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The integral below seems quite simple, but I couldn't find anywhere the result. $$ I = \int_0^\theta \cosh(a\sin x) dx$$ I tried to expand it into Taylor expansion series and successfully evaluate the integral, but it just got mess, $$ I =\sum_{k=0}^{\infty} \frac{a^{2k}}{(2k)!} \left[ \frac{1}{2^{2k}}\binom{2k}{k}\theta + \frac{(-1)^k}{2^{2k-1}}\sum_{n=0}^{k-1}(-1)^n\binom{2k}{n} \frac{\sin[(2k-2n)\theta]}{2k-2n}\right]. $$

Is there any simpler form of this integral? Any helps or hints will be appreciated!

Edited: $\theta$ can only have value of $0 < \theta < \pi/2$.

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  • $\begingroup$ A related question. $\endgroup$
    – Lucian
    May 22, 2015 at 17:52
  • $\begingroup$ Do you have any reason to expect a closed form? $\endgroup$
    – clathratus
    Dec 3, 2019 at 6:09

3 Answers 3

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This antiderivative is not an elementary function. However, at $\theta = \pi$ the integral is $\pi I_0(a)$ where $I_0$ is a modified Bessel function.

EDIT: With the substitution $x = \arcsin(t)$, the integral becomes $$ I = \int_0^{\sin(\theta)} \dfrac{\cosh(at)\; dt}{\sqrt{1-t^2}} = \sum_{k=0}^\infty {2k \choose k} 4^{-k} \int_0^{\sin(\theta)} t^{2k} \cosh(at)\; dt $$ Let $\sin(\theta) = s$. Now $$ \eqalign{\int_0^s t^{2k} \cosh(at)\; dt &= \dfrac{d^{2k}}{da^{2k}} \int_0^s \cosh(at)\; dt =\dfrac{d^{2k}}{da^{2k}} \dfrac{\sinh(as)}{a}\cr &= \dfrac{1}{2a^{2k+1}} \left(\Gamma(2k+1,-as) - \Gamma(2k+1,as)\right)}$$ so $$ I = \sum_{k=0}^\infty {2k \choose k} \dfrac{\Gamma(2k+1,-as) - \Gamma(2k+1,as)}{(2a)^{2k+1}} $$ but I don't know a closed form for that sum (nor does Maple).

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  • $\begingroup$ can you prove this? $\endgroup$ May 22, 2015 at 15:43
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    $\begingroup$ @Dr.SonnhardGraubner from the expansion above the $\sin(2(k-n)\theta)$ vanishes at $\theta = \pi$. so things simplify. $\endgroup$
    – Chinny84
    May 22, 2015 at 15:56
  • $\begingroup$ @Dr.SonnhardGraubner Did you mean the value at $\pi$, or the fact the antiderivative is non-elementary? The value at $\pi$ is "well-known", the non-elementary antiderivative comes from the Risch theory. $\endgroup$ May 22, 2015 at 21:20
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$\int_0^\theta\cosh(a\sin x)~dx=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}~dx=\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$

For $n$ is any natural number,

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

This result can be done by successive integration by parts.

$\therefore\int_0^\theta\left(1+\sum\limits_{n=1}^\infty\dfrac{a^{2n}\sin^{2n}x}{(2n)!}\right)~dx$

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

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

$=\theta I_0(a)-\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{a^{2n}((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}$

$=\theta I_0(a)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{a^{2n+2k+2}(k!)^2\sin^{2k+1}\theta\cos\theta}{4^{n+1}((n+k+1)!)^2(2k+1)!}$

Or you can express in terms of Incomplete Bessel Functions:

Consider $$\begin{align}J_0(ia,w)&=\dfrac{2}{\pi}\int_0^w\cos(ia\cos x)~dx \\&=\dfrac{2}{\pi}\int_0^w\cosh(a\cos x)~dx \\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cosh\left(a\cos\left(x-\dfrac{\pi}{2}\right)\right)~d\left(x-\dfrac{\pi}{2}\right) \\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cosh(a\sin x)~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_\frac{\pi}{2}^0\cosh\left(a\sin\left(\dfrac{\pi}{2}-x\right)\right)~d\left(\dfrac{\pi}{2}-x\right) \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_\pi^\frac{\pi}{2}\cosh(a\cos(\pi-x))~d(\pi-x) \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cosh(a\cos x)~dx-\dfrac{1}{\pi}\int_\frac{\pi}{2}^\pi\cosh(a\cos x)~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-\dfrac{1}{\pi}\int_0^\pi\cosh(a\cos x)~dx \\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx-I_0(a)\end{align} $$

Then $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cosh(a\sin x)~dx=I_0(a)+I_0(a,w)$

$\therefore\int_0^\theta\cosh(a\sin x)~dx=\dfrac{\pi}{2}\left(I_0(a)+I_0\left(a,\theta-\dfrac{\pi}{2}\right)\right)$

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As we see in the other answers, there is no closed-form for this indefinite integral.
However there is a closed form for certain definite integrals. For example $$ \int_0^{\pi/2} \cosh(a\sin \theta)\;d\theta = \frac{\pi}{2}\;I_0(a) $$ where $I_0$ is the hyperbolic Bessel function of order $0$


Similar answers:
https://math.stackexchange.com/a/2829346/442
https://math.stackexchange.com/a/1844794/442

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