integration
RG
I have been trying to simplify(integrate) the following function, but
M6 seems to give a complex answer which i cann't understand.. please
help.
x[s_]=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[
\*FractionBox[\(r\ t\ \((\(-\[Kappa]0\) + \[Kappa]1 + r\ \[Kappa]1)\)
+ \((1 + r)\)\ S\ \((\[Kappa]0 - \[Kappa]1)\)\ \((\(-Log[S]\) + Log[S
+ r\ t])\)\),
SuperscriptBox[\(r\), \(2\)]]] \[DifferentialD]t\)\)
Regards,
RG
Jean-Marc Gulliet
> First, notice that if we use the *InputForm* of the above expression, we
> can easily add assumptions on the parameters of the integral (or we
> could use *Assuming*), for instance that S, r, and s are real and r != 0
> or s > 0.
>
> However, it seems that the above integral has no solution if the
> parameter S is positive. On the other hand, ff we allow S to be negative
> (or complex) then the integral has a symbolic complex solution.
>
> In[49]:= Integrate[
> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
> S*(\[Kappa]0 - \[Kappa]1)*
> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
> Assumptions -> S > 0]
>
> Out[49]= Integrate[
> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2], {t,
> 0, s}, Assumptions -> S > 0]
>
> In[46]:= Integrate[
> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
> S*(\[Kappa]0 - \[Kappa]1)*
> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
> Assumptions -> {Element[{S, r, s}, Reals], r != 0, s > 0}]
>
> Out[46]= If[r S > 0 || s + S/r <= 0, (1/(
> 2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1)))
> r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>
> [... output partially deleted ...]
>
> r^2)] Sin[(S \[Kappa]0)/r^2 - (S \[Kappa]1)/r^2 - (
> S \[Kappa]1)/r]),
> Integrate[
> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/
> r^2], {t, 0, s},
> Assumptions ->
> r != 0 && s > 0 && r S <= 0 && r (r s + S) > 0 &&
> S \[Element] Reals]]
>
>
> You can manipulate further the integral thanks to *FullSimplify* and
> some assumptions on the parameters.
>
> Assuming[r S > 0 || s + S/r <= 0,
> FullSimplify[
> 1/(2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1))
> r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>
> [... input partially deleted ...]
>
> S \[Kappa]1)/r])]]
It took a long time, but the last expression returned the following
result (which is valid only for r S > 0 || s + S/r <= 0):
(1/(2 r))E^(-((I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/
r^2)) (S ExpIntegralE[-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2), -((I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2)] -
S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(
1 + (I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)
ExpIntegralE[-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2), -((
I (r s + S) (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2)] +
E^((2 I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/
r^2) (S ExpIntegralE[(I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2, (
I S (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2] - (S^3)^((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2) (r s + S)^(
1 - (I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2) ((\[Kappa]0 - (1 + r) \[Kappa]1)^2)^((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2) (S^2 (r s + S)^2 (\[Kappa]0 - (1 + r) \[Kappa]1)^2)^(-((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/
r^2)) ((r s + S) (S + r Conjugate[s]))^((
I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)
ExpIntegralE[(I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2, (
I (r s + S) (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1))/r^2]))
Regards,
-- Jean-Marc
sjoerd.c...@gmail.com
Which part of the result do you have difficulties with? The result is
an expression which is true if certain conditions (listed in the first
part of the output) are true.
The expression is indeed complex, but there isn't a law against the
existence of complex expressions. Nor is it the case that they can
always be simplified. Have you tried some values for the various
constants in the expression to see if the result make any sense?
Cheers -- Sjoerd
Alexei Boulbitch
I have been trying to simplify(integrate) the following function, but
M6 seems to give a complex answer which i cann't understand.. please
help.
x[s_]=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[
\*FractionBox[\(r\ t\ \((\(-\[Kappa]0\) + \[Kappa]1 + r\ \[Kappa]1)\)
+ \((1 + r)\)\ S\ \((\[Kappa]0 - \[Kappa]1)\)\ \((\(-Log[S]\) + Log[S
+ r\ t])\)\),
SuperscriptBox[\(r\), \(2\)]]] \[DifferentialD]t\)\)
Regards,
RG
Hi, RG,
your problem seems to be in Mathematics, rather than in Mathematica. The integral you need (as it is now) is too cumbersome.
Make the integrand simpler by hiding and simplifying your notations, and the result will become more understandable.
For example, below is your integral along with the result in which I denoted some terms by a, b and c to make it shorter:
In[2]:= \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[\
t\ a + b + c*Log[S + r\ t]] \[DifferentialD]t\)\)
Out[2]= If[(Re[S/(r s)] >= 0 && S/(r s) != 0) || Re[S/(r s)] <= -1 ||
Im[S/(r s)] != 0, -(1/(2 a))
S^(-\[ImaginaryI] c) (-((\[ImaginaryI] a S)/
r))^(-\[ImaginaryI] c) (S^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a S)/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 S^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a S)/
r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])) + (1/(
2 a))(r s + S)^(-\[ImaginaryI] c) (-((\[ImaginaryI] a (r s + S))/
r))^(-\[ImaginaryI] c) ((r s + S)^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 (r s + S)^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/
r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])),
Integrate[Cos[b + a t + c Log[t r + S]], {t, 0, s},
Assumptions -> ! ((Re[S/(r s)] >= 0 && S/(r s) != 0) ||
Re[S/(r s)] <= -1 || Im[S/(r s)] != 0)]]
This is your result after simplification:
In[3]:= -(1/(2 a))
S^(-\[ImaginaryI] c) (-((\[ImaginaryI] a S)/
r))^(-\[ImaginaryI] c) (S^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a S)/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 S^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a S)/
r] (\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r])) +
1/(2 a)(r s + S)^(-\[ImaginaryI] c) (-((\[ImaginaryI] a (r s + S))/
r))^(-\[ImaginaryI] c) ((r s + S)^(2 \[ImaginaryI] c)
Gamma[1 + \[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
r)] (-\[ImaginaryI] Cos[b - (a S)/r] + Sin[b - (a S)/r]) + ((
a^2 (r s + S)^2)/r^2)^(\[ImaginaryI] c)
Gamma[1 - \[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/
r] (\[ImaginaryI] Cos[b - (a S)/r] +
Sin[b - (a S)/r])) // FullSimplify
Out[3]= (1/(2 r))\[ExponentialE]^(-((\[ImaginaryI] (b r + a S))/
r)) (\[ExponentialE]^(2 \[ImaginaryI] b) S^(1 + \[ImaginaryI] c)
ExpIntegralE[-\[ImaginaryI] c, -((\[ImaginaryI] a S)/
r)] - \[ExponentialE]^(2 \[ImaginaryI] b) (r s + S)^(
1 + \[ImaginaryI] c)
ExpIntegralE[-\[ImaginaryI] c, -((\[ImaginaryI] a (r s + S))/
r)] + \[ExponentialE]^((2 \[ImaginaryI] a S)/
r) (S^(1 - \[ImaginaryI] c)
ExpIntegralE[\[ImaginaryI] c, (\[ImaginaryI] a S)/
r] - (r s + S)^(1 - \[ImaginaryI] c)
ExpIntegralE[\[ImaginaryI] c, (\[ImaginaryI] a (r s + S))/r]))
It is already comprehensible. Probably you may go still further by checking properties of the integral exponents.
Have a look into the book: Abramowitz, M. & Stegun, I. A. Handbook of Mathematical Functions with formulas,
graphs and mathematical tables. (National Bureau of Standards, 1964). Another way I would go is to try to transform it first,
and to integrate afterwards.
Success, Alexei
--
Alexei Boulbitch, Dr., Habil.
Senior Scientist
IEE S.A.
ZAE Weiergewan
11, rue Edmond Reuter
L-5326 Contern
Luxembourg
Phone: +352 2454 2566
Fax: +352 2454 3566
Website: www.iee.lu
This e-mail may contain trade secrets or privileged, undisclosed or otherwise confidential information. If you are not the intended recipient and have received this e-mail in error, you are hereby notified that any review, copying or distribution of it is strictly prohibited. Please inform us immediately and destroy the original transmittal from your system. Thank you for your co-operation.
Jean-Marc Gulliet
First, notice that if we use the *InputForm* of the above expression, we
r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/
r^2], {t, 0, s},
Assumptions ->
r != 0 && s > 0 && r S <= 0 && r (r s + S) > 0 &&
S \[Element] Reals]]
You can manipulate further the integral thanks to *FullSimplify* and
some assumptions on the parameters.
Assuming[r S > 0 || s + S/r <= 0,
FullSimplify[
1/(2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1))
r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
[... input partially deleted ...]
S \[Kappa]1)/r])]]
HTH,
-- Jean-Marc
RG
wrote:
> RG wrote:
> > I have been trying to simplify(integrate) the following function, but
> > M6 seems to give a complex answer which i cann't understand.. please
> > help.
>
> > x[s_]=\!\(
> > \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(s\)]\(Cos[
> > \*FractionBox[\(r\ t\ \((\(-\[Kappa]0\) + \[Kappa]1 + r\ \[Kappa]1)\)
> > + \((1 + r)\)\ S\ \((\[Kappa]0 - \[Kappa]1)\)\ \((\(-Log[S]\) + Log[S
> > + r\ t])\)\),
> > SuperscriptBox[\(r\), \(2\)]]] \[DifferentialD]t\)\)
>
> First, notice that if we use the *InputForm* of the above expression, we
> can easily add assumptions on the parameters of the integral (or we
0
> or s > 0.
>
> However, it seems that the above integral has no solution if the
> parameter S is positive. On the other hand, ff we allow S to be negative
> (or complex) then the integral has a symbolic complex solution.
>
> In[49]:= Integrate[
> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
> S*(\[Kappa]0 - \[Kappa]1)*
> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
> Assumptions -> S > 0]
>
> Out[49]= Integrate[
> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
Dear All, i think M6 would be able to give a simplified answer, which
is more understandable without the appearance of imaginary numbers in
the answer. The assumption of the integral should be:
[1]{k1,k2,r,s,S} are real numbers
[2] r > -1
[3] S > 0
[4] 0<= s<= S
I tried doing with these assumption, but the imaginary part still
exists.. Is there anyway to ask M6 to give the right assumption for
imaginary-free answer? Thank you very much sirs...
Jean-Marc Gulliet
> Greetings...
> Thanks Jean-Marc Gulliet,
> i think M6 would be able to give a simplified answer, which is more
> understandable without the appearance of imaginary numbers in the answer.
> The assumption of the integral should be:
> [1]{k1,k2,r,s,S} are real numbers
> [2] r > -1
> [3] S > 0
> [4] 0<= s<= S
> I tried doing with these assumption, but the imaginary part still exists..
> Is there anyway to ask M6 to give the right assumption for imaginary-free
>
> Gobithaasan
Please, could you post the expression you used and its result. On my
system, using the above assumptions, Mathematica returns the integral
unevaluated, which is in agreement with what I already noticed when S
> 0:
>>> [...] However, it seems that the above integral has no solution if the
>>> parameter S is positive. On the other hand, ff we allow S to be negative (or
>>> complex) then the integral has a symbolic complex solution.
>>>
>>> In[49]:= Integrate[
>>> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
>>> S*(\[Kappa]0 - \[Kappa]1)*
>>> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
>>> Assumptions -> S > 0]
>>>
>>> Out[49]= Integrate[
>>> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
>>> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2], {t,
>>> 0, s}, Assumptions -> S > 0]
Regards,
-- Jean-Marc
Gobithaasan
Thanks Jean-Marc Gulliet,
I think Mathematica 6 would be able to give a simplified answer,
which is more understandable without the appearance of imaginary
numbers in the answer. The assumption of the integral should be:
[1]{k1,k2,r,s,S} are real numbers
[2] r > -1
[3] S > 0
[4] 0<= s<= S
I tried doing with these assumption, but the imaginary part still
Gobithaasan
Jean-Marc Gulliet wrote:
> Jean-Marc Gulliet wrote:
>
>> RG wrote:
>>
>> First, notice that if we use the *InputForm* of the above expression,
>> we can easily add assumptions on the parameters of the integral (or
>> we could use *Assuming*), for instance that S, r, and s are real and
>> r != 0 or s > 0.
>>
>> However, it seems that the above integral has no solution if the
>> parameter S is positive. On the other hand, ff we allow S to be
>> negative (or complex) then the integral has a symbolic complex solution.
>>
>> In[49]:= Integrate[
>> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
>> S*(\[Kappa]0 - \[Kappa]1)*
>> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
>> Assumptions -> S > 0]
>>
>> Out[49]= Integrate[
>> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
>> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2], {t,
>> 0, s}, Assumptions -> S > 0]
>>
>> In[46]:= Integrate[
>> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
>> S*(\[Kappa]0 - \[Kappa]1)*
>> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
>> Assumptions -> {Element[{S, r, s}, Reals], r != 0, s > 0}]
>>
>> Out[46]= If[r S > 0 || s + S/r <= 0, (1/(
>> 2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1)))
>> r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>>
>> [... output partially deleted ...]
>>
>> r^2)] Sin[(S \[Kappa]0)/r^2 - (S \[Kappa]1)/r^2 - (
>> S \[Kappa]1)/r]),
>> Integrate[
>> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
>> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/
>> r^2], {t, 0, s},
>> Assumptions ->
>> r != 0 && s > 0 && r S <= 0 && r (r s + S) > 0 &&
>> S \[Element] Reals]]
>>
>>
>> You can manipulate further the integral thanks to *FullSimplify* and
>> some assumptions on the parameters.
>>
>> Assuming[r S > 0 || s + S/r <= 0,
>> FullSimplify[
>> 1/(2 (\[Kappa]0 - \[Kappa]1 - r \[Kappa]1))
>> r S^(-((I (1 + r) S (\[Kappa]0 - \[Kappa]1))/r^2)) (r s + S)^(-((
>>
>> [... input partially deleted ...]
>>
>> S \[Kappa]1)/r])]]
>
Jean-Marc Gulliet
<jeanmarc...@gmail.com> wrote:
> On Wed, Oct 15, 2008 at 5:19 AM, Gobithaasan <gobit...@gmail.com> wrote:
>> Greetings...
>> Thanks Jean-Marc Gulliet,
>> understandable without the appearance of imaginary numbers in the answer.
>> The assumption of the integral should be:
>> [1]{k1,k2,r,s,S} are real numbers
>> [2] r > -1
>> [3] S > 0
>> [4] 0<= s<= S
>> I tried doing with these assumption, but the imaginary part still exists..
>> answer? Thank you very much Jean...
>>
>> Gobithaasan
>
> system, using the above assumptions, Mathematica returns the integral
> unevaluated, which is in agreement with what I already noticed when S
>> 0:
>
>>>> parameter S is positive. On the other hand, ff we allow S to be negative (or
>>>> complex) then the integral has a symbolic complex solution.
>>>>
>>>> In[49]:= Integrate[
>>>> Cos[(r*t*(-\[Kappa]0 + \[Kappa]1 + r*\[Kappa]1) + (1 + r)*
>>>> S*(\[Kappa]0 - \[Kappa]1)*
>>>> (-Log[S] + Log[S + r*t]))/r^2], {t, 0, s},
>>>> Assumptions -> S > 0]
>>>>
>>>> Out[49]= Integrate[
>>>> Cos[(r t (-\[Kappa]0 + \[Kappa]1 + r \[Kappa]1) + (1 +
>>>> r) S (\[Kappa]0 - \[Kappa]1) (-Log[S] + Log[S + r t]))/r^2], {t,
>>>> 0, s}, Assumptions -> S > 0]
In addition, do not forget that a symbolic expression with imaginary
parts does not necessarily yields complex values (depending, of
course, on the choice of parameter values). For instance,
In[10]:= sol =
Integrate[
Cos[(r t (-k0 + k1 + r k1) + (1 + r) S (k0 - k1) (-Log[S] +
Log[S + r t]))/r^2], {t, 0, s},
Assumptions -> {Im[k0] == 0, Im[k1] == 0, Im[r] == 0, Im[s] == 0,
Im[S] == 0, r > 0, s > 0}]
Out[10]= If[S > 0 || r s + S <= 0, (1/(2 (k0 - k1 - k1 r)))
r^(1 - (2 I k0 S)/r^2 + (2 I k1 S)/r^2 - (2 I k0 S)/r + (2 I k1 S)/r)
S^(-((I (k0 - k1) (1 + r) S)/
r^2)) (-((I (-k0 + k1 + k1 r) S)/r^2))^(-((I (k0 - k1) (1 + r) S)/
r^2)) (r s + S)^(-((I (k0 - k1) (1 + r) S)/
r^2)) (I (k0 - k1 (1 + r)) (r s + S))^(-((I (k0 - k1) (1 + r) S)/
r^2)) (-I r^((2 I (k0 - k1) (1 + r) S)/r^2) S^((
I (k0 - k1) (1 + r) S)/r^2) (r s + S)^((I (k0 - k1) (1 + r) S)/
r^2) (I (k0 - k1 (1 + r)) (r s + S))^((I (k0 - k1) (1 + r) S)/
r^2) Cos[(k0 S)/r^2 - (k1 S)/r^2 - (k1 S)/r] Gamma[(
I (-I r^2 + (k0 - k1) S + (k0 - k1) r S))/
r^2, -((I (-k0 + k1 + k1 r) S)/r^2)] +
I r^((4 I (k0 - k1) (1 + r) S)/
r^2) (-((I (-k0 + k1 + k1 r) S)/r^2))^((I (k0 - k1) (1 + r) S)/
r^2) (r s + S)^((2 I (k0 - k1) (1 + r) S)/r^2)
Cos[(k0 S)/r^2 - (k1 S)/r^2 - (k1 S)/r] Gamma[(
I (-I r^2 + (k0 - k1) S + (k0 - k1) r S))/
r^2, -((I (-k0 + k1 + k1 r) (r s + S))/r^2)] +
I r^((2 I (k0 - k1) (1 + r) S)/r^2) S^((I (k0 - k1) (1 + r) S)/
r^2) (((-k0 + k1 + k1 r)^2 S^2)/r^4)^((I (k0 - k1) (1 + r) S)/
r^2) (r s + S)^((I (k0 - k1) (1 + r) S)/
r^2) (I (k0 - k1 (1 + r)) (r s + S))^((I (k0 - k1) (1 + r) S)/
r^2) Cos[(k0 S)/r^2 - (k1 S)/r^2 - (k1 S)/r] Gamma[-((
I (I r^2 + (k0 - k1) S + (k0 - k1) r S))/r^2), (
I (-k0 + k1 + k1 r) S)/r^2] -
I S^((2 I (k0 - k1) (1 + r) S)/
r^2) (-((I (-k0 + k1 + k1 r) S)/r^2))^((I (k0 - k1) (1 + r) S)/
r^2) Abs[-k0 + k1 + k1 r]^((2 I (k0 - k1) (1 + r) S)/r^2)
Abs[r s + S]^((2 I (k0 - k1) (1 + r) S)/r^2)
Cos[(k0 S)/r^2 - (k1 S)/r^2 - (k1 S)/r] Gamma[-((
I (I r^2 + (k0 - k1) S + (k0 - k1) r S))/r^2), (
I (-k0 + k1 + k1 r) (r s + S))/r^2] +
r^((2 I (k0 - k1) (1 + r) S)/r^2) S^((I (k0 - k1) (1 + r) S)/
r^2) (r s + S)^((I (k0 - k1) (1 + r) S)/
r^2) (I (k0 - k1 (1 + r)) (r s + S))^((I (k0 - k1) (1 + r) S)/
r^2) Gamma[(I (-I r^2 + (k0 - k1) S + (k0 - k1) r S))/
r^2, -((I (-k0 + k1 + k1 r) S)/r^2)] Sin[(k0 S)/r^2 - (k1 S)/
r^2 - (k1 S)/r] -
r^((4 I (k0 - k1) (1 + r) S)/
r^2) (-((I (-k0 + k1 + k1 r) S)/r^2))^((I (k0 - k1) (1 + r) S)/
r^2) (r s + S)^((2 I (k0 - k1) (1 + r) S)/r^2)
Gamma[(I (-I r^2 + (k0 - k1) S + (k0 - k1) r S))/
r^2, -((I (-k0 + k1 + k1 r) (r s + S))/r^2)] Sin[(k0 S)/r^2 - (
k1 S)/r^2 - (k1 S)/r] +
r^((2 I (k0 - k1) (1 + r) S)/r^2) S^((I (k0 - k1) (1 + r) S)/
r^2) (((-k0 + k1 + k1 r)^2 S^2)/r^4)^((I (k0 - k1) (1 + r) S)/
r^2) (r s + S)^((I (k0 - k1) (1 + r) S)/
r^2) (I (k0 - k1 (1 + r)) (r s + S))^((I (k0 - k1) (1 + r) S)/
r^2) Gamma[-((I (I r^2 + (k0 - k1) S + (k0 - k1) r S))/r^2), (
I (-k0 + k1 + k1 r) S)/
r^2] Sin[(k0 S)/r^2 - (k1 S)/r^2 - (k1 S)/r] -
S^((2 I (k0 - k1) (1 + r) S)/
r^2) (-((I (-k0 + k1 + k1 r) S)/r^2))^((I (k0 - k1) (1 + r) S)/
r^2) Abs[-k0 + k1 + k1 r]^((2 I (k0 - k1) (1 + r) S)/r^2)
Abs[r s + S]^((2 I (k0 - k1) (1 + r) S)/r^2)
Gamma[-((I (I r^2 + (k0 - k1) S + (k0 - k1) r S))/r^2), (
I (-k0 + k1 + k1 r) (r s + S))/
r^2] Sin[(k0 S)/r^2 - (k1 S)/r^2 - (k1 S)/r]),
Integrate[
Cos[(r (-k0 + k1 + k1 r) t + (k0 - k1) (1 + r) S (-Log[S] +
Log[S + r t]))/r^2], {t, 0, s},
Assumptions ->
k0 \[Element] Reals && k1 \[Element] Reals && S <= 0 && s > 0 &&
r > -(S/s)]]
In[11]:= sol /. {S -> 1, k0 -> 0, k1 -> 1, r -> 1, s -> 1}
Out[11]= -(-I)^(4 I) 4^(-1 +
4 I) (-I (-8 I)^(-2 I) Cos[2] Gamma[1 - 2 I, -2 I] +
I (-I)^(-2 I) 2^(-6 I) Cos[2] Gamma[1 - 2 I, -4 I] +
I (-32 I)^(-2 I) Cos[2] Gamma[1 + 2 I, 2 I] -
I (-I)^(-2 I) 2^(-10 I) Cos[2] Gamma[1 + 2 I, 4 I] - (-8 I)^(-2 I)
Gamma[1 - 2 I, -2 I] Sin[2] + (-I)^(-2 I) 2^(-6 I)
Gamma[1 - 2 I, -4 I] Sin[2] - (-32 I)^(-2 I)
Gamma[1 + 2 I, 2 I] Sin[2] + (-I)^(-2 I) 2^(-10 I)
Gamma[1 + 2 I, 4 I] Sin[2])
In[12]:= % // N
Out[12]= 0.957467+ 0. I
In[13]:= % // Chop
Out[13]= 0.957467
Regards,
-- Jean-Marc