Why won't mathematica evaluate the integral I've defined?

Question

I defined

Integrate[Exp[p_. Cos[x_] + q_. Sin[x_]]*Sin[a_. Cos[x_] + b_. Sin[x_] - m_. x_],
{x_, 0, 2*Pi}] := Sqrt[-1]*Pi*((b - p)^2 + (a + q)^2)^(-m/2)*(((p^2 - q^2 + a^2 - b^2) 
+ Sqrt[-1]*(2*(p*q + a*b)))^(m/2)*BesselI[m, Sqrt[(p^2 + q^2 - a^2 - b^2) -     
Sqrt[-1]*(2 (a*p + b*q))]] - ((p^2 - q^2 + a^2 - b^2) -
Sqrt[-1]*(2 (p*q + a*b)))^(m/2)*BesselI[m, Sqrt[(p^2 + q^2 - a^2 - b^2) +       
Sqrt[-1]*(2 (a*p + b*q))]])

In accordance with the examples given in the mathematica documentation. However, the integral only evaluates for the exact symbolic values p,q,a,b,m. If I try to evaluate any other form of the integral, e.g.

Integrate[Exp[Cos[x]+Sin[x]]*Sin[Sin[x]+Cos[x]-x],{x,0,2*Pi}]

Mathematica is unable to evaluate the integral. Why doesn't the integral, the way I've defined it, evaluate for any given values of p,q,a,b,m?

Answer 1

One problem is that Exp[y] evaluates to Power[E, y], so that the integral does not match the (held) pattern with Exp. Another is that other functions sometimes evaluate to other forms, such as Sin:

Sin[Sin[x] + Cos[x] - x]
(* -Sin[x - Cos[x] - Sin[x]] *)

Here is a fix that works on the example. I added a constant factor c_ to take care of the -1 factoring out of Sin[], and changed -m_ to +m_, mutatis mutandis. Other examples may simplify in other ways, so some testing may be necessary to cover all possible use-cases.

Internal`InheritedBlock[{Integrate},
 Unprotect[Integrate];
 Integrate[
   c_. Power[E, p_. Cos[x_] + q_. Sin[x_]]*
    Sin[a_. Cos[x_] + b_. Sin[x_] + m_. x_], {x_, 0, 2*Pi}] := 
  c (Sqrt[-1]*
     Pi*((b - p)^2 + (a + q)^2)^(m/2) * 
       (((p^2 - q^2 + a^2 - b^2) + Sqrt[-1]*(2*(p*q + a*b)))^(-m/2) * 
        BesselI[-m, Sqrt[(p^2 + q^2 - a^2 - b^2) - Sqrt[-1]*(2 (a*p + b*q))]] - 
      ((p^2 - q^2 + a^2 - b^2) - Sqrt[-1]*(2 (p*q + a*b)))^(-m/2)*
        BesselI[-m, Sqrt[(p^2 + q^2 - a^2 - b^2) + Sqrt[-1]*(2 (a*p + b*q))]]));
 Protect[Integrate];
 Integrate[Exp[Cos[x] + Sin[x]]*Sin[Sin[x] + Cos[x] - x], {x, 0, 2*Pi}]
 ]
(*
 -2 I π (-(1/2) (-1)^(3/4) BesselI[1, 2 (-1)^(1/4)] + 
    1/2 (-1)^(1/4) BesselI[1, 2 (-1)^(3/4)])
*)

See here for Internal`InheritedBlock.

Answer 2

Michael E2 and Bob have solved the integral. Here is an alternative method which might be of interest as well. I have used the similar method already in How to solve this integration?.

We solve the integral transforming it into a complex contour integral which, after a simple binomial expansion, can easily be soved by the Cauchy theorem. The remaining infinte sum can be expressed by a hypergeometric function which reduces to a modified Bessel function of the first kind.

Let

f := Exp[Cos[x] + Sin[x]]*Sin[Sin[x] + Cos[x] - x]
g := Integrate[f, {x, 0, 2 \[Pi]}]

First we shall use the exponential form of the Sin[] outside the Exp[] function in f (in the end we take the imaginary part)

Sin[Sin[x] + Cos[x] - x] == Im[Exp[I (Sin[x] + Cos[x] - x)]];
ComplexExpand[%]

(* Out[812]= True *)

so that the argument of the Exp[] function becomes

Cos[x] + Sin[x] + I (Cos[x] + Sin[x] - x)

which can be written as

Exp[I x] + I Exp[-I x] - I x

The integral g then becomes

g := Integrate[Exp[Exp[I x] + I Exp[-I x] - I x], {x, 0, 2 \[Pi]}]

Now with the substitution

{z -> Exp[-I x], dz = -I z dx};

g simplifies to a contour integral around the origin

h = I  Integrate[Exp[1/z + I z], {z, 1, I, -1, -I, 1}]

(* Out[814]= Integrate[E^(1/z + I z), {z, 1, I, -1, -I, 1}] *)

As this integral is returned unevaluated we expand the Exp[] function in a power series.

A typical term can in turn be expanded into a binomial sum

(1/z + I z)^n == Sum[Binomial[n, k] z^-k (I z)^(n - k), {k, 0, n}] == 
  Sum[Binomial[n, k] z^(n - 2 k) I^(n - k), {k, 0, n}];

Now, by the Cauchy theorem, the contour integral of integer powers of z is only different from zero for the first negative power. Examples are

Table[{m, Integrate[z^m, {z, 1, I, -1, -I, 1}]}, {m, -2, 1}]

(* Out[820]= {{-2, 0}, {-1, 2 I \[Pi]}, {0, 0}, {1, 0}} *)

Hence we have the condition

Reduce[n - 2 k == -1, Integers ] /. C[1] -> m

(* Out[827]= m \[Element] Integers && k == m && n == -1 + 2 m *)

Taking into account the factor 2 \[Pi] I from the integration and the 1/n! from the Exp[] function we arrive at the sum which is immediatey evaluated by Mathematica:

h1 = 2 \[Pi] Sum[
   1/(2 m - 1)! Binomial[2 m - 1, m] I^(m - 1), {m, 0, \[Infinity]}]

(* Out[843]= -2 (-1)^(3/4) \[Pi] BesselI[1, 2 (-1)^(1/4)] *)

The original integral is the imaginary part of h1.

Numerically

h1 // N

(* Out[868]= 5.76177 + 3.09803 I *)

Another form of h1 is obtained by expressing (-1) in exponential form

h1 /. (-1)^c_ -> Exp[I \[Pi] c]

(* Out[865]= -2 E^((3 I \[Pi])/4) \[Pi] BesselI[1, 2 E^((I \[Pi])/4)] *)

Which can be simplified to

FullSimplify[%]

(* Out[866]= 2 \[Pi] Hypergeometric0F1Regularized[2, I] *)

% // N

(* Out[867]= 5.76177 + 3.09803 I *)

Answer 3

Rather than define a function, you can define a replacement rule using Hold and RuleDelayed

rule = Hold[Integrate[
     Exp[p_. Cos[x_] + q_. Sin[x_]]*
      Sin[a_. Cos[x_] + b_. Sin[x_] - m_. x_], {x_, 0, 2 Pi}]] :>
   I*Pi*((b - p)^2 + (a + q)^2)^(-m/
       2)*(((p^2 - q^2 + a^2 - b^2) + I*(2*(p*q + a*b)))^(m/2)*
       BesselI[m, 
        Sqrt[(p^2 + q^2 - a^2 - b^2) - 
          I*(2 (a*p + b*q))]] - ((p^2 - q^2 + a^2 - b^2) - 
          I*(2 (p*q + a*b)))^(m/2)*
       BesselI[m, Sqrt[(p^2 + q^2 - a^2 - b^2) + I*(2 (a*p + b*q))]]);

Hold[Integrate[
    Exp[Cos[x] + Sin[x]]*Sin[Sin[x] + Cos[x] - x], {x, 0, 2 Pi}]] /. 
  rule // Simplify

(*  (-(-1)^(1/4))*Pi*
   (BesselI[1, 2*(-1)^(1/4)] + 
      I*BesselI[1, 2*(-1)^(3/4)])  *)

With inexact numerical values and using Chop to remove the negligible imaginary artifact

% // N // Chop

(*  3.09803  *)

Verifying with NIntegrate

NIntegrate[Exp[Cos[x] + Sin[x]]*Sin[Sin[x] + Cos[x] - x], {x, 0, 2 Pi}]

(*  3.09803  *)