Can the integral of ##e^{\cos x}dx## be expressed as a unique Function?

[chwala]

Homework Statement:

Find the indefinite integral of $e^{\cos x}dx$

Relevant Equations:

integration

I just came across this and it seems we do not have a definite answer...there are those who have attempted using integration by parts; see link below...i am aware that $\cos x$ has no closed form...same applies to the exponential function.

https://math.stackexchange.com/questions/2468863/what-is-the-integral-of-e-cos-x

would appreciate insight...

cheers!

 

[anuttarasammyak]

y=e^(cos x) is a periodic function as we see in https://www.wolframalpha.com/input?i=graph+of+y＝e^(cos+x) .
$$\int_0^{2\pi} e^{\cos x}dx := 2\pi a \approx 7.95493$$
$$\int_0^{2\pi} (e^{\cos x}-a) dx = 0$$
This modified integral is periodic, i.e.,
$$\int_0^{X_1} (e^{\cos x}-a) dx = \int_0^{X_2} (e^{\cos x}-a) dx$$
where
$$X_2=X_1-2n\pi ,0<X_2<2\pi$$
Any of the modified integral is reduced to the integral in region $[0,2\pi]$. Surely there exists the integral but I do not expect that it can be expressed by ordinary functions.

$$\int_{X_1}^{X_2} e^{\cos x}dx=I(x_2)-I(x_1)+a(X_2-X_1)$$
where
$$x_1=X_1-2n_1\pi, 0<x_1<2\pi$$
$$x_2=X_2-2n_2\pi, 0<x_2<2\pi$$
$$I(x):=\int _0^x (e^{\cos t}-a)dt,\ 0<x<2\pi$$

 

[pasmith]

I will amend the question my bad ; didn't get the English correctly...I meant being expressed as a function.

I think perhaps you are looking for "expressed in terms of elementary functions"; it is trivial, but uninformative, to define $$F(t) = \int_0^t e^{\cos u}\,du.$$

 

[Ssnow]

I think you can try to have a infinite sum expanding by Taylor the exponential:

$\int e^{\cos(x)}dx=\int 1+\cos{x}+\frac{\cos^2{x}}{2!}+\frac{\cos^3{x}}{3!} dx$

now by linearity:

$\int e^{\cos(x)}dx=x+\sin{x}+\int\frac{\cos^2{x}}{2!} dx+\int \frac{\cos^3{x}}{3!} dx + ...$

If you have a closed form for $\int \cos^n{x}dx$ I think you can find an expansion for $\int e^{\cos{x}}dx$.

Ssnow

 

[anuttarasammyak]

I think you can try to have a infinite sum expanding by Taylor the exponential:

∫ecos⁡(x)dx=∫1+cos⁡x+cos2⁡x2!+cos3⁡x3!dx

now by linearity:

∫ecos⁡(x)dx=x+sin⁡x+∫cos2⁡x2!dx+∫cos3⁡x3!dx+...

If you have a closed form for ∫cosn⁡xdx I think you can find an expansion for ∫ecos⁡xdx.

Ssnow

we may do further reduction making use of

from https://socratic.org/questions/how-do-you-find-the-integral-of-cos-n-x .
I am not patient and see it messy.

 

[chwala]

interesting ...how would we attempt to find then the limits of $e^{-\cos x}$ and $e^{cos x}$ as $x$ tends to infinity? my interest is on the approach, i can tell from the graph that the limits tend to $±∞$.

 

[anuttarasammyak]

They are periodical functions as well as cos x is. They have no limits for x=$\pm \infty$.

$$e^{-\cos x}=e^{\cos(x+\pi)}$$

 

[chwala]

They are periodical functions as well as cos x is. They have no limits for x=$\pm \infty$.View attachment 322205

$$e^{-\cos x}=e^{\cos(x+\pi)}$$
View attachment 322206

I need to look at this, can we say that it is bounded ...supremum, infimum? need to check on this man!

 

[anuttarasammyak]

$$e^{-1} \le e^{\cos x} \le e$$

 

[chwala]

$$e^{-1} \le e^{\cos x} \le e$$

Nice, yes, If $y=\cos x$ then we have the maximum at $y=1$ and minimum at $y=-1$... then how comes that this function has no limit? given
$\lim_{x \rightarrow +\infty} {e^{cos x}}$

am i missing something here...

 

[chwala]

They are periodical functions as well as cos x is. They have no limits for x=$\pm \infty$.View attachment 322205

$$e^{-\cos x}=e^{\cos(x+\pi)}$$
View attachment 322206

No limits or undefined?
...just seen this explanation;

https://socratic.org/questions/what...es-infinity-of-cosx#:~:text=There is no limit.

i guess i need to read more on this area ...

 

[dextercioby]

By "no limits", we usually mean "it diverges". So what is the limit of sin x when tends to infinity? It does not exist. We say sin x has no limit to infinity, or that at infinity sin x is undefined. Just a matter of wording, mathematicians once they write a formula they know exactly what it means, irrespective of which exact words are suited best to describe the formula.

"So a function/sequence/series does not converge to a limit, i.e. it diverges, or its limit does not exist". This should be clear. Let us not turn mathematics into semantics, a discipline of linguistics.