If we use the series defintion of the Bessel function (first kind)$$J_0(z)=\sum_{m=0}^{\infty}{\frac{(-1)^m}{(m!)^2}\left(\frac{z}{2}\right)^{2m}}$$
Can you prove that
$$\lim_{z\to\infty}{J_0(z)=\lim_{z\to\infty} \sum_{m=0}^{\infty}{\frac{(-1)^m}{(m!)^2}{\left(\frac{z}{2}\right)}^{2m}} = 0}$$
Or can this only be done with the integral representation and the asymptotic expansion here?
Limits of First Kind Bessel Function
asymptoticsbessel functionssequences-and-seriesspecial functions
Related Solutions
One can try to do the following derivations. $$\mathrm{Int}=\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx$$ You can simplify the power of the exponent: $$-\left(\frac{x}{\alpha}+\frac{1}{2w^2}(x-\hat{x})^2\right)=-\left(\frac{(x-\mu)^2}{2w^2}+\gamma \right),$$ where $\gamma=\left(\frac{\hat{x}}{\alpha}-\frac{w^2}{2\alpha^2}\right)$, $\mu=\hat{x}-\frac{w^2}{\alpha^2}$. So: $$\mathrm{Int}=\frac{1}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{(-\frac{x}{\alpha}-\frac{1}{2w^2}(x-\hat{x})^2)} \, I_0\left(\frac{x}{\beta}\right)\,dx=\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, x \, e^{-\frac{(x-\mu)^2}{2w^2}} \, I_0\left(\frac{x}{\beta}\right)\,dx$$ Then you can change the variable of integration $y=x-\mu$: $$\mathrm{Int}=\frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty} \, (y+\mu) \, e^{-\frac{y^2}{2w^2}} \, I_0\left(\frac{y+\mu}{\beta}\right)\,dx$$ and make use of the Neumann’s addition theorem: $$\mathop{I_{{\nu}}}\nolimits\!\left(u\pm v\right)=\sum _{{k=-\infty}}^{\infty}(\pm 1)^{k}\mathop{I_{{\nu+k}}}\nolimits\!\left(u\right)\mathop{I_{{k}}}\nolimits\!\left(v\right)$$ setting $\nu=0$, assuming that $y,\mu,\beta\in\mathbb{R}$ and using connection formulas one will get: $$\mathop{I_{0}}\nolimits\!\left(\frac{y+\mu}{\beta}\right)=\sum _{{k=-\infty}}^{\infty}\mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{k}}}\nolimits\!\left(\frac{\mu}{\beta}\right)=\mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{0}}}\nolimits\!\left(\frac{\mu}{\beta}\right)+2\sum _{{k=1}}^{\infty}\mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\mathop{I_{{k}}}\nolimits\!\left(\frac{\mu}{\beta}\right)$$
$$\mathrm{Int}=\! \frac{\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)e^{-\gamma}}{\sqrt{2\pi w^2}}\!\int_{-\infty}^{+\infty} \, (y+\mu)\! e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx\! +\!\frac{2 e^{-\gamma}}{\sqrt{2\pi w^2}}\int_{-\infty}^{+\infty}\! (y+\mu) \, e^{-\frac{y^2}{2w^2}} \! \sum _{k=1}^{\infty}\!\mathop{I_k}\!\!\left(\frac{y}{\beta}\right)\!\!\mathop{I_k}\nolimits\!\left(\frac{\mu}{\beta}\right) \,dx$$ Or reorganising terms: $\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)$
$$\mathrm{Int}=\! \frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\!\left(\mathop{I_0}\nolimits\!\left(\frac{\mu}{\beta}\right)\int_{-\infty}^{+\infty} \, (y\!+\!\mu)\! e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx\! +\!2\!\sum _{k=1}^{\infty}\!\mathop{I_k}\!\left(\frac{\mu}{\beta}\right)\!\int_{-\infty}^{+\infty}\! (y\!+\!\mu) \, e^{-\frac{y^2}{2w^2}} \! \!\mathop{I_k}\!\left(\frac{y}{\beta}\right)\!\!\,dx \right)$$ Let's set $$\mathrm{Int}_k(\mu,\alpha,\beta)\!=\!\int_{-\infty}^{+\infty} \, (y\!+\!\mu) e^{-\frac{y^2}{2w^2}}\! \mathop{I_{k}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx$$ $$\mathrm{Int}_0(\mu,\alpha,\beta)\!=\!\int_{-\infty}^{+\infty} \, (y\!+\!\mu) e^{-\frac{y^2}{2w^2}}\! \mathop{I_{0}}\nolimits\!\left(\frac{y}{\beta}\right)\!dx$$ Here one can play with the evenness/unevenness of the functions under integrals (as I did at first), or just use CAS and get:
$$\mathrm{Int}_k(\mu,\alpha,\beta)\!=\!\sqrt{\frac{\pi \alpha ^2}{8 \beta ^2}}\! e^{\frac{\alpha ^2}{4 \beta ^2}} \left(\!\!2 \beta \left(\!\!(-1)^k+1\!\right)\! \mu I_{\frac{k}{2}}\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)+\alpha ^2 \left(\!1-(-1)^k\!\right)\!\! \left(\!\!I_{\frac{k-1}{2}}\!\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)+I_{\frac{k+1}{2}}\!\left(\!\!\frac{\alpha ^2}{4 \beta ^2}\!\!\right)\!\!\right)\!\!\right)$$ $$\mathrm{Int}_0(\mu,\alpha,\beta)\!=\! \sqrt{2 \pi } \alpha \mu e^{\frac{\alpha ^2}{4 \beta ^2}} I_0\left(\frac{\alpha ^2}{4 \beta ^2}\right)$$ So the original integral can be represented in the following way: $$\mathrm{Int}(\mu,\alpha,\beta)\!=\! \frac{e^{-\gamma}}{\sqrt{2\pi w^2}}\left(\mathrm{Int}_0(\mu,\alpha,\beta)+ 2\!\sum _{k=1}^{\infty}\mathop{I_k}\left(\frac{\mu}{\beta}\right) \mathrm{Int}_k(\mu,\alpha,\beta) \right)$$ And I am not shure that there is any chance to simplify it futher (meaning finding the sum of the series).
The integral does not admit a closed form, I am afraid. It is discussed in the vol. 2 of Bateman's "Higher Transcendental functions", section 7.7.3, formula 15, which gives $$ 2^{\mu + \nu} \alpha^{-\mu} \beta^{-\nu} \gamma^{\lambda + \mu+\nu} \Gamma\left(\nu+1\right) \int_0^\infty J_{\mu}(\alpha t) J_\nu(\beta t) \mathrm{e}^{-\gamma t} t^{\lambda -1} \mathrm{d} t = \sum_{m=0}^\infty \frac{\Gamma\left(2m+\lambda+\mu+\nu\right)}{m! \Gamma(m+\mu+1)} \cdot {}_2F_1\left(-m,-m-\mu; \nu+1; \frac{\beta^2}{\alpha^2}\right) \left( -\frac{\alpha^2}{4 \gamma^2} \right)^m $$ In your case, $\lambda = 1$, $\mu=0$, $\nu=1$, $\gamma=1$: $$ \frac{2}{\beta} \int_0^\infty J_0(\alpha t) J_1(\beta t) \mathrm{e}^{-t} \mathrm{d}t = \sum_{m=0}^\infty \frac{\Gamma\left(2m+2\right)}{m! \Gamma(m+1)} \cdot {}_2F_1\left(-m,-m; 2; \frac{\beta^2}{\alpha^2}\right) \left( -\frac{\alpha^2}{4} \right)^m $$
By expanding the Bessel function $J_1(b t)$ in its defining series and integrating term-wise we can find other series representations: $$ \int_0^\infty J_0(a t) J_1(b t) \mathrm{e}^{-t} \mathrm{d}t = \frac{b}{2} \sum_{m=0}^\infty \binom{2m+1}{m} \frac{\left(-\frac{b^2}{4}\right)^m}{(1+a^2)^{2m+3/2}} \cdot {}_2F_1\left(-m, -m-\frac{1}{2}; 1; -a^2\right) $$ where, additionally, the Euler's transformation of the Gauss' hypergeometric function had been used.
Best Answer
This is not an answer but a comment on why this isn't a good idea even if it might be doable. The series representation is a Taylor series around $z = 0$; it's well-suited to computations for small $z$ but there's no reason to expect it to be particularly well-suited to computations for large $z$, given all the signs. This situation already occurs for the simpler series
$$\exp(-z) = \sum_{n=0}^{\infty} (-1)^n \frac{z^n}{n!}$$
where we have $\lim_{z \to \infty} \exp(-z) = 0$; can you tell that the limit is zero from this series? The problem is that for $z$ large a tremendous amount of cancellation needs to be happening among the terms; the largest term occurs when $n \approx z$ and grows like $\exp(z)$, yet the final result ends up converging to zero (and quite rapidly) as $z \to \infty$. This is really not clear at all from the series, and I don't know a direct argument from the series at all. It's just not well-suited to understanding what happens when $z$ is large even if it does technically converge and we should use something else.
Similarly, in this case for $z$ large a tremendous amount of cancellation needs to be happening. The largest term occurs when $m \approx \frac{z}{2}$ and again grows like $\exp(z)$, and again the final result needs to end up converging to zero.