Trigonometric functions expressed as definite integrals with Bessel functions

Question

Prove that $$\frac{\sin(x)}{x}=\int_0^\frac{\pi}{2}J_0(x\cos(\theta))\cos(\theta)\,d\theta \tag{a}$$ $$\frac{1-\cos(x)}{x}=\int_0^\frac{\pi}{2}J_1(x\cos(\theta))\,d\theta \tag{b}$$ Hint: $$\int_0^\frac{\pi}{2}\cos^{2s+1}(\theta)\,d\theta = \frac{2\cdot 4\cdot6\cdots(2s)}{1\cdot3\cdot5\cdots(2s+1)}$$

I have no idea how to approach this problem. Any suggestions? I did express sine function in exponential form but then I have no clue where to go from there so that I can end up with the integral as an answer indicated above.

Answer 1

Since $$J_0(x)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{4^m\,m!^2}x^{2m}\tag{1}$$ we have: $$\begin{eqnarray*}\int_{0}^{\pi/2}J_0(x\cos\theta)\cos\theta\,d\theta&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{4^m\,m!^2}\int_{0}^{\pi/2}(\cos\theta)^{2m+1}d\theta\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{4^m\,m!^2}\cdot\frac{4^m\,m!^2}{(2m+1)!}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m}}{(2m+1)!}=\frac{\sin x}{x}.\end{eqnarray*}$$ In a similar way $$ J_1(x)=\sum_{m=0}^{+\infty}\frac{(-1)^m}{2\cdot 4^m\, (m+1)\,m!^2}x^{2m+1}\tag{2}$$ gives: $$\begin{eqnarray*}\int_{0}^{\pi/2}J_1(x\cos\theta)\,d\theta&=&\sum_{m=0}^{+\infty}\frac{(-1)^m\,x^{2m+1}}{2\cdot 4^m\, (m+1)\,m!^2}\int_{0}^{\pi/2}(\cos\theta)^{2m+1}d\theta\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m\,x^{2m+1}}{2\cdot 4^m\, (m+1)\,m!^2}\cdot\frac{4^m\,m!^2}{(2m+1)!}\\&=&\sum_{m=0}^{+\infty}\frac{(-1)^m x^{2m+1}}{(2m+2)!}=\frac{1-\cos x}{x}.\end{eqnarray*}$$ $(1)$ and $(2)$ can be derived from the integral representations given in the Wikipedia page for Bessel functions.

Answer 2

The first integral is the case $\alpha = \frac{1}{2}$ of the integral formula

$$\frac{2^{1/2-\alpha} \, \Gamma(1- \alpha)}{\sqrt{\pi}} \, \sin(x) \, x^{\alpha-3/2} = \int_{0}^{\pi/2} J_{\alpha-1/2}(x \cos \theta) \cos^{\alpha+1/2}(\theta) \sin^{1- 2 \alpha}(\theta) \, \mathrm d \theta, \tag{1}$$

where $x>0$ and $0 < \alpha <1$.

I stumbled upon $(1)$ while messing around with the Abel transform. (It can, of course, be proven in a more direct manner using the beta function.)

A variant of the Abel transform states that if $$f(x) = \int_{0}^{x} \frac{g(t)}{(x^{2}-t^{2})^{\alpha}} \, \mathrm dt, \quad 0 < \alpha <1, $$

then $$g(t) = \frac{2 \sin(\pi \alpha)}{\pi} \frac{\mathrm d}{\mathrm dt} \int_{0}^{t} \frac{x f(x)}{(t^{2}-x^{2})^{1- \alpha}} \, \mathrm dx.$$

I'll post a proof at the end.

If we let $f(x) = \sin(x)$, then $$g(t) = \frac{2 \sin(\pi \alpha)}{\pi} \frac{\mathrm d}{\mathrm dt} \int_{0}^{t} \frac{x \sin (x)}{(t^{2}-x^{2})^{1 - \alpha}} \, \mathrm dx. $$

Integrating by parts and using Poisson's integral representation of the Bessel function of the first kind, we get $$ \begin{align} g(t) &= \frac{\sin(\pi \alpha)}{\pi \alpha } \frac{\mathrm d}{\mathrm dt} \int_{0}^{t} (t^{2}-x^{2})^{\alpha} \cos(x) \, \mathrm dx \\ &= \frac{\sin(\pi \alpha)}{\pi \alpha } \left( (t^{2}-t^{2})^{\alpha} \cos(t) + 2t \alpha\int_{0}^{t} (t^{2}-x^{2})^{\alpha -1} \cos(x) \, \mathrm dx \right) \\ &= \frac{2 t \sin(\pi \alpha)}{\pi} \int_{0}^{t} (t^{2}-x^{2})^{\alpha -1} \cos(x) \, \mathrm dx \\ &= \frac{2 t \sin(\pi \alpha)}{\pi} \int_{0}^{t} t^{2(\alpha -1)} \left(1- \left(\frac{x}{t}\right)^{2} \right)^{\alpha-1} \cos(x) \, \mathrm dx \\ &= \frac{2t^{2 \alpha} \sin(\pi \alpha)}{\pi} \int_{0}^{1} (1-u^{2})^{\alpha -1} \cos(tu) \, \mathrm du \\ &= \frac{2t^{2 \alpha} \sin(\pi \alpha)}{\pi} \frac{\sqrt{\pi} \, \Gamma(\alpha)}{2 \left(\frac{t}{2}\right)^{\alpha -1/2}} \, J_{\alpha - 1/2}(t) \\&= \frac{\sqrt{\pi} \, 2^{1/2-\alpha} \, t^{\alpha+1/2}}{\Gamma(1-a)} \, J_{\alpha-1/2}(t) . \end{align} $$

Therefore, $$\begin{align} \sin(x) &= \int_{0}^{x} \frac{g(t)}{(x^{2}-t^{2})^{\alpha}} \, \mathrm dt \\ &= \int_{0}^{x} \frac{1}{x^{2 \alpha}} \frac{g(t)}{\left(1- \left(\frac{t}{x} \right)^{2} \right)^{\alpha}} \, \mathrm dt \\ &= x^{1- 2 \alpha} \int_{0}^{1} \frac{g(xv)}{(1-v^{2})^{\alpha}} \, \mathrm dv \\ &= x^{1- 2 \alpha} \int_{0}^{1} \frac{1}{(1-v^{2})^{\alpha}}\frac{\sqrt{\pi} \, 2^{\alpha-1/2} \, (xv)^{\alpha+1/2}}{\Gamma(1-a)} \, J_{\alpha-1/2}(xv) \, \mathrm dv \\ &= \frac{ \sqrt{\pi} \, 2^{\alpha-1/2} \, x^{3/2- \alpha} }{\Gamma(1-\alpha)} \int_{0}^{1} \frac{v^{\alpha +1/2}}{(1-v^{2})^{\alpha}} \, J_{\alpha-1/2}(xv) \, \mathrm dv \\ &= \frac{ \sqrt{\pi} \, 2^{\alpha-1/2} \, x^{3/2- \alpha}}{\Gamma(1-\alpha)} \int_{0}^{\pi/2 } J_{\alpha-1/2}(x \cos \theta) \cos^{\alpha+1/2}(\theta) \sin^{1- 2 \alpha}(\theta) \, \mathrm d \theta. \end{align}$$

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Proof of the Abel transform:

Assume that $$f(x) = \int_{0}^{x} \frac{g(t)}{(x^{2}-t^{2})^{\alpha}} \, \mathrm dt, \quad 0 < \alpha < 1. $$

Then $$ \begin{align} \int_{0}^{t} \frac{x}{(t^{2}-x^{2})^{1- \alpha}} \, f(x) \, \mathrm dx &= \int_{0}^{t} \frac{x}{(t^{2}-x^{2})^{1- \alpha}} \int_{0}^{x} \frac{g(y)}{(x^{2}-y^{2})^{\alpha}} \, \mathrm dy \, \mathrm dx \\ &= \int_{0}^{t} g(y) \int_{y}^{t} x (t^{2}-x^{2})^{\alpha-1} (x^{2}-y^{2})^{-\alpha} \, \mathrm dx \, \mathrm dy \\ &= \frac{1}{2} \int_{0}^{t} g(y) \int_{y^{2}}^{t^{2}} (t^{2}-u)^{\alpha-1} (u-y^{2})^{-\alpha} \, \mathrm du \, \mathrm dy \\ &\overset{(2)}{=} \frac{1}{2} \int_{0}^{t} g(y) \int_{0}^{1} (1-w)^{\alpha-1} w^{-\alpha} \, \mathrm dw \, \mathrm dy \\&= \frac{1}{2} \int_{0}^{t} g(y) \, B \left(1- \alpha, \alpha \right) \mathrm dy \\ &= \frac{\pi}{2} \csc(\pi \alpha) \int_{0}^{t} g(y) \, \mathrm dy. \end{align}$$

Now differentiate both sides of the above equation with respect to $t$.

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$(2)$ Let $w = \frac{u-y^2}{t^2-y^2}$.