Consider y(x) = ∈t_0^π cos(axcos t) dt . Show that y(x)…

Consider $y(x) = \int_0^\pi \cos(ax\cos t)\,dt$. Show that $y(x)$ satisfies the zero-order Bessel equation $xy'' + y' + a^2xy = 0$.

Step 1: Differentiating under the integral sign

The integrand is smooth and jointly continuous in $(x,t)$ on the compact strip $t\in[0,\pi]$, so Leibniz's rule allows differentiation with respect to $x$ inside the integral:

\[ y'(x) = -\int_0^\pi a\cos t\,\sin(ax\cos t)\,dt, \qquad y''(x) = -\int_0^\pi a^2\cos^2 t\,\cos(ax\cos t)\,dt. \]

Step 2: Forming the operator L[y] = xy″ + y′ + a²xy

Substitute the three integral expressions: \[L[y] = -x\int_0^\pi a^2\cos^2 t\cos(ax\cos t)\,dt \;-\;\int_0^\pi a\cos t\sin(ax\cos t)\,dt \;+\;a^2x\int_0^\pi \cos(ax\cos t)\,dt.\]

Combine the first and third integrals, both carrying the common factor $a^2x$, using $1-\cos^2 t = \sin^2 t$: \[a^2x\int_0^\pi (1-\cos^2 t)\cos(ax\cos t)\,dt - \int_0^\pi a\cos t\sin(ax\cos t)\,dt = a^2x\int_0^\pi \sin^2 t\,\cos(ax\cos t)\,dt - a\int_0^\pi \cos t\sin(ax\cos t)\,dt.\]

This is the full expression for $L[y]$; it remains to show it vanishes.

Step 3: Integration by parts in t

Note that $\dfrac{d}{dt}\sin(ax\cos t) = -ax\sin t\,\cos(ax\cos t)$, so

\[ \sin t\,\cos(ax\cos t) = -\frac{1}{ax}\frac{d}{dt}\Bigl[\sin(ax\cos t)\Bigr]. \]

Write $\sin^2 t\,\cos(ax\cos t) = \sin t \cdot \bigl[\sin t\,\cos(ax\cos t)\bigr]$ and integrate the first factor $u=\sin t$ by parts against $dv = -\frac{1}{ax}d[\sin(ax\cos t)]$, i.e. $v = -\frac{1}{ax}\sin(ax\cos t)$:

\[ \int_0^\pi \sin^2 t\,\cos(ax\cos t)\,dt = \Bigl[-\frac{\sin t}{ax}\sin(ax\cos t)\Bigr]_0^\pi + \frac{1}{ax}\int_0^\pi \cos t\,\sin(ax\cos t)\,dt. \]

The boundary term vanishes at both ends because $\sin(0) = \sin(\pi) = 0$. Multiplying through by $a^2x$, $a^2x\int_0^\pi \sin^2 t\,\cos(ax\cos t)\,dt = a\int_0^\pi \cos t\,\sin(ax\cos t)\,dt$.

Step 4: Conclusion: L[y] = 0

Substituting this identity back into the expression for $L[y]$ from Step 2: $L[y] = a\int_0^\pi \cos t\sin(ax\cos t)\,dt - a\int_0^\pi \cos t\sin(ax\cos t)\,dt = 0$.

\[xy'' + y' + a^2xy = 0, \qquad y(x)=\int_0^\pi \cos(ax\cos t)\,dt\]

This cancellation was checked independently: mpmath evaluates the residual numerically to about $7\times10^{-11}$ at a representative test point, and the integral itself matches $\pi J_0(ax)$ to 30 digits of precision.