Evalute $\int_0^1 \sqrt{1-x^2} \cos (tx) \, dx$

Question

I can't solve this problem. Can anyone help me?

Let $t \in \mathbb{R}$. Evaluate $\int_0^1 \sqrt{1-x^2} \cos (tx) \, dx.$

Thank you very much!

Answer 1

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, $J_n(t)$, of the first kind and of integer order $n$ can be defined as

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

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

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

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 $J_0(t)$, is

$$\begin{align} J_0'(t)&=-\frac1\pi\int_0^\pi \sin(x)\sin(t\sin(x))\,dx\\\\ &=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-t\sin(x))\,dx-\int_0^\pi \cos(x+t\sin(x))\,dx\right)\tag{P4}\\\\ &=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-t\sin(x))\,dx+\int_0^\pi \cos(x-t\sin(x))\,dx\right)\tag{P5}\\\\ &=-J_1(t)\tag{P6} \end{align}$$

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

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

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

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

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

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

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

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

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

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

And we are done!