Indefinite and definite integral of e^sin(x) dx

[Emmanuel_Euler]

Look to this indefinite integral →∫e^(sin(x))dx
Antiderivative or integral could not be found.and impossible to solve.

Look to this definite integral
∫e^(sin(x))dx (Upper bound is π and Lower bound is zero)=??

my question is : can we find any solution for this integral (definite integral) ??

 

[axmls]

There is no closed-form solution for the antiderivative, but we can still approximate the definite integral.

 

[pasmith]

It may be possible to use contour integration to find an analytic value for [itex]\int_0^\pi e^{\sin x}\,dx[/itex].

 

[Cgty]

Assume that we have a solution like that int(y dy)=int(e^sinx dx). It is clear we must find y^2/2=int(e^sinx dx). In order to equality, int[ln(y) dy]=int(sinx dx). Due to int(lny dy) is equal to y(lny-1); y(lny-1)=-cosx+c and y=[-cosx+c]/[lny-1]. We need to find y^2/2 therefore, y^2/2=[(cosx+c)/(lny-1)]^2/2. This is the solution of int(e^sinx dx) and we have a non-linear euation.

 

[lurflurf]

lets consider
$$\frac{1}{\pi}\int_0^\pi\!e^{\sin(x)}\,\mathrm{d}x$$
I flipped through some books and did not find much on that, but I did find that
$$\frac{1}{\pi}\int_0^\pi\!e^{\cos(x)}\,\mathrm{d}x=\operatorname{I}_0(1)$$
I is the modified Bessel function of the first kind.
Also we know that
$$\operatorname{I}_0(1)=\frac{1}{\pi}\int_0^\pi\!\cosh(\sin(x))\,\mathrm{d}x\\
\operatorname{I}_0(1)\sim1.26606587775201$$
http://people.math.sfu.ca/~cbm/aands/page_376.htm
and
$$\operatorname{L}_0(1)=\frac{1}{\pi}\int_0^\pi\!\sinh(\sin(x))\,\mathrm{d}x\\
\operatorname{L}_0(1)\sim0.710243185937891$$
L is the Modified Struve Function
http://people.math.sfu.ca/~cbm/aands/page_498.htm
so
$$\frac{1}{\pi}\int_0^\pi\!e^{\sin(x)}\,\mathrm{d}x=\operatorname{I}_0(1)+\operatorname{L}_0(1)\sim1.97630906368990$$

 

[Emmanuel_Euler]

can you give me the name of the books please, because i need them and thank you so much for help

 

[lurflurf]

I found that in the famous Handbook of Mathematical Functions edited by M. Abramowitz and I. A. Stegun a "work for hire performed for the US Government" thus freely available.
For example here
http://people.math.sfu.ca/~cbm/aands/toc.htm
It is also of course available in print if you prefer.

 

[Emmanuel_Euler]

thank you so much for help...

 

[alexpeter_pen]

The indefinite integral has a nice shape as well, still using special function but it depicts precisely the shape of the actual integral.

https://mathoverflow.net/questions/303391/is-frac-pi4l-0z-sum-limits-n-1-infty-1n1-fraci-2n-1

 

[jostpuur]

By using the formulas

[tex]
\sin(\alpha)\sin(\beta) = \frac{1}{2}\big(\cos(\alpha - \beta) - \cos(\alpha + \beta)\big)
[/tex]
[tex]
\sin(\alpha)\cos(\beta) = \frac{1}{2}\big(\sin(\alpha - \beta) + \sin(\alpha + \beta)\big)
[/tex]

it is possible to write the powers [itex](\sin(x))^n[/itex] in a form where non-trivial powers do not appear. By using this approach we get a series that starts as

[tex]
\int\limits_0^{\pi} e^{\sin(x)}dx = \pi + 2 + \frac{1}{2!}\frac{\pi}{2} + \frac{1}{3!}\frac{4}{3} + \frac{1}{4!}\frac{3\pi}{8} + \frac{1}{5!}\frac{16}{15} + \cdots
[/tex]

It is unfortunate of course that it might be impossible to get a nice formula for these terms, but it's not obvious if that's the way it's going to be. It could be that there exists some theory for the coefficients in the formula for [itex](\sin(x))^n[/itex].

 

[alexpeter_pen]

Just to add to my previous answer the actual formula so one does not have to follow the site

$$\displaystyle \int e^{\sin(x)} dx=I_0(1)x + \frac{\pi}{2}L_0(1) + 2\sum_{n=1}^{+\infty} \frac{I_n(1)}{n} \sin \left ( nx - \frac{n\pi}{2} \right ) $$

Another nice way of solving definite integral apart for simply stating its value through Struve and Bessel (which is the shortest possible known expression at the moment) goes like this:

First let us get rid of ##\sin(x)##, introducing ##u=\sin(x), du=\cos(x)dx## This leads to

$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=2\int_{0}^{1} \frac{e^u}{\sqrt{1-u^2}} du$$

Notice that we have taken it twice from ##0## to ##\frac{\pi}{2}## as ##e^{\sin(x)}## is symmetrical.

Now we use expansion of ##e^u## reducing it all to the sum of integrals

$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=2 \sum_{k=0}^{\infty} \int_{0}^{1} \frac{u^k}{k!\sqrt{1-u^2}} du$$

Now

$$\displaystyle \int_{0}^{1} \frac{u^k}{k!\sqrt{1-u^2}} du= \frac{1}{k!}\frac{\sqrt{\pi}\Gamma(\frac{k+1}{2})}{2\Gamma(\frac{k}{2}+1)}$$

coming from the connection between Beta and Gamma function, making it all

$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=\sum_{k=0}^{\infty} \frac{\sqrt{\pi}\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2}+1)k!}$$

or in a split form

$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=\sum_{k=0}^{\infty}\frac{\pi}{4^n(n!)^2} + \sum_{k=0}^{\infty} \frac{2^{n+1}n!}{(2n+1)!(2n+1)!}$$

First 10 terms are giving 20 digit precision already.

I am happy with 4 terms 4 digit precision

$$\displaystyle \frac{328}{147} + \frac{2917 π}{2304} \approx 6.2087$$

Just to make the connection

$$\displaystyle \pi L_0(1)=\sum_{k=0}^{\infty} \frac{2^{n+1}n!}{(2n+1)!(2n+1)!}$$
$$\displaystyle \pi I_0(1)=\sum_{k=0}^{\infty}\frac{\pi}{4^n(n!)^2}$$

 

[jostpuur]

$$\displaystyle \int_{0}^{\pi} e^{\sin(x)} dx=\sum_{k=0}^{\infty} \frac{\sqrt{\pi}\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2}+1)k!}$$

Yes, this is the same series whose first terms I wrote down with their more explicit values. Very nice, thanks.