The function indicated in the following equation represents the Bessel function of the first kind of order ''$\nu$''.
\begin{eqnarray}
J_{\nu}(x)=\sum_{k=0}^{+\infty} (-1)^k\frac{(\frac{x}{2})^{\nu+2k}}{k!\;\Gamma(\nu+k+1)}
\end{eqnarray}
The ''$\Gamma(s)$'' above is called as the gamma function which is indicated in the equation (2).
\begin{eqnarray}
\Gamma(s)=\int_0^{+\infty}e^{-x}x^{s-1}dx\;\;\;(s>0)
\end{eqnarray}
Applying the integration by parts, this equation can be converted to the equation (3).
\begin{eqnarray}
\Gamma(s+1)=\int_0^{+\infty}e^{-x}x^sdx=[-e^{-x}x^s]_0^{+\infty}+s\int_0^{+\infty}x^{s-1}e^{-x}dx=s\Gamma(s) \nonumber
\end{eqnarray}
Accordingly, the following equation can be indicated.
\begin{eqnarray}
\Gamma(s)=(s-1)\Gamma(s-1)=(s-1)(s-2)\Gamma(s-2)=\cdots \nonumber
\end{eqnarray}
This equation can be generally indicated as shown in the equation (4).
\begin{eqnarray}
\Gamma(s)=(s-1)(s-2)\cdots(s-k)\Gamma(s-k)\;\;\;(k < s)
\end{eqnarray}
Then, in terms of the natural number, ''$n$'', the following equation can be indicated.
\begin{eqnarray}
\Gamma(n+1)=n(n-1)\cdots2\cdot1\Gamma(1)\;\;\;\;\;;\;\Gamma(1)=\int_0^{+\infty}e^{-x}dx=[-e^{-x}]_0^{+\infty}=1 \nonumber
\end{eqnarray}
Thus, the equation (5) can be derived.
\begin{eqnarray}
\Gamma(n+1)=1\cdot2\cdot3\cdots\cdots n=n!
\end{eqnarray}
In case that the order, ''$\nu$'' is the nonnegative integer, ''$n$'' in the equation (1), the following equation (6) can be shown, taking account of the equation (5).
\begin{eqnarray}
J_n(x)=\sum_{k=0}^{+\infty}(-1)^k\frac{(\frac{x}{2})^{n+2k}}{k!(n+k)!}
\end{eqnarray}
In case of the ''$\nu = - n$'' which is the negative integer, the function to be defined in the equation (1) takes the following shape of the equation (7), taking the account that ''$1/\Gamma(s)$'' equals ''0'' with reference to ''$s$=0, −1, −2, …''.
\begin{eqnarray}
J_{-n}(x)=\sum_{k=n}^{+\infty}(-1)^k\frac{(\frac{x}{2})^{-n+2k}}{k!\Gamma(-n+k+1)}=\sum_{k=n}^{+\infty} (-1)^k\frac{(\frac{x}{2})^{-n+2k}}{k!(-n+k)!}
\end{eqnarray}
Substituting ''$k$'' with ''$p+n$'' here, the equation (8) can be indicated.
\begin{eqnarray}
J_{-n}(x)=(-1)^n\sum_{l=0}^{+\infty} (-1)^l\frac{(\frac{x}{2})^{n+2l}}{(l+n)!l!}=(-1)^nJ_n(x)
\end{eqnarray}
Then, using the Taylor series expansion next, the equations (9) can be derived.
\begin{eqnarray}
e^{-x/(2z)}=\sum_{k=0}^\infty\frac{(-\frac{x}{2z})^k}{k!}\;\;\;\;\;\;\;\;\;e^{xz/2}=\sum_{l=0}^\infty \frac{(\frac{xz}{2})^l}{l!}
\end{eqnarray}
Expanding the right side of the both equations of the equation (9), multiplying what was expanded in one equation by what was expanded in another equation with respect to each term, and sorting out the terms of the order of the identical ''$z$'', the equation (10) can be obtained.
\begin{eqnarray}
e^{(z-\frac{1}{z})x/2}&=&\sum_{k,l} \frac{(-1)^k(\frac{x}{2})^{k+1}z^{l-k}}{k!\;l!} \nonumber \\
&=&\sum_{n=-\infty}^{+\infty}z^n \sum_{k \ge 0 , l \ge 0 , l-k=n}\frac{(-1)^k(\frac{x}{2})^{k+l}}{k!\;l!}=\sum_{n=-\infty}^{+\infty}z^n\sum_k \frac{(-1)^k(\frac{x}{2})^{n+2k}}{k!(n+k)!} \nonumber
\end{eqnarray}
In addition, in the internal sum of the equation which is the last equation but one, it has the ''$l - k= n$'' relationship. Accordingly, ''$l=n + k$'' can be placed by obtaining the sum in terms of the either index, ''$k$''. In terms of the internal sum of the last equation, the calculation to obtain the sum is executed with reference to the all integers, ''$k$'' which fulfill the conditions,
''$k \ge 0$'' \& ''$n + k \ge 0$''.
Accordingly, in case of ''$n \ge 0$'', it becomes ''$\sum_{k=0}^{+\infty}$'', in case of
''$n = - m < 0$'', it becomes ''$\sum_{k=m}^{+\infty}$''.
Thus, through the all of the cases, the internal sum of the equation can be obtained as ''$J_n(x)$'' of the equations (6) & (7). Herewith, the following equation (11) can be obtained.
\begin{eqnarray}
e^{(z-\frac{1}{z})x/2}=\sum_{-\infty}^{+\infty} J_n(x)z^n
\end{eqnarray}
Assuming that ''$z=e^{i\phi}$'', the equation (12) can be obtained.
\begin{eqnarray}
e^{ixsin\phi}=\sum_{-\infty}^{+\infty} J_n(x)e^{in\phi}
\end{eqnarray}
Herewith, substituting that ''$x= A$'', ''$\phi= \omega t$'' & ''$\alpha = A sin \omega t$'', the equation (13) can be obtained.
\begin{eqnarray}
e^{i\alpha}=cos(Asin\omega t)+isin(Asin\omega t)=\sum_{n=-\infty}^{+\infty} J_n(A)e^{in\omega t}
\end{eqnarray}
Considering that it becomes ''$J_{-n}(A)=(-1)^nJ_n(A)$'' when this is divided into the real & imaginary sections, the following equations (14) & (15) can be obtained.
\begin{eqnarray}
cos(Asin\omega t)&=&\sum_{-\infty}^{+\infty}J_n(A)cos\;n\omega t \nonumber \\
&=&J_0(A)+\sum_{n=1}^{+\infty}[J_n(A)+J_{-n}(A)]cos\;n\omega t \nonumber \\
&=&J_0(A)+2\sum_{m=1}^{+\infty} J_{2m}(A)cos\;2m\omega t \\
sin(Asin\omega t)&=&\sum_{-\infty}^{+\infty}J_n(A)sin\;n\omega t \nonumber \\
&=&\sum_{n=1}^{+\infty}[J_n(A)-J_{-n}(A)]sin\;n\omega t \nonumber \\
&=&2\sum_{m=0}^{+\infty} J_{2m+1}(A)sin\;(2m+1)\omega t
\end{eqnarray}
These two equations can be respectively expanded as follows:
\begin{eqnarray}
cos\alpha&=&cos(Asin\omega t)=J_0(A)+2J_2(A)cos2\omega t+2J_4(A)cos4\omega t+\cdots \nonumber \\
sin\alpha&=&sin(Asin\omega t)=2J_1(A)sin \omega t+2J_3(A)sin3\omega t +2J_5(A)sin5\omega t+\cdots \nonumber
\end{eqnarray}
With the use of these equations and the sum of the products formula of trigonometric function (Section 5 List of Mathematical Formulas 5-1-7), the following formula expansions can be done.
\begin{eqnarray}
\int_0^T&\alpha&\;cos \alpha dt=\int_0^T Asin\omega t\;cos(Asin\omega t) dt \nonumber \\
&=&A\int_0^Tsin\omega t \{J_0(A)+2J_2(A)cos2\omega t+2J_4(A)cos4\omega t+\cdots\}dt \nonumber \\
&=&A\{J_0(A)\int_0^Tsin\omega t dt+2J_2(A)\int_0^Tsin\omega t cos2\omega t dt \nonumber \\
&\;&+2J_4(A)\int_0^Tsin\omega t cos4\omega_t dt+\cdots\} \nonumber \\
&=&A[J_0(A)\int_0^Tsin\omega t dt +J_2(A)\int_0^T\{sin3\omega t-sin\omega t\}dt \nonumber \\
&\;&+J_4(A)\int_0^T\{sin5\omega t-sin3\omega t\}dt+\cdots] \nonumber \\
&=&A\{\frac{J_0(A)}{\omega}[-cos\omega t]_0^T+J_2(A)(\frac{1}{3\omega}[-cos3\omega t]_0^T-\frac{1}{\omega}[-cos\omega t]_0^T) \nonumber \\
&\;&+J_4(A)(\frac{1}{5\omega}[-cos5\omega t]_0^T-\frac{1}{3\omega_t}[-cos3\omega t]_0^T)+\cdots\} \nonumber \\
&=&0 \nonumber
\end{eqnarray}
\begin{eqnarray}
\int_0^T&\alpha& sin\alpha dt=\int_0^T Asin\omega t\;sin(Asin\omega t)dt \nonumber \\
&=&A\int_0^Tsin\omega t \{2J_1(A)sin\omega t+2J_3(A)sin3\omega t+2J_5(A)sin5\omega t+\cdots\}dt \nonumber \\
&=&2A\{\int_0^TJ_1(A)sin^2\omega tdt+J_3(A)\int_0^Tsin\omega t\;sin3\omega tdt \nonumber \\
&\;&+J_5(A)\int_0^Tsin\omega t\;sin5\omega tdt+\cdots\} \nonumber \\
&=&A\{J_1(A)\int_0^T(1-cos2\omega t)dt +J_3(A)(\int_0^Tcos2\omega tdt-\int_0^Tcos4\omega tdt) \nonumber \\
&\;&+J_5(A)(\int_0^Tcos4\omega tdt-\int_0^Tcos6\omega tdt)+\cdots\} \nonumber \\
&=&A\{J_1(A)([t]_0^T-\frac{1}{2\omega}[sin2\omega t]_0^T)+J_3(A)(\frac{1}{2\omega}[sin2\omega t]_0^T-\frac{1}{4\omega}[sin4\omega t]_0^T) \nonumber \\
&\;&+J_5(A)(\frac{1}{4\omega}[sin4\omega t]_0^T-\frac{1}{6\omega}[sin6\omega t]_0^T)+\cdots\} \nonumber \\
&=&ATJ_1(A) \nonumber
\end{eqnarray}