# Infinite series in integration

How can I find integrals (for example recurring integrals) in terms of infinte sum notation??

Example.. What is the integral of e^(cos x) dx ???
How to find the answer in terms of infinte power series?

Thanks

### Re: Infinite series in integration

Jussi,
may you check out the derivatives of y=e^\cos x? As the general solution of y''+y'+y=0 has the form y=Ae^{s_1x}+Be^{s_2x},
where s_1, s_2 are the roots of s^2+s+1=0.
perucho

### Re: Infinite series in integration

Yes, Perucho, I had wrong y''. It's difficult to use the undetermined coefficients in the power series. Better that one differentiates several times successively:
y' = exp(cos{x})(-sin{x}) = -y sin{x},
y'' = -y'sin{x}-y cos{x},
y''' = -y''sin{x}-2y'cos{x}+y sin{x},
y'''' = -y'''sin{x}-3y''cos{x}+3y'sin{x}+y cos{x},
etc. Then one can determine the values of the derivatives in the origin by substituting x = 0, beginning from that y(0) = e. One obtains the values 0, -e, 0, 4e, ...; these are substituted to the Taylor series. The result (if not errors!) is
y = e-ex^2/2!+4ex^4/4!+...

Jussi

### Re: Infinite series in integration

Yes my dear friend, is a lot of work as it can be seen on
e^{1-x^2/2!+O(x^4)}, and expanding a few terms. This guy must use some powerful math software.
Greetings,
Pedro
PS. I would want make some comments about you interesting entry parallelogram principle''.
I see this principle as a consequence of a closedness of a space vector under sum operation. (lowercase letters will be vectors here, except mass m and \lambda parameter). So,
w=u+v over the obliqueangle triangle ABC, AB=u, BC=v, AC=w, the resultant or sum. But as you well know, one also may use triangle AB'C, AB'=v', B'C=u', thus w=v'+u', with v'=v and u'=u so we getting the geometrical conception of a parallelogram. Newton saw this by taking a particle of mass m and applyng it a force F_1=ma_1 in some direction but simultaneously another force F_2=ma_2 and composing them as F=ma, the net effect.
This principle is independent of the choice of any system of coordinates and the really definition of a vector (apart of its properties as beloging to a space vector) is based in its invariance respect to the choice of such a system, i.e. u'_i=Q_{ij}u_j. From here one can see why it is irrelevant the direction of the zero vector, that is, 0'_i=Q_{ij}0_j=0, with independency about the characterization of the transformation(direction cosines).
Apart of the magnitude and direction characterizing a vector, in mechanics is also useful speaks about the *sense* of a vector, i.e. u=\lambda v, \lambda=\pm 1. But, of course, you may handle the *sense* of -u by using the direction \theta+\pi, where \theta defines the direction of v respect to an arbitrary reference line. Nevertheless is more practical the *sense* concept when we are dealing eigenvectors with, for instance.
Also some authors avoid treating vectors as directed segments when they are discussing *finite* rotations which it are not vectors and so reserving that semantycs for such rotations (see for example Rodrigue's formula).

I think those comments could complement you interesting entry dear friend, but that it is just my opinion.

### Re: Infinite series in integration

Curious, if you differentiate y = exp(cos(x)) two times, you see that y' = -y sin(x) and y'' = -y'-y. Thus y satisfies the linear differential equation
y''+y'+y = 0 (initial conditions y(0) = e, y'(0) = e sin(0) = 0).
This solution y of this equation may be expanded to a power series
y = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...
thus y' = a_1+2a_2x+3a_3x^2+4a_4x^3+...
y'' = 2a_2+6a_3x+12a_4x^2+...
The initial conditions give from the two first series
e = a_0, 0 = a_1.
The series implies
Substitute the three series to the differential equation. Then you can solve a_2, a_3,... and also a recurrence formula for a_n.
I hope that this helps you!
Jussi