How to find $\int e^{2\cos 2x}dx$?

Question

I have found a function. I'm unable to find its integral. I have searched for it in calculus books but couldn't find answer. I just want a hint.

$$\int e^{2\cos 2x}dx.$$

Also about the function...

$$\int e^{x^2}dx.$$

Or any function with complicated power.

Answer 1

Both indefinite integrals cannot be expressed with a finite number of elementary functions.

$$\int e^{x^2}dx = \frac{1}{2} \sqrt{\pi}\:\text{erfi}(x)+constant$$ erfi(x) is a special function : http://mathworld.wolfram.com/Erfi.html

erfi$(x)$ is related to the Dawson's function Daw$(x)$ : http://mathworld.wolfram.com/DawsonsIntegral.html

Both erfi$(x)$ and Daw$(x)$ are particular hypergeometric functions.

For general information about the uses of special functions : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

The case of : $$\int e^{2\cos 2x}dx$$ is even more complicated. As far as I know, there is no convenient standard special function. Use infinite series for further calculus, and/or numerical methods of computation in practice.

Note that the above comment concerns the indefinite integrals. In some particular case of these integrals with particular bounds (particular finite integrals) closed form might exist. For example : $$\int_0^{\pi} e^{2\cos 2x}dx=\pi\: I_0(2)$$ where $I_0(x)$ is the Modified Bessel function of the first kind.

Answer 2

Not every indefinite integral has a closed form.

Answer 3

Using $$\displaystyle e^x = 1+\frac{x}{1!}+\frac{x^2}{2!}+...........\infty$$

So $$\displaystyle \int e^{x^2}dx = \int \left[1+\frac{x^2}{1!}+\frac{x^4}{2!}+......\infty\right]dx$$

So we get $$\displaystyle \int e^{x^2}dx = x+\frac{x^3}{1!\times 3}+\frac{x^5}{2!\times 5}+.......\infty+\mathcal{C} = \sum^{\infty}_{k=0}\frac{x^{2k+1}}{k!\times (2k+1)}+\mathcal{C}$$

Answer 4

For $\int e^{2\cos 2x}~dx$ ,

$\int e^{2\cos 2x}~dx$

$=\int e^{4\cos^2x-2}~dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n4^n\cos^{2n}x}{e^2n!}dx$

$=\int\left(e^{-2}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^n\cos^{2n}x}{e^2n!}\right)dx$

For $n$ is any natural number,

$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x~\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

$\therefore\int\left(e^{-2}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^n\cos^{2n}x}{e^2n!}\right)dx$

$=e^{-2}x+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!x}{e^2(n!)^3}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n4^{k-1}(2n)!((k-1)!)^2\sin x~\cos^{2k-1}x}{e^2(n!)^3(2k-1)!}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x}{e^2(n!)^3}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n4^{k-1}(2n)!((k-1)!)^2\sin x~\cos^{2k-1}x}{e^2(n!)^3(2k-1)!}+C$