Mixing
Compute $int_1^{infty} frac{1}{x^alpha + x^beta}dx$
$begingroup$
Given that $alpha,betainmathbb{R}$ such that the following integral converges I would like to find a closed form for: $$I_{alpha,beta} = int_1^{infty} frac{1}{x^alpha + x^beta}dx$$
I have found ways to represent the integral for cetain fixed cases, but is there a general representation of the integral for all $alpha$ and $beta$ ?
For example;
$$I_{1,beta} = int_1^{infty} frac{1}{x + x^{beta}}dx $$
$$= int_1^{infty} frac{1}{x(1+x^{beta-1})}$$
$$ = int_1^{infty} frac{1}{x} - frac{x^{beta - 2}}{1+x^{beta-1}}$$
$$=lnlvert xrvert-frac{1}{beta-1}cdotlnlvert 1+x^{beta-1}rvertbiggrrvert_1^{infty}$$
$$=frac{1}{beta - 1}cdotln(frac{x^{beta-1}}{1+x^{beta-1}})biggrrvert_1^{infty}$$
$$=frac{ln(1)-ln(1/2)}{beta-1}$$
$$=frac{ln2}{beta-1}$$
integration definite-integrals
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
$endgroup$
add a comment |
$begingroup$
Given that $alpha,betainmathbb{R}$ such that the following integral converges I would like to find a closed form for: $$I_{alpha,beta} = int_1^{infty} frac{1}{x^alpha + x^beta}dx$$
I have found ways to represent the integral for cetain fixed cases, but is there a general representation of the integral for all $alpha$ and $beta$ ?
For example;
$$I_{1,beta} = int_1^{infty} frac{1}{x + x^{beta}}dx $$
$$= int_1^{infty} frac{1}{x(1+x^{beta-1})}$$
$$ = int_1^{infty} frac{1}{x} - frac{x^{beta - 2}}{1+x^{beta-1}}$$
$$=lnlvert xrvert-frac{1}{beta-1}cdotlnlvert 1+x^{beta-1}rvertbiggrrvert_1^{infty}$$
$$=frac{1}{beta - 1}cdotln(frac{x^{beta-1}}{1+x^{beta-1}})biggrrvert_1^{infty}$$
$$=frac{ln(1)-ln(1/2)}{beta-1}$$
$$=frac{ln2}{beta-1}$$
integration definite-integrals
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
$endgroup$
$begingroup$
Are we assuming that $max(alpha,beta)-min(alpha,beta)$ is an integer?
$endgroup$
– Jack D'Aurizio
Jan 4 at 20:49
add a comment |
$begingroup$
Given that $alpha,betainmathbb{R}$ such that the following integral converges I would like to find a closed form for: $$I_{alpha,beta} = int_1^{infty} frac{1}{x^alpha + x^beta}dx$$
I have found ways to represent the integral for cetain fixed cases, but is there a general representation of the integral for all $alpha$ and $beta$ ?
For example;
$$I_{1,beta} = int_1^{infty} frac{1}{x + x^{beta}}dx $$
$$= int_1^{infty} frac{1}{x(1+x^{beta-1})}$$
$$ = int_1^{infty} frac{1}{x} - frac{x^{beta - 2}}{1+x^{beta-1}}$$
$$=lnlvert xrvert-frac{1}{beta-1}cdotlnlvert 1+x^{beta-1}rvertbiggrrvert_1^{infty}$$
$$=frac{1}{beta - 1}cdotln(frac{x^{beta-1}}{1+x^{beta-1}})biggrrvert_1^{infty}$$
$$=frac{ln(1)-ln(1/2)}{beta-1}$$
$$=frac{ln2}{beta-1}$$
integration definite-integrals
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
$endgroup$
Given that $alpha,betainmathbb{R}$ such that the following integral converges I would like to find a closed form for: $$I_{alpha,beta} = int_1^{infty} frac{1}{x^alpha + x^beta}dx$$
I have found ways to represent the integral for cetain fixed cases, but is there a general representation of the integral for all $alpha$ and $beta$ ?
For example;
$$I_{1,beta} = int_1^{infty} frac{1}{x + x^{beta}}dx $$
$$= int_1^{infty} frac{1}{x(1+x^{beta-1})}$$
$$ = int_1^{infty} frac{1}{x} - frac{x^{beta - 2}}{1+x^{beta-1}}$$
$$=lnlvert xrvert-frac{1}{beta-1}cdotlnlvert 1+x^{beta-1}rvertbiggrrvert_1^{infty}$$
$$=frac{1}{beta - 1}cdotln(frac{x^{beta-1}}{1+x^{beta-1}})biggrrvert_1^{infty}$$
$$=frac{ln(1)-ln(1/2)}{beta-1}$$
$$=frac{ln2}{beta-1}$$
integration definite-integrals
integration definite-integrals
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
asked Jan 4 at 20:37
Peter ForemanPeter Foreman
4177
4177
$begingroup$
Are we assuming that $max(alpha,beta)-min(alpha,beta)$ is an integer?
$endgroup$
– Jack D'Aurizio
Jan 4 at 20:49
add a comment |
$begingroup$
Are we assuming that $max(alpha,beta)-min(alpha,beta)$ is an integer?
$endgroup$
– Jack D'Aurizio
Jan 4 at 20:49
$begingroup$
Are we assuming that $max(alpha,beta)-min(alpha,beta)$ is an integer?
$endgroup$
– Jack D'Aurizio
Jan 4 at 20:49
$begingroup$
Are we assuming that $max(alpha,beta)-min(alpha,beta)$ is an integer?
$endgroup$
– Jack D'Aurizio
Jan 4 at 20:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let's say $alpha > max(1,beta)$.
$$ frac{1}{x^alpha + x^beta} = frac{1}{x^alpha (1 + x^{beta-alpha})} =
sum_{k=0}^infty frac{(-1)^k x^{kbeta}}{x^{(1+k)alpha}}$$
and integrating term-by-term
$$I_{alpha,beta} = sum_{k=0}^infty frac{(-1)^{k}}{(1+k)alpha - k beta -1}
= frac{1}{beta-1}+frac{Psileft(frac{beta - 1}{2alpha-2beta}right) - Psileft(frac{alpha-1}{2alpha-2beta}right)}{2alpha-2beta}$$
where $Psi$ is the digamma function.
edited Jan 4 at 23:32
Peter Foreman
4177
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
$endgroup$
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
1
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
1
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
add a comment |
$begingroup$
For $alpha neq beta $. Put $$t={1 over 1+ x^{beta - alpha}} .$$ Then you integral becomes
begin{align} {1 over beta -alpha } int_0^{1 over 2 } t^{alpha -1 over beta - alpha }(1-t)^{1-beta over beta -alpha} , dt,end{align}
which is ${1 over beta -alpha} B(1/2; {beta -1 over beta -alpha}, {1-alpha over beta -alpha}) $,
where $B(x;mu,nu)$ is the incomplete Beta function.
answered Jan 5 at 0:35
moutheticsmouthetics
50127
$endgroup$
add a comment |
$begingroup$
Assuming $beta > max (1,alpha)$. Enforcing a substitution of $x mapsto 1/x$ in the integral to begin with gives
$$I_{alpha, beta} = int_0^1 frac{x^{beta - 2}}{1 + x^{beta -alpha}} , dx.$$
Inside the interval of convergence, by exploiting the geometric series, namely
$$frac{1}{1 + x^{beta - alpha}} = sum_{n = 0}^infty (-1)^n x^{n beta - n alpha}, qquad |x| < 1$$
the integral can be rewritten as
begin{align}
I_{alpha, beta} &= sum_{n = 0}^infty (-1)^n int_0^1 x^{n beta - n alpha + beta - 2} , dx\
&= sum_{n = 0}^infty (-1)^n frac{1}{n beta - n alpha + beta - 1}. qquad (*)
end{align}
To handle the infinity sum that arises we will make use of the following result (see Eq. (6) in the link)
$$sum_{n = 0}^infty frac{(-1)^n}{z n + 1} = frac{1}{2z} left [psi left (frac{z + 1}{2z} right ) - psi left (frac{1}{2z} right ) right ]. qquad (**)$$
Here $psi (x)$ is the digamma function.
Rewriting the sum in ($*$) in the form of ($**$), namely
$$I_{alpha, beta} = frac{1}{beta - 1} sum_{n = 0}^infty frac{(-1)^n}{left (dfrac{beta - alpha}{beta - 1} right ) n + 1},$$
as we have $z = (beta - alpha)/(beta - 1)$, we finally arrive at
$$I_{alpha, beta} = frac{1}{2beta - 2 alpha} left [psi left (frac{2 beta - alpha - 1}{2 beta - 2alpha} right ) - psi left (frac{beta - 1}{2 beta - 2alpha} right ) right ], qquad beta > max (1,alpha).$$
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3062082%2fcompute-int-1-infty-frac1x-alpha-x-betadx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's say $alpha > max(1,beta)$.
$$ frac{1}{x^alpha + x^beta} = frac{1}{x^alpha (1 + x^{beta-alpha})} =
sum_{k=0}^infty frac{(-1)^k x^{kbeta}}{x^{(1+k)alpha}}$$
and integrating term-by-term
$$I_{alpha,beta} = sum_{k=0}^infty frac{(-1)^{k}}{(1+k)alpha - k beta -1}
= frac{1}{beta-1}+frac{Psileft(frac{beta - 1}{2alpha-2beta}right) - Psileft(frac{alpha-1}{2alpha-2beta}right)}{2alpha-2beta}$$
where $Psi$ is the digamma function.
edited Jan 4 at 23:32
Peter Foreman
4177
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
$endgroup$
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
1
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
1
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
add a comment |
$begingroup$
Let's say $alpha > max(1,beta)$.
$$ frac{1}{x^alpha + x^beta} = frac{1}{x^alpha (1 + x^{beta-alpha})} =
sum_{k=0}^infty frac{(-1)^k x^{kbeta}}{x^{(1+k)alpha}}$$
and integrating term-by-term
$$I_{alpha,beta} = sum_{k=0}^infty frac{(-1)^{k}}{(1+k)alpha - k beta -1}
= frac{1}{beta-1}+frac{Psileft(frac{beta - 1}{2alpha-2beta}right) - Psileft(frac{alpha-1}{2alpha-2beta}right)}{2alpha-2beta}$$
where $Psi$ is the digamma function.
edited Jan 4 at 23:32
Peter Foreman
4177
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
$endgroup$
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
1
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
1
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
add a comment |
$begingroup$
Let's say $alpha > max(1,beta)$.
$$ frac{1}{x^alpha + x^beta} = frac{1}{x^alpha (1 + x^{beta-alpha})} =
sum_{k=0}^infty frac{(-1)^k x^{kbeta}}{x^{(1+k)alpha}}$$
and integrating term-by-term
$$I_{alpha,beta} = sum_{k=0}^infty frac{(-1)^{k}}{(1+k)alpha - k beta -1}
= frac{1}{beta-1}+frac{Psileft(frac{beta - 1}{2alpha-2beta}right) - Psileft(frac{alpha-1}{2alpha-2beta}right)}{2alpha-2beta}$$
where $Psi$ is the digamma function.
edited Jan 4 at 23:32
Peter Foreman
4177
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
$endgroup$
Let's say $alpha > max(1,beta)$.
$$ frac{1}{x^alpha + x^beta} = frac{1}{x^alpha (1 + x^{beta-alpha})} =
sum_{k=0}^infty frac{(-1)^k x^{kbeta}}{x^{(1+k)alpha}}$$
and integrating term-by-term
$$I_{alpha,beta} = sum_{k=0}^infty frac{(-1)^{k}}{(1+k)alpha - k beta -1}
= frac{1}{beta-1}+frac{Psileft(frac{beta - 1}{2alpha-2beta}right) - Psileft(frac{alpha-1}{2alpha-2beta}right)}{2alpha-2beta}$$
where $Psi$ is the digamma function.
edited Jan 4 at 23:32
Peter Foreman
4177
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
edited Jan 4 at 23:32
Peter Foreman
4177
edited Jan 4 at 23:32
Peter Foreman
4177
edited Jan 4 at 23:32
Peter Foreman
4177
4177
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
answered Jan 4 at 20:58
Robert IsraelRobert Israel
320k23210462
320k23210462
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
1
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
1
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
add a comment |
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
1
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
1
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
How do you transform the sum into the digamma representation? Is there an identity which relates the digamma function to an infinite summation?
$endgroup$
– Peter Foreman
Jan 4 at 21:19
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
$begingroup$
+1. How about for all $alpha$ in $mathbb{R}$?
$endgroup$
– Gustavo Louis G. Montańo
Jan 4 at 21:36
1
1
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
$begingroup$
@GustavoLouisG.Montańo The integral diverges when $alpha$ < max(1,$beta$) and is trivial when $alpha=beta$.
$endgroup$
– Peter Foreman
Jan 4 at 21:42
1
1
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
This answer is actually valid for $alpha,betainmathbb{C}$ as well. As long as $mathfrak{R}(alpha) > $max(1,$mathfrak{R}(beta)$).
$endgroup$
– Peter Foreman
Jan 4 at 22:04
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
$begingroup$
Robert - To use the geometric series you need to be inside its interval of convergence of $|x| < 1$. The limits of the original integral are from 1 to $infty$, hence the reason why I made the change of $x mapsto 1/x$ to begin with in my answer.
$endgroup$
– omegadot
Jan 5 at 2:28
add a comment |
$begingroup$
For $alpha neq beta $. Put $$t={1 over 1+ x^{beta - alpha}} .$$ Then you integral becomes
begin{align} {1 over beta -alpha } int_0^{1 over 2 } t^{alpha -1 over beta - alpha }(1-t)^{1-beta over beta -alpha} , dt,end{align}
which is ${1 over beta -alpha} B(1/2; {beta -1 over beta -alpha}, {1-alpha over beta -alpha}) $,
where $B(x;mu,nu)$ is the incomplete Beta function.
answered Jan 5 at 0:35
moutheticsmouthetics
50127
$endgroup$
add a comment |
$begingroup$
For $alpha neq beta $. Put $$t={1 over 1+ x^{beta - alpha}} .$$ Then you integral becomes
begin{align} {1 over beta -alpha } int_0^{1 over 2 } t^{alpha -1 over beta - alpha }(1-t)^{1-beta over beta -alpha} , dt,end{align}
which is ${1 over beta -alpha} B(1/2; {beta -1 over beta -alpha}, {1-alpha over beta -alpha}) $,
where $B(x;mu,nu)$ is the incomplete Beta function.
answered Jan 5 at 0:35
moutheticsmouthetics
50127
$endgroup$
add a comment |
$begingroup$
For $alpha neq beta $. Put $$t={1 over 1+ x^{beta - alpha}} .$$ Then you integral becomes
begin{align} {1 over beta -alpha } int_0^{1 over 2 } t^{alpha -1 over beta - alpha }(1-t)^{1-beta over beta -alpha} , dt,end{align}
which is ${1 over beta -alpha} B(1/2; {beta -1 over beta -alpha}, {1-alpha over beta -alpha}) $,
where $B(x;mu,nu)$ is the incomplete Beta function.
answered Jan 5 at 0:35
moutheticsmouthetics
50127
$endgroup$
For $alpha neq beta $. Put $$t={1 over 1+ x^{beta - alpha}} .$$ Then you integral becomes
begin{align} {1 over beta -alpha } int_0^{1 over 2 } t^{alpha -1 over beta - alpha }(1-t)^{1-beta over beta -alpha} , dt,end{align}
which is ${1 over beta -alpha} B(1/2; {beta -1 over beta -alpha}, {1-alpha over beta -alpha}) $,
where $B(x;mu,nu)$ is the incomplete Beta function.
answered Jan 5 at 0:35
moutheticsmouthetics
50127
answered Jan 5 at 0:35
moutheticsmouthetics
50127
answered Jan 5 at 0:35
moutheticsmouthetics
50127
answered Jan 5 at 0:35
moutheticsmouthetics
50127
50127
add a comment |
add a comment |
$begingroup$
Assuming $beta > max (1,alpha)$. Enforcing a substitution of $x mapsto 1/x$ in the integral to begin with gives
$$I_{alpha, beta} = int_0^1 frac{x^{beta - 2}}{1 + x^{beta -alpha}} , dx.$$
Inside the interval of convergence, by exploiting the geometric series, namely
$$frac{1}{1 + x^{beta - alpha}} = sum_{n = 0}^infty (-1)^n x^{n beta - n alpha}, qquad |x| < 1$$
the integral can be rewritten as
begin{align}
I_{alpha, beta} &= sum_{n = 0}^infty (-1)^n int_0^1 x^{n beta - n alpha + beta - 2} , dx\
&= sum_{n = 0}^infty (-1)^n frac{1}{n beta - n alpha + beta - 1}. qquad (*)
end{align}
To handle the infinity sum that arises we will make use of the following result (see Eq. (6) in the link)
$$sum_{n = 0}^infty frac{(-1)^n}{z n + 1} = frac{1}{2z} left [psi left (frac{z + 1}{2z} right ) - psi left (frac{1}{2z} right ) right ]. qquad (**)$$
Here $psi (x)$ is the digamma function.
Rewriting the sum in ($*$) in the form of ($**$), namely
$$I_{alpha, beta} = frac{1}{beta - 1} sum_{n = 0}^infty frac{(-1)^n}{left (dfrac{beta - alpha}{beta - 1} right ) n + 1},$$
as we have $z = (beta - alpha)/(beta - 1)$, we finally arrive at
$$I_{alpha, beta} = frac{1}{2beta - 2 alpha} left [psi left (frac{2 beta - alpha - 1}{2 beta - 2alpha} right ) - psi left (frac{beta - 1}{2 beta - 2alpha} right ) right ], qquad beta > max (1,alpha).$$
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
$endgroup$
add a comment |
$begingroup$
Assuming $beta > max (1,alpha)$. Enforcing a substitution of $x mapsto 1/x$ in the integral to begin with gives
$$I_{alpha, beta} = int_0^1 frac{x^{beta - 2}}{1 + x^{beta -alpha}} , dx.$$
Inside the interval of convergence, by exploiting the geometric series, namely
$$frac{1}{1 + x^{beta - alpha}} = sum_{n = 0}^infty (-1)^n x^{n beta - n alpha}, qquad |x| < 1$$
the integral can be rewritten as
begin{align}
I_{alpha, beta} &= sum_{n = 0}^infty (-1)^n int_0^1 x^{n beta - n alpha + beta - 2} , dx\
&= sum_{n = 0}^infty (-1)^n frac{1}{n beta - n alpha + beta - 1}. qquad (*)
end{align}
To handle the infinity sum that arises we will make use of the following result (see Eq. (6) in the link)
$$sum_{n = 0}^infty frac{(-1)^n}{z n + 1} = frac{1}{2z} left [psi left (frac{z + 1}{2z} right ) - psi left (frac{1}{2z} right ) right ]. qquad (**)$$
Here $psi (x)$ is the digamma function.
Rewriting the sum in ($*$) in the form of ($**$), namely
$$I_{alpha, beta} = frac{1}{beta - 1} sum_{n = 0}^infty frac{(-1)^n}{left (dfrac{beta - alpha}{beta - 1} right ) n + 1},$$
as we have $z = (beta - alpha)/(beta - 1)$, we finally arrive at
$$I_{alpha, beta} = frac{1}{2beta - 2 alpha} left [psi left (frac{2 beta - alpha - 1}{2 beta - 2alpha} right ) - psi left (frac{beta - 1}{2 beta - 2alpha} right ) right ], qquad beta > max (1,alpha).$$
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
$endgroup$
add a comment |
$begingroup$
Assuming $beta > max (1,alpha)$. Enforcing a substitution of $x mapsto 1/x$ in the integral to begin with gives
$$I_{alpha, beta} = int_0^1 frac{x^{beta - 2}}{1 + x^{beta -alpha}} , dx.$$
Inside the interval of convergence, by exploiting the geometric series, namely
$$frac{1}{1 + x^{beta - alpha}} = sum_{n = 0}^infty (-1)^n x^{n beta - n alpha}, qquad |x| < 1$$
the integral can be rewritten as
begin{align}
I_{alpha, beta} &= sum_{n = 0}^infty (-1)^n int_0^1 x^{n beta - n alpha + beta - 2} , dx\
&= sum_{n = 0}^infty (-1)^n frac{1}{n beta - n alpha + beta - 1}. qquad (*)
end{align}
To handle the infinity sum that arises we will make use of the following result (see Eq. (6) in the link)
$$sum_{n = 0}^infty frac{(-1)^n}{z n + 1} = frac{1}{2z} left [psi left (frac{z + 1}{2z} right ) - psi left (frac{1}{2z} right ) right ]. qquad (**)$$
Here $psi (x)$ is the digamma function.
Rewriting the sum in ($*$) in the form of ($**$), namely
$$I_{alpha, beta} = frac{1}{beta - 1} sum_{n = 0}^infty frac{(-1)^n}{left (dfrac{beta - alpha}{beta - 1} right ) n + 1},$$
as we have $z = (beta - alpha)/(beta - 1)$, we finally arrive at
$$I_{alpha, beta} = frac{1}{2beta - 2 alpha} left [psi left (frac{2 beta - alpha - 1}{2 beta - 2alpha} right ) - psi left (frac{beta - 1}{2 beta - 2alpha} right ) right ], qquad beta > max (1,alpha).$$
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
$endgroup$
Assuming $beta > max (1,alpha)$. Enforcing a substitution of $x mapsto 1/x$ in the integral to begin with gives
$$I_{alpha, beta} = int_0^1 frac{x^{beta - 2}}{1 + x^{beta -alpha}} , dx.$$
Inside the interval of convergence, by exploiting the geometric series, namely
$$frac{1}{1 + x^{beta - alpha}} = sum_{n = 0}^infty (-1)^n x^{n beta - n alpha}, qquad |x| < 1$$
the integral can be rewritten as
begin{align}
I_{alpha, beta} &= sum_{n = 0}^infty (-1)^n int_0^1 x^{n beta - n alpha + beta - 2} , dx\
&= sum_{n = 0}^infty (-1)^n frac{1}{n beta - n alpha + beta - 1}. qquad (*)
end{align}
To handle the infinity sum that arises we will make use of the following result (see Eq. (6) in the link)
$$sum_{n = 0}^infty frac{(-1)^n}{z n + 1} = frac{1}{2z} left [psi left (frac{z + 1}{2z} right ) - psi left (frac{1}{2z} right ) right ]. qquad (**)$$
Here $psi (x)$ is the digamma function.
Rewriting the sum in ($*$) in the form of ($**$), namely
$$I_{alpha, beta} = frac{1}{beta - 1} sum_{n = 0}^infty frac{(-1)^n}{left (dfrac{beta - alpha}{beta - 1} right ) n + 1},$$
as we have $z = (beta - alpha)/(beta - 1)$, we finally arrive at
$$I_{alpha, beta} = frac{1}{2beta - 2 alpha} left [psi left (frac{2 beta - alpha - 1}{2 beta - 2alpha} right ) - psi left (frac{beta - 1}{2 beta - 2alpha} right ) right ], qquad beta > max (1,alpha).$$
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
answered Jan 5 at 2:17
omegadotomegadot
5,0652727
5,0652727
add a comment |
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%2f3062082%2fcompute-int-1-infty-frac1x-alpha-x-betadx%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
$begingroup$
Are we assuming that $max(alpha,beta)-min(alpha,beta)$ is an integer?
$endgroup$
– Jack D'Aurizio
Jan 4 at 20:49