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$$\int_{-1}^{1}\arctan(zx){\mathrm dx\over x\sqrt{1-x^2}}=\pi\sinh^{-1}(z)\tag1$$

$x=\sin{y}$ then $dx=\cos{y}dy$

$$\int\arctan(z\sin{y}){\mathrm dy\over \sin{y}}\tag2$$

$$\sum_{k=0}^{\infty}{(-1)^k\over z^{2k+1}}\int{\mathrm dy\over (\sin{y})^{2k+2}}\tag3$$

How do we show that $(1)=\pi\sinh^{-1}(z)?$

gymbvghjkgkjkhgfkl
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    Hint: simply differentiate both sides with respect to $z$, the resulting integral will be elementary. – pisco Dec 17 '17 at 17:32
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    @mickep After differentiating with respect to $z$, I got $$\int_{-1}^1 \frac{1}{(1+z^2x^2)\sqrt{1-x^2}} dx$$ which is convergent. – pisco Dec 18 '17 at 04:05

1 Answers1

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Setting $$I = \int^1_{-1} \tan^{-1} (zx) \frac{dx}{x \sqrt{1 - x^2}}.$$ If we let $x = \sin \theta$, then $$I = \int^{\pi/2}_{-\pi/2} \frac{\tan^{-1} (z \sin \theta)}{\sin \theta} \, d\theta. \tag1$$ Now from the Maclaurin series expansion for the inverse tangent function, namely $$\tan^{-1} t = \sum^\infty_{n = 0} \frac{(-1)^n t^{2n + 1}}{2n + 1},$$ replacing $t$ with $z \sin \theta$ one has $$\tan^{-1} (z \sin \theta) = \sum^\infty_{n = 0} \frac{(-1)^n z^{2n + 1} \sin^{2n + 1} \theta}{2n + 1}.$$ So (1) becomes $$I = \sum^\infty_{n = 0} \frac{(-1)^n z^{2n + 1}}{2n + 1} \int^{\pi/2}_{-\pi/2} \sin^{2n} \theta \, d\theta = 2 \sum^\infty_{n = 0} \frac{(-1)^n z^{2n + 1}}{2n + 1} \int^{\pi/2}_0 \sin^{2n} \theta d\theta,$$ where the summation and the integration sign have been interchanged and we note the integrand is an even function between symmetric limits.

Since (see here) $$\int^{\pi/2}_0 \sin^{2n} \theta \, d\theta = \frac{\pi (2n)!}{2^{2n + 1} (n!)^2},$$ we have $$I = \pi \sum^\infty_{n = 0} \frac{(-1)^n (2n)!}{2^{2n} (n!)^2 (2n + 1)} z^{2n + 1} = \pi \sinh^{-1} z,$$ where we recognise the infinite sum as corresponding to the Maclaurin series expansion for the inverse hyperbolic sine function (see here).

omegadot
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