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can't evaluate integral of the form $log left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right)$ times a fraction

How to evaluate $I_1=intfracleft(fracay-fracabright)^1/21-fracay mathrm d y$?Evaluate the integral $int_0^infty left(fraclog x arctan xxright)^2 dx$Evaluate $int_0^1 fraclog left( 1+x^2+sqrt3right)1+xmathrm dx$Closed form for $largeint_0^1x,operatornameli!left(frac1xright)ln^1/4!left(frac1xright)dx$Evaluate $int _0^inftydlambda left(lambda ^2 + 2blambda + cright)^-fracepsilon2$Derive $log(a+b)=log(a)-2logleft(cosleft(arctanleft(sqrtfracbaright)right)right)$How to evaluate the integral $int_0^infty e^-frac12left(x^2+ fraccx^2right)dx$?Transformation to $logleft(frac a2cos xright)$Evaluate $int_0^1 fracmathrmdxsqrt1-x^2fracx 1-k^2x^2logleft(frac1-x1+xright)$

1

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$begingroup$

I'm trying to evaluate a bunch of integrals of the form
$$int_0^infty fracsumlimits_i=0^k (c_i P^i)left(P^2+1right)^nlog left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right), dP$$
where $n$ and $k$ is are integers ($nge 1$, $m<2n$), $lambda_+>0$, $lambda_*>0$, and the polynomial $sumlimits_i=0^m(c_i P^i)$ only has even powers of $P$; for example $frac3-2P^2 +47P^4+4P^6(1+P^2)^5$.

For each particular term of the form
$$int_0^infty frac1left(P^2+1right)^klog left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right), dP$$
parital integration $int u ,dv = u v - int v,du$ (with $dv=log(ldots)$) will give me a polynomial of the order $2n-4$ divided by $1/(1+P^2)^n-1$ (which can easily be solved) plus
$$intfracarctan(P)Pleft(1-frac1sqrtlambda _* P^2+lambda _+ P+1right)dP$$
which I don't know how to solve (nor does Mathematica apparently).
In certain instances (for particular set of $c_i$ the (indefinite) integral can be solved by partial integration when the $arctan(x)$ terms cancel out (for example $frac1-16P^2 +25P^4-6P^6(1+P^2)^5$).

The queation is how do I solve the integral when $arctan(x)$ terms don't cancel out? I have tried all the variable substitutions I could think of, and couldn't figure anything out by trying the residue theorem either.

Any help would be greatly appreciated.

calculus integration logarithms

share|cite|improve this question

edited Mar 30 at 2:43

rash

576216

asked Mar 30 at 2:29

BojanBojan

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Most non-rational functions that you can write down do not have a nice closed-form anti-derivative. Since Mathematica cannot come up with one, it is unlikely (though not impossible) that yours is among the few that do. Since you only need the definite integral over $(0,infty)$ you might be able to determine that without needing an antiderivative. If $k le 2n-2$, then a suitably clever use of the Residue theorem might help you (though with one endpoint at $0$, this is doubtful). Probably, you'll have to make due with numerical approximations.
  $endgroup$
  – Paul Sinclair
  Mar 30 at 14:29
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I'm trying to evaluate a bunch of integrals of the form
$$int_0^infty fracsumlimits_i=0^k (c_i P^i)left(P^2+1right)^nlog left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right), dP$$
where $n$ and $k$ is are integers ($nge 1$, $m<2n$), $lambda_+>0$, $lambda_*>0$, and the polynomial $sumlimits_i=0^m(c_i P^i)$ only has even powers of $P$; for example $frac3-2P^2 +47P^4+4P^6(1+P^2)^5$.

For each particular term of the form
$$int_0^infty frac1left(P^2+1right)^klog left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right), dP$$
parital integration $int u ,dv = u v - int v,du$ (with $dv=log(ldots)$) will give me a polynomial of the order $2n-4$ divided by $1/(1+P^2)^n-1$ (which can easily be solved) plus
$$intfracarctan(P)Pleft(1-frac1sqrtlambda _* P^2+lambda _+ P+1right)dP$$
which I don't know how to solve (nor does Mathematica apparently).
In certain instances (for particular set of $c_i$ the (indefinite) integral can be solved by partial integration when the $arctan(x)$ terms cancel out (for example $frac1-16P^2 +25P^4-6P^6(1+P^2)^5$).

The queation is how do I solve the integral when $arctan(x)$ terms don't cancel out? I have tried all the variable substitutions I could think of, and couldn't figure anything out by trying the residue theorem either.

Any help would be greatly appreciated.

calculus integration logarithms

share|cite|improve this question

edited Mar 30 at 2:43

rash

576216

asked Mar 30 at 2:29

BojanBojan

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Most non-rational functions that you can write down do not have a nice closed-form anti-derivative. Since Mathematica cannot come up with one, it is unlikely (though not impossible) that yours is among the few that do. Since you only need the definite integral over $(0,infty)$ you might be able to determine that without needing an antiderivative. If $k le 2n-2$, then a suitably clever use of the Residue theorem might help you (though with one endpoint at $0$, this is doubtful). Probably, you'll have to make due with numerical approximations.
  $endgroup$
  – Paul Sinclair
  Mar 30 at 14:29
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to evaluate a bunch of integrals of the form
$$int_0^infty fracsumlimits_i=0^k (c_i P^i)left(P^2+1right)^nlog left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right), dP$$
where $n$ and $k$ is are integers ($nge 1$, $m<2n$), $lambda_+>0$, $lambda_*>0$, and the polynomial $sumlimits_i=0^m(c_i P^i)$ only has even powers of $P$; for example $frac3-2P^2 +47P^4+4P^6(1+P^2)^5$.

For each particular term of the form
$$int_0^infty frac1left(P^2+1right)^klog left(2+lambda _+ P+ 2 sqrtlambda _* P^2+lambda _+ P+1right), dP$$
parital integration $int u ,dv = u v - int v,du$ (with $dv=log(ldots)$) will give me a polynomial of the order $2n-4$ divided by $1/(1+P^2)^n-1$ (which can easily be solved) plus
$$intfracarctan(P)Pleft(1-frac1sqrtlambda _* P^2+lambda _+ P+1right)dP$$
which I don't know how to solve (nor does Mathematica apparently).
In certain instances (for particular set of $c_i$ the (indefinite) integral can be solved by partial integration when the $arctan(x)$ terms cancel out (for example $frac1-16P^2 +25P^4-6P^6(1+P^2)^5$).

The queation is how do I solve the integral when $arctan(x)$ terms don't cancel out? I have tried all the variable substitutions I could think of, and couldn't figure anything out by trying the residue theorem either.

Any help would be greatly appreciated.

calculus integration logarithms

share|cite|improve this question

edited Mar 30 at 2:43

rash

576216

asked Mar 30 at 2:29

BojanBojan

61

$endgroup$

share|cite|improve this question

edited Mar 30 at 2:43

rash

576216

asked Mar 30 at 2:29

BojanBojan

61

share|cite|improve this question

edited Mar 30 at 2:43

rash

576216

asked Mar 30 at 2:29

BojanBojan

61

edited Mar 30 at 2:43

rash

576216

edited Mar 30 at 2:43

rash

576216

asked Mar 30 at 2:29

BojanBojan

61

asked Mar 30 at 2:29

BojanBojan

61