Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


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

share|cite|improve this question

1 Answer 1

up vote 2 down vote accepted

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: $$ \lim_{\epsilon \downarrow 0} \frac{ \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) J_0\left(z \sin(\alpha) \right) \mathrm{d} \alpha}{ \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) \mathrm{d} \alpha} = \lim_{\epsilon \downarrow 0} \frac{ \frac{\mathrm{d}}{\mathrm{d}\epsilon}\int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) J_0\left(z \sin(\alpha) \right) \mathrm{d} \alpha}{ \frac{\mathrm{d}}{\mathrm{d}\epsilon} \int\limits_0^{\frac{\pi}{2}-\epsilon} \tan(\alpha) \mathrm{d} \alpha} = \lim_{\epsilon \downarrow 0} \frac{-\tan\left(\frac{\pi}{2}-\epsilon\right) J_0\left(z \sin\left(\frac{\pi}{2}-\epsilon\right) \right)}{ -\tan\left(\frac{\pi}{2}-\epsilon\right) } = J_0(z) $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.