# Anyone recognizing or knowing how to solve this integral?

[Cross-posted with minor changes from Math SE; see note 2 below].

I would like to find methods for closed-form (or, failing that, series) solution of the following four integrals:

$$\int^X_0 \cos(a \cos(x)) \cos(b \sin(x)) dx,$$

$$\int^X_0 \cos(a \cos(x)) \sin(b \sin(x)) dx,$$

$$\int^X_0 \sin(a \cos(x)) \cos(b \sin(x)) dx,$$

$$\int^X_0 \sin(a \cos(x)) \sin(b \sin(x)) dx,$$

where $X=\pi/2$, $\pi$ or $2\pi$ and where $a$, $b$ are constants. Substitution ($t=\tan(x/2)$, $\cos(x)$ or $\sin(x)$) or conversion of the products of trigonometric functions to their sum or difference does not seem to lead to an amenable form, as arguments become (too) complicated. I also cannot find any of them listed in tables.

NB:

1. Numerical or stochastic quadrature is not feasible in my case, as $a$ and $b$ are parameterized.

2. As opposed to the Maths SE post, my query here focuses on potential general techniques, as a core for solving these and related (more complicated) integrals of functionals. Different approaches and lines of attack are of greater interest to me than just the result itself. For example, is there a deeper reason why we have a solution for the 5th but not 6th case? Are there methods that work equally on e.g. ln(ln()) as they would on cos(cos())? Etc.

• Also post on MSE – Paul Enta Jun 5 '17 at 19:17
• BTW if both accounts (MathCub and TeXCub) belong to you, you can try to merge them so that you have better access to your questions posted from the older - unregistered - account. – Martin Sleziak Jun 6 '17 at 5:12

How about these special cases... \begin{align} \int_0^{2\pi} \cos(a \cos(x)) \cos(b \sin(x)) \;dx &= 2\pi \;J_0\big(\sqrt{a^2+b^2}\;\big) \\ \int_0^{2\pi} \cos(a \cos(x)) \sin(b \sin(x)) \;dx &= 0 \\ \int_0^{2\pi} \sin(a \cos(x)) \cos(b \sin(x)) \;dx &= 0 \\ \int_0^{2\pi} \sin(a \cos(x)) \sin(b \sin(x)) \;dx &= 0 \end{align} where $J_0$ is a Bessel function.

And \begin{align} \int_0^{\pi} \cos(a \cos(x)) \cos(b \sin(x)) \;dx &= \pi \;J_0\big(\sqrt{a^2+b^2}\;\big) \\ \int_0^{\pi} \cos(a \cos(x)) \sin(b \sin(x)) \;dx &= ? \\ \int_0^{\pi} \sin(a \cos(x)) \cos(b \sin(x)) \;dx &= 0 \\ \int_0^{\pi} \sin(a \cos(x)) \sin(b \sin(x)) \;dx &= 0 \end{align}

• Why special cases? Aren't these a complete answer? (apart the 7th) – Pietro Majer Jun 5 '17 at 20:43
• For the $2\pi$ cases, and generalizations of them, see Gradshteyn & Ryzhik 3.937 – Gerald Edgar Jun 5 '17 at 21:00

For $X = 2\pi,$ at least these are integrals of a meromorphic function over the unit circle. In fact, unless I am sadly mistaken, the only singularity of this function is at the origin, so the residue theorem is your friend (although for two of the integrals, the function is odd, so the integral is automatically zero). In the cases where the function is even, the integral from $0$ to $\pi$ is half of the integral from $0$ to $2\pi.$

• Isn't the singularity at zero essential? I'm thinking $\cos(x) = \frac{ z + z^{-1}}{2}$, $\sin(x) = \frac{z - z^{-1}}{2i}$ here, so they go to $\infty$ as $z$ goes to $0$. Then $\cos$ and $\sin$ have essential singularities at $\infty$? – Will Sawin Jun 5 '17 at 20:21
• ... and that's how you can have a Bessel function popping up. – Robert Israel Jun 5 '17 at 22:26
• @IgorRivin When you have an essential singularity, the residue theorem is no longer your friend. – Will Sawin Jun 5 '17 at 23:57
• @WillSawin Really? The author of these notes agrees with me otherwise: math.slu.edu/~lsabalka/teaching/09Spring375/Chapter9.pdf – Igor Rivin Jun 5 '17 at 23:59
• @IgorRivin I guess for me the Bessel function is defined as some integral that looks like this, so applying the residue theorem to transform the integral into a Bessel function is not needed. If for you the Bessel function is defined as some power series, then the residue theorem is needed. – Will Sawin Jun 6 '17 at 0:38