)}(a)=-int_0^{pi/2}e^{-a(sec{t})}sec^2{t}dt$$
$$I^{(6)}(a)=int_0^{pi/2}sec^3{t}e^{-a(sec{t})}dt$$
Obvioulsy $I''(0)=1$ and $I'''(0)=-pi/2$. Integrating by parts $I(a)$, with $u=cos^2(t)e^{-asec(t)},dt$ and $dv=cos(t)dt$, then $v=sin(t)$ and $du=Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt$ :
begin{align}
I(a) &= lim_{ tto pi/2^{-} } sin(t) cos^2(t)e^{-asec(t)} - int_{0}^{pi/2} sin(t) Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt \
I(a) &= 2int_{0}^{pi/2} sin^2(t)cos(t) e^{-asec(t)} dt + aint_{0}^{pi/2} sin^2(t) e^{-asec(t)} dt \
end{align}
Using the identity $sin^2(t)= 1- cos^2(t)$:
begin{align}
I(a) &= 2int_{0}^{pi/2} cos(t) e^{-asec(t)} dt - 2 int_{0}^{pi/2} cos^3(t) e^{-asec(t)} dt + a int_{0}^{pi/2} e^{-asec(t)} dt - a int_{0}^{pi/2} cos^2(t) e^{-asec(t)} dt \
I(a) &= 2 I''(a) - 2I(a) - aI''(a) + aI'(a) \
0 &= aI'''(a) - 2I''(a) - aI'(a) + 3I(a) label{1}tag{1}
end{align}
Remember that $I''(0)=1$, then if we evaluate $a=0$ in eqref{1}, we can demostrate that $I(0)=2/3$ . If you derivate eqref{1} :
$$
0 = aI^{(4)}(a) - I'''(a) -aI''(a) + 2I'(a) label{2}tag{2}
$$
If $lim_{ato 0^{+}}$ in eqref{2}, $I'(0)=-pi/4$ . Derivating eqref{2} :
$$
aI^{(5)}(a) - aI'''(a) + I''(a) = 0 label{3}tag{3}
$$
Derivating eqref{3} :
$$
aI^{(6)} + I^{(5)}(a) - a I^{(4)} = 0
$$
If $y(a)=I^{(4)}(a)$:
$$ a^2y''(a) + ay'(a) - a^2 y(a) = 0 $$
This is a modified Bessel equation and the solution is:
$$ y(a)= c_1 I_0(a) + c_2 K_0(a);quad text{with }c_1,c_2in Bbb{R} $$
The $I_0,K_0$ are the modified Bessel functions of the first and second kind with $alpha=0$, respectively. Then:
$$ I^{(5)}(a)= c_1 sum_{m=0}^{infty}frac{1}{(m!)^2}Big(frac{a}{2}Big)^{2m} + c_2 int_0^{infty}e^{-acosh{theta}}dtheta $$
If I integrate this with $a_0>0$ :
begin{align}
& I^{(4)}(a) - I^{(4)}(a_0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + c_2 int_{a_0}^{a}int_0^{infty}e^{-tcosh{theta}}dtheta dt \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty}int_{a_0}^{a} e^{-tcosh{theta}} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-acosh(theta)} }{cosh(theta)} dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I'''(a) - I'''(0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^a int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dtheta dt - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a \
end{align}
And $I'''(0)=-pi/2$, then:
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^{infty}int_0^a frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a - pi/2 \
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} -1 }{cosh^2(theta)} Bigg) dtheta
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
Then $displaystyle int_0^{infty}frac{1}{cosh^2(theta)}dtheta =bigg[tanh(theta)bigg]_0^{infty}=1$ :
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} }{cosh^2(theta)} Bigg) dtheta - c_2
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
In construction
answered Feb 3 at 8:54
El boritoEl borito
664216
$endgroup$
With a subsitution $x=sin{t}$, $dx=cos{t},dt$:
$$I(a)=int_0^1(1-x^2)e^{-a(1-x^2)^{-1/2}}dx$$
$$I(a)=int_0^{pi/2}(cos^3{t})e^{-a(sec{t})}dt$$
With the Feyman trick:
$$I'(a)=-int_0^{pi/2}(cos^2{t})e^{-a(sec{t})}dt$$
$$I''(a)=int_0^{pi/2}(cos{t})e^{-a(sec{t})}dt$$
$$I'''(a)=-int_0^{pi/2}e^{-a(sec{t})}dt$$
$$I^{(4)}(a)=int_0^{pi/2}sec{t}e^{-a(sec{t})}dt$$
$$I^{(5)}(a)=-int_0^{pi/2}e^{-a(sec{t})}sec^2{t}dt$$
$$I^{(6)}(a)=int_0^{pi/2}sec^3{t}e^{-a(sec{t})}dt$$
Obvioulsy $I''(0)=1$ and $I'''(0)=-pi/2$. Integrating by parts $I(a)$, with $u=cos^2(t)e^{-asec(t)},dt$ and $dv=cos(t)dt$, then $v=sin(t)$ and $du=Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt$ :
begin{align}
I(a) &= lim_{ tto pi/2^{-} } sin(t) cos^2(t)e^{-asec(t)} - int_{0}^{pi/2} sin(t) Big[-2sin(t)cos(t)e^{-asec(t)} - asin(t)e^{-asec(t)} Big] dt \
I(a) &= 2int_{0}^{pi/2} sin^2(t)cos(t) e^{-asec(t)} dt + aint_{0}^{pi/2} sin^2(t) e^{-asec(t)} dt \
end{align}
Using the identity $sin^2(t)= 1- cos^2(t)$:
begin{align}
I(a) &= 2int_{0}^{pi/2} cos(t) e^{-asec(t)} dt - 2 int_{0}^{pi/2} cos^3(t) e^{-asec(t)} dt + a int_{0}^{pi/2} e^{-asec(t)} dt - a int_{0}^{pi/2} cos^2(t) e^{-asec(t)} dt \
I(a) &= 2 I''(a) - 2I(a) - aI''(a) + aI'(a) \
0 &= aI'''(a) - 2I''(a) - aI'(a) + 3I(a) label{1}tag{1}
end{align}
Remember that $I''(0)=1$, then if we evaluate $a=0$ in eqref{1}, we can demostrate that $I(0)=2/3$ . If you derivate eqref{1} :
$$
0 = aI^{(4)}(a) - I'''(a) -aI''(a) + 2I'(a) label{2}tag{2}
$$
If $lim_{ato 0^{+}}$ in eqref{2}, $I'(0)=-pi/4$ . Derivating eqref{2} :
$$
aI^{(5)}(a) - aI'''(a) + I''(a) = 0 label{3}tag{3}
$$
Derivating eqref{3} :
$$
aI^{(6)} + I^{(5)}(a) - a I^{(4)} = 0
$$
If $y(a)=I^{(4)}(a)$:
$$ a^2y''(a) + ay'(a) - a^2 y(a) = 0 $$
This is a modified Bessel equation and the solution is:
$$ y(a)= c_1 I_0(a) + c_2 K_0(a);quad text{with }c_1,c_2in Bbb{R} $$
The $I_0,K_0$ are the modified Bessel functions of the first and second kind with $alpha=0$, respectively. Then:
$$ I^{(5)}(a)= c_1 sum_{m=0}^{infty}frac{1}{(m!)^2}Big(frac{a}{2}Big)^{2m} + c_2 int_0^{infty}e^{-acosh{theta}}dtheta $$
If I integrate this with $a_0>0$ :
begin{align}
& I^{(4)}(a) - I^{(4)}(a_0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + c_2 int_{a_0}^{a}int_0^{infty}e^{-tcosh{theta}}dtheta dt \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty}int_{a_0}^{a} e^{-tcosh{theta}} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I^{(4)}(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+1}}{2m+1} + c_2 int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-acosh(theta)} }{cosh(theta)} dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1} + I^{(4)}(a_0) \
& I'''(a) - I'''(0) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^a int_0^{infty} frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dtheta dt - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a \
end{align}
And $I'''(0)=-pi/2$, then:
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)} + c_2 int_0^{infty}int_0^a frac{ e^{-a_0cosh(theta)} - e^{-tcosh(theta)} }{cosh(theta)} dt dtheta - c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a + I^{(4)}(a_0)a - pi/2 \
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} -1 }{cosh^2(theta)} Bigg) dtheta
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
Then $displaystyle int_0^{infty}frac{1}{cosh^2(theta)}dtheta =bigg[tanh(theta)bigg]_0^{infty}=1$ :
begin{align}
& I'''(a) = c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a^{2m+2}}{(2m+2)(2m+1)}
+ c_2 int_0^{infty}Bigg( frac{e^{-a_0cosh(theta)}a}{cosh(theta)} + frac{ e^{-acosh(theta)} }{cosh^2(theta)} Bigg) dtheta - c_2
- c_1 sum_{m=0}^{infty}frac{1}{(m!)^2 4^m }frac{a_0^{2m+1}}{2m+1}a
+ I^{(4)}(a_0)a - pi/2 \
end{align}
In construction
answered Feb 3 at 8:54
El boritoEl borito
664216
edited Feb 9 at 20:24
answered Feb 3 at 8:54
El boritoEl borito
664216
answered Feb 3 at 8:54
El boritoEl borito
664216
answered Feb 3 at 8:54
El boritoEl borito
664216
664216
1
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
add a comment |
1
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
1
1
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Impressive, for sure ! For the summation, the result seems to be $$frac{left(a^4+7 a^2+24right) I_1(a)-2 a left(a^2+6right) I_0(a)}{a^4}$$ For the remaining integral, I am more than stuck. Waiting for your update. Cheers.
$endgroup$
– Claude Leibovici
Feb 4 at 11:48
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
Thank you, I am going to update this in a day of these
$endgroup$
– El borito
Feb 5 at 2:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
$begingroup$
@ClaudeLeibovici I was derivating $I^{(4)}$, so the answer was very bad, I am sorry
$endgroup$
– El borito
Feb 8 at 3:28
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098264%2fintegral-arising-from-qed-vacuum-polarization%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
