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

\(\int_0^x r {J_0}(r)dr = x \times {J_1}(x)\)

Short Answer

Expert verified

The obtained integral is \(\int_0^x r {J_0}(r)dr = x{J_1}(x)\).

Step by step solution

01

Define differential recurrence relation.

Recurrence formulas that relate Bessel functions 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)

02

Obtain the integration.

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

\(x{J_0}(x) = \frac{d}{{dx}}\left( {x{J_1}(x)} \right)\)

Integrate both sides of the expression.

\(\begin{array}{l}\int_0^x r {J_0}(r)dr = \int_0^x {\frac{d}{{dx}}} \left( {x{J_1}(x)} \right)\\\int_0^x r {J_0}(r)dr = x{J_1}(x)\end{array}\)

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