Integration of exponential functions and cosine function

Question

I am trying to solve the following equation; $$\int_{-1}^{1}e^{i(x+a\cos x)} \, \mathrm{d}(\cos x)$$ or $$\int_{0}^{\pi}e^{i(x+a\cos x)} \sin x \, \mathrm{d}x$$

I tried this in Wolfram Alpha, but it says that answer cannot be obtained.

Answer 1

Let $F(a)$ be defined as

$$F(a)=\int_0^{\pi}e^{i(x+a\cos x)}\sin x\,dx \tag 1$$

Now, it is easy to show that the real part of $F$ is identically $0$. To do this, we write the real part of $(1)$ as

$$\text{Re}\lbrace F(a)\rbrace=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{\pi/2}^{\pi}\cos(x+a\cos x)\sin x\,dx \tag 2$$

Then, enforcing the substitution $x\to \pi -x$ in the second integral on the right-hand side of $(2)$ reveals

$$\begin{align} \text{Re}\lbrace F(a)\rbrace&=\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx+\int_{0}^{\pi/2}\cos(\pi-x+a\cos (\pi-x))\sin (\pi-x)\,dx\\\\ & =\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx-\int_0^{\pi/2}\cos(x+a\cos x)\sin x\,dx\\\\ &=0 \tag 3 \end{align}$$

as was to be shown. We can write, therefore, $F(a)$ as

$$\begin{align} F(a)&=i\int_0^{\pi}\sin(x+a\cos x)\sin x\,dx \\\\ &=i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)+\sin(x)\cos(x)\sin(a\cos x)\,\right)dx\tag 4 \end{align}$$

where we arrived at the right-hand side of $(4)$ using a parallel development that led to $(3)$.

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The second integral on the right-hand side of $(4)$ is easily evaluated. Substituting $x\to \cos(x)$ and integrating by parts reveals that

$$\begin{align} i\int_0^{\pi}\sin(x)\cos(x)\sin(a\cos x)\,dx&=i2\int_{0}^1 x\sin(ax)\,dx\\\\ &=i2\frac{-a\cos(a)+\sin(a)}{a^2} \tag 5 \end{align}$$

Using $(5)$ in $(4)$ reduces $F(a)$ to

$$\begin{align} F(a)&=i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)\,dx\\\\ &+i2\frac{-a\cos(a)+\sin(a)}{a^2} \tag 6 \end{align}$$

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In order to evaluate the integral in $(6)$ we will enforce two successive substitutions. First, we let $a\cos x\to x$. This yields

$$i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)=\frac{i2}{a}\int_0^a \sqrt{1-\left(\frac{u}{a}\right)^2} \cos (x)\,dx$$

where we exploited the fact that the integrand is an even function. Next, we enforce the substitution $x\to a\sin x$ which reveals

$$\begin{align} i\int_0^{\pi}\left(\sin^2(x)\cos(a\cos x)\right)&=i\int_0^{\pi}\cos^2(x)\cos(a\sin x)\,dx \tag 7\\\\ &=i\int_0^{\pi}\cos(a\sin x)\,dx-i\int_0^{\pi}\sin^2(x)\cos(a\sin x)\,dx\\\\ &=i\pi J_0(a)+i\pi J_1'(a)\\\\ &=i\pi\frac{J_1(a)}{a} \tag 8 \end{align}$$

where $J_0(a$) and $J_1(a)$ are the Bessel Function of the first kind and order zero and one, respectively, while $J_1'(a)$ is the first derivative of $J_1$. In deriving $(8)$ we relied on Equations $(B5)$, $(B6)$, and $(B8)$ in the section on Bessel Functions.

Aside, we exploited the symmetry of $\cos^2(x)\cos(a\sin x)$ around $\pi/2$ to arrive at $(7)$.

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FINAL RESULT

Putting $(6)$ and $(8)$ together finally yields

$$\bbox[5px,border:2px solid #C0A000]{F(a)=i2\frac{-a\cos(a)+\sin(a)}{a^2}+i\pi\frac{J_1(a)}{a}} $$

which agrees with the empirically obtained hypothesis of @ClaudeLeibovici and @JoshBroadhurst!

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BESSEL FUNCTION ASIDE

We note an integral representation for the first kind Bessel Function of order $n$

$$\begin{align} J_n(a)&=\frac{1}{\pi}\int_0^{\pi}\cos(nx-a\sin x)dx \\\\ &=\frac{1}{\pi}\int_0^{\pi} \left(\cos (nx)\cos(a\sin x)+\sin(nx)\sin(a\sin x)\right) \,dx \tag{B1} \end{align}$$

along with recurrence relationships

$$J_n(a)=\frac{a}{2n}\left(J_{n-1}(a)+J_{n+1}(a)\right) \tag{B2}$$

and

$$J_n'(a)=\frac12 \begin{cases} J_{n-1}(a)-J_{n+1}(a),&n\ne 0\\\\ -J_1(a),&n = 0 \tag{B4} \end{cases}$$

Setting $n=0$ in $(B1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{J_0(a)=\frac{1}{\pi}\int_0^{\pi} \left(\cos(a\sin x)\right) \,dx} \tag{B5}$$

Taking the two derivatives of $J_0(a)$ and multiplying by $-1$ and using $(B4)$ yields

$$\bbox[5px,border:2px solid #C0A000]{J_1'(a)=-\frac{1}{\pi}\int_0^{\pi} \sin^2(x)\cos(a\sin x)\,dx}\tag{B6}$$

Note that using $(B3)$ and $(B4)$, we can eliminate $J_{n+1}$ and write $J_n$ as

$$J_n(a)=\frac{a}{n}\left(J_{n-1}(a)-J_n'(a)\right) \tag{B7}$$

Setting $n=1$ in $(B7)$ and rearranging we obtain

$$\bbox[5px,border:2px solid #C0A000]{\frac{J_1(a)}{a}=J_0(a)+J_1'(a)} \tag{B8}$$

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Answer 2

If you leave out the (what I assume is a) constant $a$ in your original problem and type it into WolframAlpha, it tells you that the solution involves the Bessel function of the first kind ($J_1$) and gives the following answer $$ (\pi J_1(1)-2 \cos(1)+2 \sin(1))i \approx 1.9848i$$

If you set $a=2$ as the input to WolframAlpha this gives $$ \frac{1}{2} (\pi J_1(2)-2 \cos(2)+\sin(2))i \approx 1.7767i$$

Again with $a=3$ as the input gives $$ \frac{1}{9} (3 \pi J_1(3)-6 \cos(3)+2 \sin(3))i \approx 1.0464i$$

Perhaps you can use a few more values for $a$ and generalize a pattern. The input I'm using looks like:

integrate from 0 to pi: e^(i (x + 3 cosx)) sinx dx

Hopefully this is sufficient for your needs.

Answer 3

In the same spirit as Josh Broadhurst's answer, you should arrive to $$\int_{0}^{\pi}e^{i(x+a\cos(x))} \sin(x)\,dx=\frac{\pi a J_1(a)-2 a \cos (a)+ 2 \sin (a)}{a^2}\,i$$ I do not find any way to compute the antiderivative itself.