Difference between revisions of "Integral of Bessel J for nu=1"
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==Proof== | ==Proof== | ||
Recall, from definition, that | Recall, from definition, that | ||
| + | $$J_0(t) = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k t^{2k}}{(k!)^2 2^{2k}},$$ | ||
| + | and | ||
$$J_1(t) = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kt^{2k+1}}{k! (k+1)! 2^{2k+1}}.$$ | $$J_1(t) = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kt^{2k+1}}{k! (k+1)! 2^{2k+1}}.$$ | ||
Integrating from $0$ to $z$ yields | Integrating from $0$ to $z$ yields | ||
| Line 13: | Line 15: | ||
&=\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^{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)}} \\ | &=\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}$$ | \end{array}$$ | ||
| + | completing the proof. | ||
| + | |||
==References== | ==References== | ||
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Integral of Bessel J for nu=n+1|next=findme}}: $11.1.6$ | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Integral of Bessel J for nu=n+1|next=findme}}: $11.1.6$ | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[Category:Proved]] |
| + | [[Category:Justify]] | ||
Latest revision as of 23:18, 20 February 2018
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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $11.1.6$
