Evaluating $\int_0^1 \sqrt{1-x^2}\cos(ax) dx$

Question

I want to evaluate this integral $$I=\int_0^1 \sqrt{1-x^2}\cos(ax) dx \quad (a \in \mathbb{R})$$ but I cannot find a useful strategy. Could you please give me a hint?

Answer 1

Hint: Maclaurin series in powers of $a$.

EDIT: Alternate hint: Laplace transform.

Answer 2

I thought it might be instructive to present an approach that relies on tools acquired in a first course in calculus. However, we begin with a primer on Bessel functions, a topic not typically introduced in elementary calculus.

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PRIMER $1$:

The Bessel Function, $J_n(a)$, of the first kind and of integer order $n$ can be defined as

$$J_n(a)=\frac1\pi\int_0^\pi \cos(nx-a\sin(x))\,dx \tag{P1}$$

Then, the zeroth order Bessel function, $J_0(a)$ is expressed as

$$\begin{align} J_0(a)&=\frac1\pi\int_0^\pi \cos(a\sin(x))\,dx\\\\ &=\frac1\pi\int_0^{\pi/2} \cos(a\sin(x))\,dx+\frac1\pi\int_{\pi/2}^\pi \cos(a\sin(x))\,dx\tag{P2}\\\\ &=\frac2\pi\int_0^{\pi/2}\cos(a\sin(x))\,dx\tag{P3} \end{align}$$

where we enforced the substitution $x\mapsto \pi-x$ in the second integral of $(\text{P}2)$ to arrive at $(\text{P}3)$.

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PRIMER $2$:

The derivative of $J_0(a)$, is

$$\begin{align} J_0'(a)&=-\frac1\pi\int_0^\pi \sin(x)\sin(a\sin(x))\,dx\\\\ &=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-a\sin(x))\,dx-\int_0^\pi \cos(x+a\sin(x))\,dx\right)\tag{P4}\\\\ &=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-a\sin(x))\,dx+\int_0^\pi \cos(x-a\sin(x))\,dx\right)\tag{P5}\\\\ &=-J_1(a)\tag{P6} \end{align}$$

where we enforced the substitution $x\mapsto \pi-x$ in the second integral of $(\text{P}4)$ to arrive at $(\text{P}5)$.

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Let $f$ be given by the integral

$$f(a)=\int_0^1 \sqrt{1-x^2}\,\,(a\cos(tx))\,dx\tag 1$$

Writing $t\cos(xa)$ as $a\cos(xa)=\frac{d\sin(xa)}{dx}$ in $(1)$ reveals

$$f(a)=\int_0^1 \sqrt{1-x^2}\,\frac{d\sin(xa)}{dx}\,dx\tag2$$

Integrating by parts the integral in $(2)$, we obtain

$$f(a)=\int_0^1 \frac{x}{\sqrt{1-x^2}}\sin(xa)\,dx\tag3$$

Enforcing the substation $x\mapsto \sin(x)$ in $(3)$ yields

$$\begin{align} f(t)&=\int_0^{\pi/2} \sin(x)\sin(a\sin(x))\,dx\\\\ &=-\frac{d}{da}\int_0^{\pi/2} \cos(a\sin(x))\,dx\tag 4\\\\ &=-\frac{d}{da}\left(\frac{\pi}{2}J_0(a)\right)\tag5\\\\ &=\frac\pi2 J_1(a)\tag6 \end{align}$$

In going from $(4)$ to $(5)$, we used $(\text{P}3)$ and in going from $(5)$ to $(6)$, we used $(\text{P}6)$.

Finally, we have

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

And we are done!

Answer 3

Let $$I = \int_0^1 \sqrt{1 - x^2} \cos (ax) \, dx.$$ Integrating by parts we have $$I = \frac{1}{a} \int_0^1 \frac{x \sin (ax)}{\sqrt{1 - x^2}} \, dx.$$

Now, using a result I show here, namely $$J_0 (a) = \frac{2}{\pi} \int_0^1 \frac{\cos (ax)}{\sqrt{1 - x^2}} \, dx,$$ where $J_0 (x)$ is the Bessel function of the first kind of order zero, differentiating this result with respect to the parameter $a$ we have $$J'_0 (a) = - \frac{2}{\pi} \int_0^1 \frac{x \sin (ax)}{\sqrt{1 - x^2}} \, dx.$$ Now from the reasonably well-known result of $J'_0 (x) = - J_1 (x)$ where $J_1(x)$ is the Bessel function of the first kind of order one, we have $$J_1 (a) = \frac{2}{\pi} \int_0^1 \frac{x \sin (ax)}{\sqrt{1 - x^2}} \, dx,$$ and we conclude $$I = \frac{\pi}{2a} J_1 (a).$$