Expressing Sine and Cosine as Integrals of Bessel Functions

Sine and cosine can be written as definite integrals involving Bessel functions. The convenient orders are $-\tfrac12$ and $\tfrac12$, because Bessel functions of these orders reduce to cosine and sine multiplied by a simple power of their argument.

For $x\geq 0$, the identities are

$$\sin x=\sqrt{\frac{\pi}{2}}\int_0^x \sqrt{t}\,J_{-1/2}(t)\,dt,$$

and

$$\cos x=1-\sqrt{\frac{\pi}{2}}\int_0^x \sqrt{t}\,J_{1/2}(t)\,dt.$$

Here $J_\nu$ denotes the Bessel function of the first kind of order $\nu$. The factor $\sqrt{t}$ matters: these are not formulas for the unweighted integrals $\int_0^x J_\nu(t)\,dt$.

Why half-integer orders work

For positive $t$, the two relevant Bessel functions have elementary forms:

$$J_{-1/2}(t)=\sqrt{\frac{2}{\pi t}}\cos t, \qquad J_{1/2}(t)=\sqrt{\frac{2}{\pi t}}\sin t.$$

Multiplication by $\sqrt{t}$ cancels the factor $1/\sqrt{t}$:

$$\sqrt{t}\,J_{-1/2}(t)=\sqrt{\frac{2}{\pi}}\cos t, \qquad \sqrt{t}\,J_{1/2}(t)=\sqrt{\frac{2}{\pi}}\sin t.$$

The desired identities then follow by ordinary integration. In particular, the integrands after multiplication by $\sqrt t$ remain finite as $t$ approaches zero.

Deriving the formulas

For the sine identity, substitute the expression for $J_{-1/2}$ into the integral:

$$\begin{aligned} \int_0^x \sqrt{t}\,J_{-1/2}(t)\,dt &=\sqrt{\frac{2}{\pi}}\int_0^x \cos t\,dt\\ &=\sqrt{\frac{2}{\pi}}\sin x. \end{aligned}$$

Multiplying both sides by $\sqrt{\pi/2}$ gives the stated formula.

For cosine, use the order $1/2$ instead:

$$\begin{aligned} \int_0^x \sqrt{t}\,J_{1/2}(t)\,dt &=\sqrt{\frac{2}{\pi}}\int_0^x \sin t\,dt\\ &=\sqrt{\frac{2}{\pi}}(1-\cos x). \end{aligned}$$

Rearranging gives the expression for $\cos x$.

Worked example: evaluate the sine integral at $x=\pi/2$

Set $x=\pi/2$ in the first identity. Since $\sin(\pi/2)=1$,

$$\int_0^{\pi/2}\sqrt{t}\,J_{-1/2}(t)\,dt =\sqrt{\frac{2}{\pi}}.$$

You can also check directly using the elementary form of the Bessel function:

$$\int_0^{\pi/2}\sqrt{t}\,J_{-1/2}(t)\,dt =\sqrt{\frac{2}{\pi}}\int_0^{\pi/2}\cos t\,dt =\sqrt{\frac{2}{\pi}}.$$

The same check for the cosine formula gives

$$\int_0^{\pi/2}\sqrt{t}\,J_{1/2}(t)\,dt=\sqrt{\frac{2}{\pi}},$$

so $1-\sqrt{\pi/2}\,\sqrt{2/\pi}=0=\cos(\pi/2)$.

What changes for negative x?

The displayed real integrals use $\sqrt{t}$, so they are written for $x\geq0$. For $x<0$, recover the values from parity: $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$. Equivalently, apply the integral formulas to $|x|$ and then use those parity rules.

Common mistakes

• Dropping the weight: The formulas involve $\sqrt{t}\,J_\nu(t)$, not just $J_\nu(t)$. Removing the weight changes the integral.
• Swapping the orders: $J_{-1/2}$ is proportional to $\cos t$, while $J_{1/2}$ is proportional to $\sin t$.
• Forgetting the constant in the cosine formula: Integrating sine from $0$ to $x$ gives $1-\cos x$, so the result must be rearranged as $\cos x=1-\cdots$.
• Ignoring the lower limit: The choice of zero makes the sine integral start at zero and fixes the constant in the cosine identity.

These formulas are useful when a problem specifically calls for a Bessel-function representation. They are direct consequences of the elementary half-integer Bessel formulas, rather than special evaluations of general unweighted Bessel integrals.