# Problem with ratios of integrals

I have the following integral from a paper I'm reading:

$$f(z)=\frac{\displaystyle\int_0^{\pi/2}\,\tan \alpha\, J_0(z \sin\alpha)\, d\alpha}{\displaystyle \int_0^{\pi/2}\tan\alpha\,d\alpha}$$

where $J_0$ is the Bessel function of first kind and zeroth order. The authors claim that although the integrals in the numerator and the denominator are infinite, their ratio is finite. They then proceed to give the value (no proof) as

$$f(z)=J_0(z)$$

How can this be proven (or is this correct)?

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Let $$f(z, \epsilon) = \frac{ \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) J_0\left(z \sin(\alpha) \right) \mathrm{d} \alpha}{ \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) \mathrm{d} \alpha}$$ I guess the paper defines $f(z)$ as the right limit: $$f(z) \stackrel{\text{def}}{=} \lim_{\epsilon \downarrow 0} f(z,\epsilon)$$ The limit can be evaluated by a L'Hospital's rule: $$\lim_{\epsilon \downarrow 0} \frac{ \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) J_0\left(z \sin(\alpha) \right) \mathrm{d} \alpha}{ \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) \mathrm{d} \alpha} = \lim_{\epsilon \downarrow 0} \frac{ \frac{\mathrm{d}}{\mathrm{d}\epsilon}\int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) J_0\left(z \sin(\alpha) \right) \mathrm{d} \alpha}{ \frac{\mathrm{d}}{\mathrm{d}\epsilon} \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) \mathrm{d} \alpha} = \lim_{\epsilon \downarrow 0} \frac{-\tan\left(\frac{\pi}{2}-\epsilon\right) J_0\left(z \sin\left(\frac{\pi}{2}-\epsilon\right) \right)}{ -\tan\left(\frac{\pi}{2}-\epsilon\right) } = J_0(z)$$