Finding the following definite and indefinite integrals

Question

I want to calculate the integral $$\int_0^{\frac{\pi}{2}} e ^{ \sin t}\, dt.$$ Can we find a primitive function for $f(t) = e ^{\sin t}$?

Answer 1

No, this one is among the functions that do not admit primitives expressible as a composition of elementary functions.

The definite integral, though, can still be computed, through other means than the Leibniz-Newton theorem. Nevertheless, the result is ugly, since it is expressed with the aid of special functions, so just forget about it.

Answer 2

For $\int e^{\sin t}~dt$ ,

$\int e^{\sin t}~dt$

$=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n)!}dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$

For $n$ is any natural number,

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

This result can be done by successive integration by parts.

For $n$ is any non-negative integer,

$\int\sin^{2n+1}t~dt$

$=-\int\sin^{2n}t~d(\cos t)$

$=-\int(1-\cos^2t)^n~d(\cos t)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}t~d(\cos t)$

$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{k!(n-k)!(2k+1)}+C$

$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$

$=t+\sum\limits_{n=1}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}+C$

$\therefore$ For $\int_0^\frac{\pi}{2}e^{\sin t}~dt$ ,

$\int_0^\frac{\pi}{2}e^{\sin t}~dt$

$=\left[\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}\right]_0^\frac{\pi}{2}$

$=\sum\limits_{n=0}^\infty\dfrac{\pi}{2^{2n+1}(n!)^2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n+1)!k!(n-k)!(2k+1)}$

Answer 3

We recall the following Jacobi-Anger expansion (DLMF 10.35.3):

$$e^{z\sin t}=I_0(z) + 2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\sin((2k+1)t)+2\sum_{k=1}^\infty (-1)^k I_{2k}(z)\cos(2kt)$$

Integrating both sides from $t=0$ to $\pi/2$:

\begin{align} \int_0^{\pi/2}e^{z\sin t}\,dt &= \frac{\pi}{2}I_0(z)+2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\frac{1+\sin k\pi}{2k+1}+2\sum_{k=1}^\infty (-1)^k I_{2k}(z)\frac{\sin \pi k}{2k}\\ &= \frac{\pi}{2}I_0(z)+2\sum_{k=0}^\infty (-1)^k I_{2k+1}(z)\frac{1}{2k+1} \end{align}

There's a few ways we could deal with this infinite sum. First, we can expand the modified Bessel functions in powers of $z$ and confirm agreement with the infinite sums found previously. Second, if we're satisfied with numerical results, we may note that when $z=1$ this alternating sum converges rapidly: To five decimals, the first five partial sums are $$1.98873,\;3.11905,\;3.10427,\;3.10438 , \; 3.10438 $$ which already has converged to the desired precision.