Question

In Problems 29 and 30 use (22) or (23) to obtain the given result.

\({J_0}(x) = {J_{ - 1}}(x) = {J_1}(x)\)

Short answer

The obtained integral is \(J_0'(x) = - {J_1}(x) = {J_{ - 1}}(x)\).

Step-by-step solution

Step 1:Define differential recurrence relation.

Recurrence formulas that relate Besselfunctions of different orders are important in theory and in applications.

\(\frac{d}{{dx}}\left( {{x^{ - v}}{J_v}(x)} \right) = - {x^{ - v}}{J_{v + 1}}(x)\)… (1)

\(\frac{d}{{dx}}\left( {{x^v}{J_v}(x)} \right) = {x^v}{J_{v - 1}}(x)\) … (2)

Obtain the integration.

Substitute the value\(v = 0\)in the equation (1).

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^0}{J_0}(x)} \right) = - {x^0}{J_{0 + 1}}(x)\\\frac{d}{{dx}}\left( {{J_0}(x)} \right) = {J_1}(x)\end{aligned}\)

\(J_0'(x) = - {J_1}(x)\) … (3)

Substitute the value\(v = 0\)in the equation (2).

\(\begin{aligned}\frac{d}{{dx}}\left( {{x^0}{J_0}(x)} \right) = {x^0}{J_{0 - 1}}(x)\\\frac{d}{{dx}}\left( {{J_0}(x)} \right) = {J_{ - 1}}(x)\end{aligned}\)

\(J_0'(x) = {J_{ - 1}}(x)\) … (4)

From (3) and (4) yields the result,

\(J_0'(x) = - {J_1}(x) = {J_{ - 1}}(x)\)