Evaluation of $int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$
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For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
Now how can i proceed further, Help required, Thanks
definite-integrals
$endgroup$
add a comment |
$begingroup$
For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
Now how can i proceed further, Help required, Thanks
definite-integrals
$endgroup$
1
$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06
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To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08
add a comment |
$begingroup$
For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
Now how can i proceed further, Help required, Thanks
definite-integrals
$endgroup$
For $0<a,b<1.$
Evaluation of $$int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
$bf{My; Try::}$ Let $$I = int_{0}^{1}frac{x^{a-1}-x^{b-1}}{1-x}dx+int_{1}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx$$
Now how can i proceed further, Help required, Thanks
definite-integrals
definite-integrals
asked Jan 30 '17 at 19:39
juantheronjuantheron
34.2k1147142
34.2k1147142
1
$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06
$begingroup$
To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08
add a comment |
1
$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06
$begingroup$
To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08
1
1
$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06
$begingroup$
Are you familiar with techniques from complex analysis?
$endgroup$
– Michael Seifert
Jan 30 '17 at 20:06
$begingroup$
To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08
$begingroup$
To Michael Saifert. not to much, elementry knowledge.
$endgroup$
– juantheron
Jan 31 '17 at 3:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
end{align}
In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
,{dd x over -x^{2}}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
\[5mm] & =
bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
end{align}
$ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.
Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
$$
bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
picotpars{pi a} - picotpars{pi b}}}
$$
$endgroup$
$begingroup$
I knew the result MyGlasses derived looked familiar!
$endgroup$
– J.G.
Jan 30 '17 at 22:52
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@J.G. Yes. It's true. Thanks.
$endgroup$
– Felix Marin
Jan 30 '17 at 23:15
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Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
$endgroup$
– juantheron
Jan 31 '17 at 3:11
1
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
$endgroup$
– Felix Marin
Jan 31 '17 at 3:20
$begingroup$
Thanks Felix Marin got it.
$endgroup$
– juantheron
Jan 31 '17 at 4:33
add a comment |
$begingroup$
For $0<a,b<1$, with $x=e^u$
begin{eqnarray}
int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
&=&
sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
end{eqnarray}
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which gives us what?
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– tired
Jan 30 '17 at 21:17
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I'm stuck here. maybe diverge!
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– Nosrati
Jan 30 '17 at 21:21
1
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no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
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– tired
Jan 30 '17 at 21:29
add a comment |
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You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.
Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
end{align}
In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
,{dd x over -x^{2}}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
\[5mm] & =
bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
end{align}
$ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.
Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
$$
bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
picotpars{pi a} - picotpars{pi b}}}
$$
$endgroup$
$begingroup$
I knew the result MyGlasses derived looked familiar!
$endgroup$
– J.G.
Jan 30 '17 at 22:52
$begingroup$
@J.G. Yes. It's true. Thanks.
$endgroup$
– Felix Marin
Jan 30 '17 at 23:15
$begingroup$
Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
$endgroup$
– juantheron
Jan 31 '17 at 3:11
1
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
$endgroup$
– Felix Marin
Jan 31 '17 at 3:20
$begingroup$
Thanks Felix Marin got it.
$endgroup$
– juantheron
Jan 31 '17 at 4:33
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
end{align}
In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
,{dd x over -x^{2}}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
\[5mm] & =
bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
end{align}
$ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.
Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
$$
bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
picotpars{pi a} - picotpars{pi b}}}
$$
$endgroup$
$begingroup$
I knew the result MyGlasses derived looked familiar!
$endgroup$
– J.G.
Jan 30 '17 at 22:52
$begingroup$
@J.G. Yes. It's true. Thanks.
$endgroup$
– Felix Marin
Jan 30 '17 at 23:15
$begingroup$
Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
$endgroup$
– juantheron
Jan 31 '17 at 3:11
1
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
$endgroup$
– Felix Marin
Jan 31 '17 at 3:20
$begingroup$
Thanks Felix Marin got it.
$endgroup$
– juantheron
Jan 31 '17 at 4:33
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
end{align}
In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
,{dd x over -x^{2}}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
\[5mm] & =
bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
end{align}
$ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.
Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
$$
bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
picotpars{pi a} - picotpars{pi b}}}
$$
$endgroup$
$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x & =
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x
end{align}
In the RHS second integral I'll perform the change $ds{x mapsto 1/x}$:
begin{align}
&bbox[10px,#ffd]{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x +
int_{1}^{0}{pars{1/x}^{a - 1} - pars{1/x}^{1 - b} over 1 - 1/x}
,{dd x over -x^{2}}
\[5mm] = &
int_{0}^{1}{x^{a - 1} - x^{b - 1} over 1 - x},dd x -
int_{0}^{1}{x^{-a} - x^{-b} over 1 - x},dd x
\[5mm] & =
bbx{ds{H_{b - 1} - H_{a - 1} + H_{-a} - H_{-b}}}
quadmbox{with} Repars{a},, Repars{b} in pars{0,1}
end{align}
$ds{H_{n}}$ is a Harmonic Number and I used a well known identity ( as given by Euler ): $ds{H_{z} = int_{0}^{1}{1 - t^{z} over 1 - t},dd t}$ with $ds{Repars{z} > -1}$.
Note that $ds{H_{b - 1} - H_{-b} = -picotpars{pi b}}$ such that:
$$
bbx{ds{int_{0}^{infty}{x^{a - 1} - x^{b - 1} over 1 - x},dd x =
picotpars{pi a} - picotpars{pi b}}}
$$
edited Dec 22 '18 at 16:19
answered Jan 30 '17 at 21:11
Felix MarinFelix Marin
67.8k7107142
67.8k7107142
$begingroup$
I knew the result MyGlasses derived looked familiar!
$endgroup$
– J.G.
Jan 30 '17 at 22:52
$begingroup$
@J.G. Yes. It's true. Thanks.
$endgroup$
– Felix Marin
Jan 30 '17 at 23:15
$begingroup$
Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
$endgroup$
– juantheron
Jan 31 '17 at 3:11
1
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
$endgroup$
– Felix Marin
Jan 31 '17 at 3:20
$begingroup$
Thanks Felix Marin got it.
$endgroup$
– juantheron
Jan 31 '17 at 4:33
add a comment |
$begingroup$
I knew the result MyGlasses derived looked familiar!
$endgroup$
– J.G.
Jan 30 '17 at 22:52
$begingroup$
@J.G. Yes. It's true. Thanks.
$endgroup$
– Felix Marin
Jan 30 '17 at 23:15
$begingroup$
Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
$endgroup$
– juantheron
Jan 31 '17 at 3:11
1
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
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– Felix Marin
Jan 31 '17 at 3:20
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Thanks Felix Marin got it.
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– juantheron
Jan 31 '17 at 4:33
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I knew the result MyGlasses derived looked familiar!
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– J.G.
Jan 30 '17 at 22:52
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I knew the result MyGlasses derived looked familiar!
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– J.G.
Jan 30 '17 at 22:52
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@J.G. Yes. It's true. Thanks.
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– Felix Marin
Jan 30 '17 at 23:15
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@J.G. Yes. It's true. Thanks.
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– Felix Marin
Jan 30 '17 at 23:15
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Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
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– juantheron
Jan 31 '17 at 3:11
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Thanks Felix Marin, plz explain me how you get $H_{b-1}-H_{-b} = -pi cot(pi b)$
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– juantheron
Jan 31 '17 at 3:11
1
1
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
$endgroup$
– Felix Marin
Jan 31 '17 at 3:20
$begingroup$
@juantheron $H_{m}$ is related to the Digamma Function $Psi$. Namely, $H_{m} = Psileft(m + 1right) + gamma$. $gamma$ is the Euler-Mascheroni Constant. Then, $H_{b - 1} - H_{-b} = Psileft(bright) - Psileft(-b + 1right) = -picotleft(pi bright)$. The last one is the Euler Reflection Formula.
$endgroup$
– Felix Marin
Jan 31 '17 at 3:20
$begingroup$
Thanks Felix Marin got it.
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– juantheron
Jan 31 '17 at 4:33
$begingroup$
Thanks Felix Marin got it.
$endgroup$
– juantheron
Jan 31 '17 at 4:33
add a comment |
$begingroup$
For $0<a,b<1$, with $x=e^u$
begin{eqnarray}
int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
&=&
sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
end{eqnarray}
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$begingroup$
which gives us what?
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– tired
Jan 30 '17 at 21:17
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I'm stuck here. maybe diverge!
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– Nosrati
Jan 30 '17 at 21:21
1
$begingroup$
no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
$endgroup$
– tired
Jan 30 '17 at 21:29
add a comment |
$begingroup$
For $0<a,b<1$, with $x=e^u$
begin{eqnarray}
int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
&=&
sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
end{eqnarray}
$endgroup$
$begingroup$
which gives us what?
$endgroup$
– tired
Jan 30 '17 at 21:17
$begingroup$
I'm stuck here. maybe diverge!
$endgroup$
– Nosrati
Jan 30 '17 at 21:21
1
$begingroup$
no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
$endgroup$
– tired
Jan 30 '17 at 21:29
add a comment |
$begingroup$
For $0<a,b<1$, with $x=e^u$
begin{eqnarray}
int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
&=&
sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
end{eqnarray}
$endgroup$
For $0<a,b<1$, with $x=e^u$
begin{eqnarray}
int_{0}^{infty}frac{x^{a-1}-x^{b-1}}{1-x}dx
&=&
int_{-infty}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{e^{au}-e^{bu}}{1-e^u}dx+int_{0}^{infty}frac{e^{(1-a)u}-e^{(1-b)u}}{1-e^u}dx\
&=&
int_{0}^{infty}frac{-e^{-u}}{1-e^{-u}}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
int_{0}^{infty}sum_{n=1}^infty e^{-nu}Big(e^{au}-e^{bu}+e^{(1-a)u}-e^{(1-b)u}Big)dx\
&=&
sum_{n=1}^inftyint_{0}^{infty}Big(e^{(a-n)u}-e^{(b-n)u}+e^{(1-a-n)u}-e^{(1-b-n)u}Big)dx\
&=&
sum_{n=1}^inftyBig(frac{1}{n-a}-frac{1}{n-b}+frac{1}{n+a-1}-frac{1}{n+b-1}Big)
end{eqnarray}
answered Jan 30 '17 at 20:51
NosratiNosrati
26.5k62354
26.5k62354
$begingroup$
which gives us what?
$endgroup$
– tired
Jan 30 '17 at 21:17
$begingroup$
I'm stuck here. maybe diverge!
$endgroup$
– Nosrati
Jan 30 '17 at 21:21
1
$begingroup$
no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
$endgroup$
– tired
Jan 30 '17 at 21:29
add a comment |
$begingroup$
which gives us what?
$endgroup$
– tired
Jan 30 '17 at 21:17
$begingroup$
I'm stuck here. maybe diverge!
$endgroup$
– Nosrati
Jan 30 '17 at 21:21
1
$begingroup$
no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
$endgroup$
– tired
Jan 30 '17 at 21:29
$begingroup$
which gives us what?
$endgroup$
– tired
Jan 30 '17 at 21:17
$begingroup$
which gives us what?
$endgroup$
– tired
Jan 30 '17 at 21:17
$begingroup$
I'm stuck here. maybe diverge!
$endgroup$
– Nosrati
Jan 30 '17 at 21:21
$begingroup$
I'm stuck here. maybe diverge!
$endgroup$
– Nosrati
Jan 30 '17 at 21:21
1
1
$begingroup$
no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
$endgroup$
– tired
Jan 30 '17 at 21:29
$begingroup$
no the divergencies of the single terms are all of the same magnitude ($sim log(n)$)and therefore cancel out...what is left is most likely a so called digamma function
$endgroup$
– tired
Jan 30 '17 at 21:29
add a comment |
$begingroup$
You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.
Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
$endgroup$
add a comment |
$begingroup$
You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.
Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
$endgroup$
add a comment |
$begingroup$
You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.
Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
$endgroup$
You have to study what happens near $x=1$ and when $xto infty$, as I suppose you have notice.
Defining $f(x)=frac{x^{a-1}-x^{b-1}}{1-x}$, $lim_{xto 1} f(x)$ exists and it's finite (why?). So you can define $f(1)$ by continuity, and $int_0^1 f(x) dx$ exists and it's a number.
I do not know how to proceed from here. Any help would be appreciated.
edited Jan 30 '17 at 19:53
answered Jan 30 '17 at 19:47
A. Salguero-AlarcónA. Salguero-Alarcón
3,267319
3,267319
add a comment |
add a comment |
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1
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Are you familiar with techniques from complex analysis?
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– Michael Seifert
Jan 30 '17 at 20:06
$begingroup$
To Michael Saifert. not to much, elementry knowledge.
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– juantheron
Jan 31 '17 at 3:08