Integral of Bessel J for nu=1

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Theorem

The following formula holds: $$\displaystyle\int_0^z J_1(t) \mathrm{d}t = 1-J_0(z),$$ where $J_1$ denotes the Bessel function of the first kind.

Proof

Recall, from definition, that $$J_0(t) = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k t^{2k}}{(k!)^2 2^{2k}},$$ and $$J_1(t) = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kt^{2k+1}}{k! (k+1)! 2^{2k+1}}.$$ Integrating from $0$ to $z$ yields $$\begin{array}{ll} \displaystyle\int_0^z J_1(t) \mathrm{d}t &= \displaystyle\int_0^z \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kt^{2k+1}}{k! (k+1)! 2^{2k+1}} \mathrm{d}t \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k! (k+1)! 2^{2k+1}} \displaystyle\int_0^z t^{2k+1} \mathrm{d}t \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k t^{2k+2}}{(k+1)!^2 2^{2k+2}} \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k t^{2(k+1)}}{(k+1)!^2 2^{2(k+1)}} \\ &\stackrel{\mathrm{reindex}}{=} \displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1} t^{2k}}{(k!)^2 2^{2k}} \\ &=1-\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^{k} t^{2k}}{(k!)^2 2^{2k}} \\ &=1-J_0(z), \end{array}$$ completing the proof.

References

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