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Wolfram Alpha says that this integral does not converge

$$\int^{\pi/2}_{-\pi/2}\frac{\mathrm{d}x}{a+b\sin(x)},$$ $a,b \in \mathbb{R}^+$ with also the condition $b \geq a$.

Is it true? Making a fast calculation using the residue theorem I have found that it is equal to zero.

I have used the fact that $$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$ and the substitution $z = e^{ix}$.

projectilemotion
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Nisba
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    I think it converges if $a > b$. Otherwise, there is a simple pole at $x = \arcsin(- a/b)$. Maybe you want to compute the principal value? – Jennifer Jan 16 '17 at 20:00
  • Mmm, I noticed that the value of the integral is equal to half the value on $-\pi, \pi$. Using the substitution I proposed the poles belongs to the unit circle. If we integrate inside the circle we get zero so shouldn't the integration on the "limit circle" (i.e the unit circle) be zero? – Nisba Jan 16 '17 at 20:03
  • The point is that even if you integrate inside the circle you can't neglect a pole that's right on the contour. – Jennifer Jan 16 '17 at 20:18
  • @jennifer so what to do in such a case? – Nisba Jan 16 '17 at 20:19
  • See http://math.stackexchange.com/questions/721475/general-method-of-integration-when-poles-on-contour or http://math.stackexchange.com/search?q=pole+on+contour – Jennifer Jan 16 '17 at 20:28

2 Answers2

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If $b \ge a > 0$ there is a singularity at $x = \arcsin(-a/b)$, and the integral diverges. However, if $b > a > 0$ it will have a Cauchy principal value.

Robert Israel
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Robert Israel provided the good explanations.

Concerning the antiderivative, using the tangent half-angle substitution, you should arrive to $$I=\int \frac{dx}{a+b\sin(x)}=\int\frac{dt}{at^2+2bt+a}$$ If $b \neq a$, then $$I=\frac{2 \tan ^{-1}\left(\frac{a \tan \left(\frac{x}{2}\right)+b}{\sqrt{a^2-b^2}}\right)}{\sqrt{a^2-b^2}}$$ from which (if $a>b$)$$J=\int^{\pi/2}_{-\pi/2}\frac{dx}{a+b\sin(x)}=\frac{2 \left(\tan ^{-1}\left(\frac{a-b}{\sqrt{a^2-b^2}}\right)+\tan ^{-1}\left(\frac{a+b}{\sqrt{a^2-b^2}}\right)\right)}{\sqrt{a^2-b^2}}$$ Now, using $$\tan^{-1}(p)+\tan^{-1}(q)=\tan ^{-1}\left(\frac{p+q}{1-p\, q}\right)$$ should give (since, here, $p\,q=1$) $$J=\frac \pi{\sqrt{a^2-b^2}} $$

Claude Leibovici
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