can this be written as an elementary function? if so, what is the elementary function?

Question

integrate (cos(x)+1)/x^5 from 10 to infinity
Can it be written as an elementary function?
If so, what is the elementry function?
My CAS can't figure it out.
I got the CAS for free...you get what you pay for.
Convergence is obvious but i am uncertain about it's indefinite integral.
I have the feeling as though it cannot be written as an elementary function.

Thanks

Answer 1

There are 2 fractions in it: cos(x)/x^5 and 1/x^5. The latter is easy to integrate. Try integration by parts for the first one.
Let u=cos(x), dv/dx = 1/x^5, thus, du/dx=-sin(x) and v=-1/4x^4.
int(cos(x)/x^5) = -cos(x)/4x^4 - int(sin(x)/4x^4)
Again, let u=sin(x), dv/dx = 1/4x^4, thus, du/dx=cos(x) and v=-1/12x^3, so
int(cos(x)/x^5) = -cos(x)/4x^4 - int(sin(x)/4x^4)
int(cos(x)/x^5) = -cos(x)/4x^4 - [-sin(x)/12x^3 - int(-cos(x)/12x^3)]
int(cos(x)/x^5) = -cos(x)/4x^4 + sin(x)/12x^3 - int(cos(x)/12x^3)
Similarly,
int(cos(x)/12x^3) = 1/12 int(cos(x)/x^3)
= 1/12 * [-cos(x)/2x^2 + sin(x)/2x - int(cos(x)/2x)]

In a way, you are right because:

From MatLab: int(cos(x)/x^5) = -1/4*cos(x)/x^4+1/12*sin(x)/x^3+1/24*cos(x)/x^2-1/24*sin(x)/x+1/24*cosint(x)
where cosint(x) = Euler's constant + ln(x) + int_0_x((cos(t)-1)/t)dt

With this, we should have separated it into (cos(x)-1)/x * 1/x^4.

And fyi, the answer is:
-47/120000*cos(10) - 1/24*cosint(10) - 1/40000 + 49/12000*sin(10) = -2.3765e-005

Answer 2

Hi. I tried to write the indefinite integral but failed. So I used Matlab. The answer is

-1/4*cos(x)/x^4
+1/12*sin(x)/x^3
+1/24*cos(x)/x^2
-1/24*sin(x)/x
+1/24*cosint(x)
-1/4/x^4

You note that there is a "cosint" involved. Because the cosine integral function is non-elementary, I think the integral can't be written as an elementary function.

Answer 3

No, it can't be written in terms of elementary functions.

Repeated integration by parts will get you

-(cos x + 1)/(4x^4) + (sin x)/(12x^3) + (cos x)/(24x^2) + (sin x)/(24x) -(1/24)∫(10 to ∞) (cos x dx) / x

Some information about this last integral is available at the source below. Using Mathematica's evaluator gives Ci(10) as -0.04545643300 to 10 s.f., so we'd get
0.04545643300 / 24 = 0.001894018042 for the integral term. But no closed form. ;-)

Answer 4

this would sound like an dazzling answer, yet i'd say that the respond is in simple terms approximately arbitrary. evaluate probably uncomplicated results of antiderivatives: cos(x) e^(x) ln(x) yet are those quite 'uncomplicated'? Are they hassle-free? i'd say the respond is sure -using fact- we've names for those applications and make contact with them hassle-free. there isn't any algebraic thank you to calculate those applications; they are all trascendental applications. So what makes them "hassle-free"? We provide them names, and that does. To calculate e^x for extremely nearly all values of x, you are able to evaluate a limiteless sequence or use a given approximation of e^x. the activity of comparing an "hassle-free" quintessential is probably no longer any less demanding than calculating an approximation of the quintessential at as quickly as. If one defines a function using fact the tip results of an quintessential (as is in many situations performed), then that would desire to be seen an hassle-free function. i do no longer think of that maximum would evaluate it that in spite of the incontrovertible fact that, whether it takes as plenty processing to evaluate it as many different applications do. which you will say that in case you "do no longer comprehend" a function to represent it, then the tip result isn't "hassle-free" to you. in case you do comprehend a function to calculate it, then it would desire to or is probably no longer hassle-free, reckoning on the way you sense approximately it. e.g. The gamma function (the single on the topic of factorials) is defined in terms of an quintessential, yet is it hassle-free or no longer?