P.T. $\int\limits_0^b xJ_0(ax) dx =\dfrac ta J

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P.T. $\int\limits_0^b xJ_0(ax) dx =\dfrac ta J_1(ab) $

written 9.3 years ago by

teamques10 ★ 70k

modified 4.2 years ago by

krithikkm200 • 10

Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : DEC 2015

engineering mathematics

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written 9.3 years ago by

teamques10 ★ 70k

$$I=\int\limits_0^b xJ_0 (ax) dx, put \space ax=t, dx=dt/a$$ enter image description here

$I=\int\limits_{t=0}^{ab} \dfrac ta J_0 (t) \dfrac {dt}a =\dfrac 1{a^2} \int\limits_{t=0}^{ab} tJ_0(t) dt =\dfrac 1{a^2}[tJ_1(t)]_0^{ab} \\ =\dfrac 1{a^2}[ab.J_1(ab)] =\dfrac ba J_1(ab) $

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