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I will be happy if someone checks my solution to $$I=\int_{-1}^1 \frac{1}{x-z}\frac{dx}{\sqrt{1-x^2}}=\frac{- \pi}{\sqrt{z^2-1}}$$ where $z \in \mathbb{C} \setminus[-1,1]$.

Here is my computation:

Substituting $x=\cos(t)$, we get $I=\frac{1}{2}\int_{0}^{2 \pi} \frac{dt}{\cos(t)-z}=\frac{1}{i}\int_{|w|=1} \frac{dw}{w^2-2zw+1} $. We know that we have poles at $z \pm \sqrt{z^2-1}$ and since $z - \sqrt{z^2-1}$ is the only pole that lies inside the unite disk, by residue theorem, we get $I= \frac{- \pi}{\sqrt{z^2-1}}$.

Aleph-null
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  • @Dr.SonnhardGraubner, since $z$ is nonzero, I think we get the same answer. – Aleph-null Oct 14 '17 at 07:45
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    @Dr.SonnhardGraubner: $\frac{1}{z^2}z$ is quite a awkward way for writing $\frac{1}{z}$. Should I repeat that simply pasting the output of a CAS is not really helpful? – Jack D'Aurizio Oct 14 '17 at 15:34

1 Answers1

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Substituting $x=\sin(u)\to u=\arcsin(x), dx=\cos(u)du$ and using the formula $1-\sin^2(u)=\cos^2(u) :$

$$ {\displaystyle\int}\dfrac{1}{\left(x-z\right)\sqrt{1-x^2}}\,\mathrm{d}x ={\displaystyle\int}\dfrac{\cos\left(u\right)}{\left(\sin\left(u\right)-z\right)\sqrt{1-\sin^2\left(u\right)}}\,\mathrm{d}u $$

Preparing and performing Weierstrass substitution and $v=\tan\left(\dfrac{u}{2}\right) \to \mathrm{d}u=\dfrac{2}{\sec^2\left(\frac{u}{2}\right)}\,\mathrm{d}v = =\dfrac{2}{v^2+1}\,\mathrm{d}v$ :

$$ ={\displaystyle\int}\dfrac{1}{\frac{2\tan\left(\frac{u}{2}\right)}{\tan^2\left(\frac{u}{2}\right)+1}-z}\,\mathrm{d}u = =-\class{steps-node}{\cssId{steps-node-1}{2}}{\displaystyle\int}\dfrac{1}{zv^2-2v+z}\,\mathrm{d}v $$

Let's solve :

$${\displaystyle\int}\dfrac{1}{zv^2-2v+z}\,\mathrm{d}v = {\displaystyle\int}\dfrac{1}{\left(\sqrt{z}v-\frac{1}{\sqrt{z}}\right)^2+z-\frac{1}{z}}\,\mathrm{d}v$$

Substituting $w=\dfrac{zv-1}{\sqrt{z}\sqrt{z-\frac{1}{z}}} \to \mathrm{d}v=\dfrac{\sqrt{z-\frac{1}{z}}}{\sqrt{z}}\,\mathrm{d}w$

$$=\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{\sqrt{z}\sqrt{z-\frac{1}{z}}}}}{\displaystyle\int}\dfrac{1}{w^2+1}\,\mathrm{d}w =\dfrac{\arctan\left(w\right)}{\sqrt{z}\sqrt{z-\frac{1}{z}}}$$

By reversing substitution for $ w=\dfrac{zv-1}{\sqrt{z}\sqrt{z-\frac{1}{z}}},v=\tan\left(\dfrac{u}{2}\right),\tan\left(\dfrac{\class{steps-node}{\cssId{steps-node-5}{\arcsin\left(x\right)}}}{2}\right)=\dfrac{1-\sqrt{1-x^2}}{x} $ and plugging in the solved integrals :

$$-\class{steps-node}{\cssId{steps-node-4}{2}}{\displaystyle\int}\dfrac{1}{zv^2-2v+z}\,\mathrm{d}v = =-\dfrac{2\arctan\left(\frac{zv-1}{\sqrt{z}\sqrt{z-\frac{1}{z}}}\right)}{\sqrt{z}\sqrt{z-\frac{1}{z}}} =-\dfrac{2\arctan\left(\frac{\frac{z\left(1-\sqrt{1-x^2}\right)}{x}-1}{\sqrt{z}\sqrt{z-\frac{1}{z}}}\right)}{\sqrt{z}\sqrt{z-\frac{1}{z}}} $$

Now, we have the solution to the indefinite integral :

$${\displaystyle\int}\dfrac{1}{\left(x-z\right)\sqrt{1-x^2}}\,\mathrm{d}x =\dfrac{2\arctan\left(\frac{z\sqrt{1-x^2}+x-z}{\sqrt{z^2-1}x}\right)}{\sqrt{z^2-1}}+C$$

Since your one is a definite, let's plug in numbers - You can carry out the calculation yourself :

$$\int_{-1}^1 \frac{1}{x-z}\frac{dx}{\sqrt{1-x^2}} = \Bigg[ \dfrac{2\arctan\left(\frac{z\sqrt{1-x^2}+x-z}{\sqrt{z^2-1}x}\right)}{\sqrt{z^2-1}} \Bigg]_{-1}^1 $$

Rebellos
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