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How to integrate the multiplication of a polynonmial, a fraction and an exponential

0

$begingroup$

EDITED:

Any ideas on how to do the integral of this function?

$int_0^1fracx^2(x+2)(x-2)e^x/(x+2)e^x/(x-2)dt$

integration fourier-analysis exponential-function

share|cite|improve this question

edited Mar 29 at 12:57

DrIbraComms

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for you reply Moo. It is $intop_0^pi/2fracsin^2(x)(sin^2(x)+2)(sin^2(x)-2)e^fracsin^2(x)(sin^2(x)+2)e^fracsin^2(x)(sin^2(x)-2)dx$
  $endgroup$
  – DrIbraComms
  Mar 26 at 18:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is right, my bad. x^2 translates to sin^4(x).
  $endgroup$
  – DrIbraComms
  Mar 26 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is actually an integral of an iterated multiplication of the same function but with different constants as $intop_0^pi/2prod_i=1^2fracsin^2(x)(sin^2(x)+c_i)e^fracsin^2(x)(sin^2(x)+c_i)dx$
  $endgroup$
  – DrIbraComms
  Mar 26 at 18:11
  
  

  
  
  
  

add a comment | 

0

$begingroup$

EDITED:

Any ideas on how to do the integral of this function?

$int_0^1fracx^2(x+2)(x-2)e^x/(x+2)e^x/(x-2)dt$

integration fourier-analysis exponential-function

share|cite|improve this question

edited Mar 29 at 12:57

DrIbraComms

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for you reply Moo. It is $intop_0^pi/2fracsin^2(x)(sin^2(x)+2)(sin^2(x)-2)e^fracsin^2(x)(sin^2(x)+2)e^fracsin^2(x)(sin^2(x)-2)dx$
  $endgroup$
  – DrIbraComms
  Mar 26 at 18:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is right, my bad. x^2 translates to sin^4(x).
  $endgroup$
  – DrIbraComms
  Mar 26 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is actually an integral of an iterated multiplication of the same function but with different constants as $intop_0^pi/2prod_i=1^2fracsin^2(x)(sin^2(x)+c_i)e^fracsin^2(x)(sin^2(x)+c_i)dx$
  $endgroup$
  – DrIbraComms
  Mar 26 at 18:11
  
  

  
  
  
  

add a comment | 

$begingroup$

EDITED:

Any ideas on how to do the integral of this function?

$int_0^1fracx^2(x+2)(x-2)e^x/(x+2)e^x/(x-2)dt$

integration fourier-analysis exponential-function

share|cite|improve this question

edited Mar 29 at 12:57

DrIbraComms

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

$endgroup$

share|cite|improve this question

edited Mar 29 at 12:57

DrIbraComms

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

share|cite|improve this question

edited Mar 29 at 12:57

DrIbraComms

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

asked Mar 26 at 17:55

DrIbraCommsDrIbraComms

32

1 Answer
1

active

oldest

votes

0

$begingroup$

This is not a complete answer:

Substitution of

$z=frac2 x^2(x-2) (x+2)$

and the settings

$kappa =-frac13 sqrt3$,
$u=-frac23$,
$nu =frac32$ and $mu =-frac12$

leads to the more general form of the integral above:

$I = kappa int_0^1 e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz$.

Now you can figure out the solution. In fact you have a special function (Hypergeometric, Meijer G or H-Fox-function with two variables). Reduction takes place for $u/2->u$ or in the case:

$I =kappa int_0^infty e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz=2^nu (-u)^-nu textGamma[nu ], textHypergeometricU[nu,mu +nu ,2]$

EDIT

$e^fracx(x-2),e^fracx(x+2)=e^frac2,x^2(x-2) (x+2)$

share|cite|improve this answer

edited Mar 27 at 16:26

answered Mar 27 at 13:12

stochastocha

31138

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This would work if there are only one exponential product. However, in this case the powers of the two exponentials will add up. Hence, the first assumption is invalid for this problem.
  $endgroup$
  – DrIbraComms
  Mar 27 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function is originally in the following form $int_0^pi/2prod_i=1^2fracsin^2left(xright)sin^2left(xright)+c_ie^mfracsin^2left(xright)sin^2left(xright)+c_idx$ When the product has a single iteration (i=1 only) your approach works perfectly.
  $endgroup$
  – DrIbraComms
  Mar 27 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did not understand your comment, because I just simplify the two exponentials and did the substitution (See EDIT). I calculate the integral numerical under Mathematica and it worked, fine!
  $endgroup$
  – stocha
  Mar 27 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's true, your answer is right. That was one particular case of the main problem. I just edited the main question. Sorry for the inconvenience
  $endgroup$
  – DrIbraComms
  Mar 27 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would suggest to solve the particular case first and then the general case. It should work the same way.
  $endgroup$
  – stocha
  Mar 27 at 16:47
  

  
  
  
  

 | 
show 1 more comment

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0

$begingroup$

This is not a complete answer:

Substitution of

$z=frac2 x^2(x-2) (x+2)$

and the settings

$kappa =-frac13 sqrt3$,
$u=-frac23$,
$nu =frac32$ and $mu =-frac12$

leads to the more general form of the integral above:

$I = kappa int_0^1 e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz$.

Now you can figure out the solution. In fact you have a special function (Hypergeometric, Meijer G or H-Fox-function with two variables). Reduction takes place for $u/2->u$ or in the case:

$I =kappa int_0^infty e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz=2^nu (-u)^-nu textGamma[nu ], textHypergeometricU[nu,mu +nu ,2]$

EDIT

$e^fracx(x-2),e^fracx(x+2)=e^frac2,x^2(x-2) (x+2)$

share|cite|improve this answer

edited Mar 27 at 16:26

answered Mar 27 at 13:12

stochastocha

31138

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This would work if there are only one exponential product. However, in this case the powers of the two exponentials will add up. Hence, the first assumption is invalid for this problem.
  $endgroup$
  – DrIbraComms
  Mar 27 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function is originally in the following form $int_0^pi/2prod_i=1^2fracsin^2left(xright)sin^2left(xright)+c_ie^mfracsin^2left(xright)sin^2left(xright)+c_idx$ When the product has a single iteration (i=1 only) your approach works perfectly.
  $endgroup$
  – DrIbraComms
  Mar 27 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did not understand your comment, because I just simplify the two exponentials and did the substitution (See EDIT). I calculate the integral numerical under Mathematica and it worked, fine!
  $endgroup$
  – stocha
  Mar 27 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's true, your answer is right. That was one particular case of the main problem. I just edited the main question. Sorry for the inconvenience
  $endgroup$
  – DrIbraComms
  Mar 27 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would suggest to solve the particular case first and then the general case. It should work the same way.
  $endgroup$
  – stocha
  Mar 27 at 16:47
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

This is not a complete answer:

Substitution of

$z=frac2 x^2(x-2) (x+2)$

and the settings

$kappa =-frac13 sqrt3$,
$u=-frac23$,
$nu =frac32$ and $mu =-frac12$

leads to the more general form of the integral above:

$I = kappa int_0^1 e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz$.

Now you can figure out the solution. In fact you have a special function (Hypergeometric, Meijer G or H-Fox-function with two variables). Reduction takes place for $u/2->u$ or in the case:

$I =kappa int_0^infty e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz=2^nu (-u)^-nu textGamma[nu ], textHypergeometricU[nu,mu +nu ,2]$

EDIT

$e^fracx(x-2),e^fracx(x+2)=e^frac2,x^2(x-2) (x+2)$

share|cite|improve this answer

edited Mar 27 at 16:26

answered Mar 27 at 13:12

stochastocha

31138

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This would work if there are only one exponential product. However, in this case the powers of the two exponentials will add up. Hence, the first assumption is invalid for this problem.
  $endgroup$
  – DrIbraComms
  Mar 27 at 15:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function is originally in the following form $int_0^pi/2prod_i=1^2fracsin^2left(xright)sin^2left(xright)+c_ie^mfracsin^2left(xright)sin^2left(xright)+c_idx$ When the product has a single iteration (i=1 only) your approach works perfectly.
  $endgroup$
  – DrIbraComms
  Mar 27 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did not understand your comment, because I just simplify the two exponentials and did the substitution (See EDIT). I calculate the integral numerical under Mathematica and it worked, fine!
  $endgroup$
  – stocha
  Mar 27 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's true, your answer is right. That was one particular case of the main problem. I just edited the main question. Sorry for the inconvenience
  $endgroup$
  – DrIbraComms
  Mar 27 at 16:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would suggest to solve the particular case first and then the general case. It should work the same way.
  $endgroup$
  – stocha
  Mar 27 at 16:47
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

This is not a complete answer:

Substitution of

$z=frac2 x^2(x-2) (x+2)$

and the settings

$kappa =-frac13 sqrt3$,
$u=-frac23$,
$nu =frac32$ and $mu =-frac12$

leads to the more general form of the integral above:

$I = kappa int_0^1 e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz$.

Now you can figure out the solution. In fact you have a special function (Hypergeometric, Meijer G or H-Fox-function with two variables). Reduction takes place for $u/2->u$ or in the case:

$I =kappa int_0^infty e^u, z z^nu -1 left(1-fracu2 ,zright)^mu -1 , dz=2^nu (-u)^-nu textGamma[nu ], textHypergeometricU[nu,mu +nu ,2]$

EDIT

$e^fracx(x-2),e^fracx(x+2)=e^frac2,x^2(x-2) (x+2)$

share|cite|improve this answer

edited Mar 27 at 16:26

answered Mar 27 at 13:12

stochastocha

31138

$endgroup$

share|cite|improve this answer

edited Mar 27 at 16:26

answered Mar 27 at 13:12

stochastocha

31138

answered Mar 27 at 13:12

stochastocha

31138

answered Mar 27 at 13:12

stochastocha

31138