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closed form of the following integral :$int_0^infty- sqrtx+ sqrtxcoth (x) dx$?

Showing that $2 Gamma(a) zeta(a) left(1-frac12^a right) = int_0^inftyleft( fracx^a-1sinh x - x^a-2 right) , dx$Do these integrals have a closed form? $I_1 = int_-infty ^infty fracsin (x)x cosh (x) , dx$Closed form of $int_0^infty fractanh(x),tanh(2x)x^2;dx$Help evaluating the integral $int_0^infty omega cos(omega t) coth(alpha omega) textd omega$.Can this integral be computed in closed-form?How do I find a closed form of $pi^2nover zeta(2n)int_-1^1x^2n-2over pi^2+(2tanh^-1x)^2dx$?Is there a closed form for this integral $int_0^inftye^x-1over e^ax-1 dx?$On a closed form for $int_-infty^inftyfracdxleft(1+x^2right)^p$Evaluate: $int_0^inftyfracln^n(x)ln(1+x)x(1+x^2)mathrm dx$Closed-form of $int_0^infty exp (-ax^2)log x mathrm d x$ without using Laplace Transform?

2

1

$begingroup$

I have tried to evaluate this:$int_0^infty- sqrtx+ sqrtxcoth (x)$ using the the following formula

$$2 Gamma(a) zeta(a) left(1-frac12^a right) = int_0^inftyBig( fracx^a-1sinh x - x^a-2Big) , dx , colorbleu-1 <textRe(a) <1. tag1$$
in order to present the preceding integral in closed form but I didn't succeed , This
$int_0^infty- sqrtx+ sqrtxcoth (x)$ close to $1.63$ using wolfram alpha and I really bielive that integral could be represented in a closed form value using $(1)$ since Coth function has a relationship with Sinh function , Now Is there a general formula for:$$int_0^inftyBig( fracx^a-1tanh x - x^a-1Big) , dx , colorbleu-1 <textRe(a) <1. tag2$$ if we assume $a$ as a complex varible with real part lie in $(-1,1)$ ?

integration closed-form gamma-function zeta-functions

share|cite|improve this question

edited Mar 29 at 21:34

zeraoulia rafik

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The definite integral that you linked on Wolfram Alpha has a $cothleft(sqrtxright)$ rather than $cothleft(xright)$.
  $endgroup$
  – Jake
  Mar 29 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks , I edited the link for Wolfram alpha , the result is 1.63
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:35
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I have tried to evaluate this:$int_0^infty- sqrtx+ sqrtxcoth (x)$ using the the following formula

$$2 Gamma(a) zeta(a) left(1-frac12^a right) = int_0^inftyBig( fracx^a-1sinh x - x^a-2Big) , dx , colorbleu-1 <textRe(a) <1. tag1$$
in order to present the preceding integral in closed form but I didn't succeed , This
$int_0^infty- sqrtx+ sqrtxcoth (x)$ close to $1.63$ using wolfram alpha and I really bielive that integral could be represented in a closed form value using $(1)$ since Coth function has a relationship with Sinh function , Now Is there a general formula for:$$int_0^inftyBig( fracx^a-1tanh x - x^a-1Big) , dx , colorbleu-1 <textRe(a) <1. tag2$$ if we assume $a$ as a complex varible with real part lie in $(-1,1)$ ?

integration closed-form gamma-function zeta-functions

share|cite|improve this question

edited Mar 29 at 21:34

zeraoulia rafik

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The definite integral that you linked on Wolfram Alpha has a $cothleft(sqrtxright)$ rather than $cothleft(xright)$.
  $endgroup$
  – Jake
  Mar 29 at 21:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks , I edited the link for Wolfram alpha , the result is 1.63
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:35
  

  
  
  
  

add a comment | 

$begingroup$

I have tried to evaluate this:$int_0^infty- sqrtx+ sqrtxcoth (x)$ using the the following formula

$$2 Gamma(a) zeta(a) left(1-frac12^a right) = int_0^inftyBig( fracx^a-1sinh x - x^a-2Big) , dx , colorbleu-1 <textRe(a) <1. tag1$$
in order to present the preceding integral in closed form but I didn't succeed , This
$int_0^infty- sqrtx+ sqrtxcoth (x)$ close to $1.63$ using wolfram alpha and I really bielive that integral could be represented in a closed form value using $(1)$ since Coth function has a relationship with Sinh function , Now Is there a general formula for:$$int_0^inftyBig( fracx^a-1tanh x - x^a-1Big) , dx , colorbleu-1 <textRe(a) <1. tag2$$ if we assume $a$ as a complex varible with real part lie in $(-1,1)$ ?

integration closed-form gamma-function zeta-functions

share|cite|improve this question

edited Mar 29 at 21:34

zeraoulia rafik

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

$endgroup$

share|cite|improve this question

edited Mar 29 at 21:34

zeraoulia rafik

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

share|cite|improve this question

edited Mar 29 at 21:34

zeraoulia rafik

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

asked Mar 29 at 21:18

zeraoulia rafikzeraoulia rafik

2,37711133

1 Answer
1

active

oldest

votes

3

$begingroup$

An easier approach is to rewrite the integral as $$int_0^inftyfrac2sqrtxe^-2xdx1-e^-2x=2sum_nge 1int_0^inftysqrtxe^-2nxdx=2Gammaleft(frac32right)sum_n=1(2n)^-3/2=sqrtfracpi8zetaleft(frac32right).$$This zeta constant isn't known, it seems, to have a nice closed form (but see also its discussion here, which up to $x=t^2$ reports the same integral representation).

share|cite|improve this answer

edited Mar 29 at 21:37

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks , I edited the link for Wolfram alpha , the result is 1.63
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @zeraouliarafik Well, 1.63706 etc. Thanks for clarifying. FWIW the problem in your previous WA link would have also been solvable the same way, viz. $$int_0^inftyfrac2sqrtxe^-2sqrtxdx1-e^-2sqrtx=4sum_nge 1int_0^infty t^2e^-2ntdt=zeta(3)$$ (assuming I haven't made any mistakes in my arithmetic).
  $endgroup$
  – J.G.
  Mar 29 at 21:38
  
  

  
  
  
  


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  $begingroup$
  In fact, more generally I find $$int_0^inftyx^aleft(coth x^b-1right)dx=frac2^1-fraca+1bbGammaleft(fraca+1bright)zetaleft(fraca+1bright).$$
  $endgroup$
  – J.G.
  Mar 29 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice closed form
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zeraouliarafik Well, insofar as we consider special functions such as $Gamma,,zeta$ closed forms. That's part of what we like about them.
  $endgroup$
  – J.G.
  Mar 29 at 22:16
  

  
  
  
  

add a comment | 

Your Answer

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3

$begingroup$

An easier approach is to rewrite the integral as $$int_0^inftyfrac2sqrtxe^-2xdx1-e^-2x=2sum_nge 1int_0^inftysqrtxe^-2nxdx=2Gammaleft(frac32right)sum_n=1(2n)^-3/2=sqrtfracpi8zetaleft(frac32right).$$This zeta constant isn't known, it seems, to have a nice closed form (but see also its discussion here, which up to $x=t^2$ reports the same integral representation).

share|cite|improve this answer

edited Mar 29 at 21:37

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks , I edited the link for Wolfram alpha , the result is 1.63
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @zeraouliarafik Well, 1.63706 etc. Thanks for clarifying. FWIW the problem in your previous WA link would have also been solvable the same way, viz. $$int_0^inftyfrac2sqrtxe^-2sqrtxdx1-e^-2sqrtx=4sum_nge 1int_0^infty t^2e^-2ntdt=zeta(3)$$ (assuming I haven't made any mistakes in my arithmetic).
  $endgroup$
  – J.G.
  Mar 29 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, more generally I find $$int_0^inftyx^aleft(coth x^b-1right)dx=frac2^1-fraca+1bbGammaleft(fraca+1bright)zetaleft(fraca+1bright).$$
  $endgroup$
  – J.G.
  Mar 29 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice closed form
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zeraouliarafik Well, insofar as we consider special functions such as $Gamma,,zeta$ closed forms. That's part of what we like about them.
  $endgroup$
  – J.G.
  Mar 29 at 22:16
  

  
  
  
  

add a comment | 

3

$begingroup$

An easier approach is to rewrite the integral as $$int_0^inftyfrac2sqrtxe^-2xdx1-e^-2x=2sum_nge 1int_0^inftysqrtxe^-2nxdx=2Gammaleft(frac32right)sum_n=1(2n)^-3/2=sqrtfracpi8zetaleft(frac32right).$$This zeta constant isn't known, it seems, to have a nice closed form (but see also its discussion here, which up to $x=t^2$ reports the same integral representation).

share|cite|improve this answer

edited Mar 29 at 21:37

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks , I edited the link for Wolfram alpha , the result is 1.63
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @zeraouliarafik Well, 1.63706 etc. Thanks for clarifying. FWIW the problem in your previous WA link would have also been solvable the same way, viz. $$int_0^inftyfrac2sqrtxe^-2sqrtxdx1-e^-2sqrtx=4sum_nge 1int_0^infty t^2e^-2ntdt=zeta(3)$$ (assuming I haven't made any mistakes in my arithmetic).
  $endgroup$
  – J.G.
  Mar 29 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, more generally I find $$int_0^inftyx^aleft(coth x^b-1right)dx=frac2^1-fraca+1bbGammaleft(fraca+1bright)zetaleft(fraca+1bright).$$
  $endgroup$
  – J.G.
  Mar 29 at 21:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Nice closed form
  $endgroup$
  – zeraoulia rafik
  Mar 29 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @zeraouliarafik Well, insofar as we consider special functions such as $Gamma,,zeta$ closed forms. That's part of what we like about them.
  $endgroup$
  – J.G.
  Mar 29 at 22:16
  

  
  
  
  

add a comment | 

$begingroup$

An easier approach is to rewrite the integral as $$int_0^inftyfrac2sqrtxe^-2xdx1-e^-2x=2sum_nge 1int_0^inftysqrtxe^-2nxdx=2Gammaleft(frac32right)sum_n=1(2n)^-3/2=sqrtfracpi8zetaleft(frac32right).$$This zeta constant isn't known, it seems, to have a nice closed form (but see also its discussion here, which up to $x=t^2$ reports the same integral representation).

share|cite|improve this answer

edited Mar 29 at 21:37

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250

$endgroup$

share|cite|improve this answer

edited Mar 29 at 21:37

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250

answered Mar 29 at 21:33

J.G.J.G.

32.9k23250