오늘의 계산 62 - A Bizarre Integral

While calculating a specific problem, I succeeded in proving a more general problem.

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Proposition. For $0<r<1$ and $r<s$, the following holds:[1]

$${\int }_{-1}^1\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log \left|\frac{1+2rsx+(r^2+s^2-1)x^2}{1-2rsx+(r^2+s^2-1)x^2}\right|dx=4\pi \arcsin r.$$

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Proof. We divide the proof into several steps.

Step 1. (Case reduction by analytic continuation)

We first remark that given $r$ and $s$, we always have

$$\begin{array}{}\text{(2)}& \underset{-1\le x\le 1}{min}\{1\pm 2rsx+(r^2+s^2-1)x^2\}>0.\end{array}$$

Indeed, it is not hard to check if we utilize the following equality

$$1\pm 2rsx+(r^2+s^2-1)x^2=(1\pm rsx{)}^2-(1-r^2)(1-s^2)x^2.$$

Then $\text{(2)}$ allows us to extend $s\mapsto I(r,s)$ as a holomorphic function on some open set containing the line segment $(r,\mathrm{\infty })\subset \mathbb{C}$. Then by the principle of analytic continuation, it is sufficient to prove that $\text{(1)}$ holds for $r<s<1$.

Step 2. (Integral representation of $I$)

Put $r=\sin \alpha$ and $s=\sin \beta$, where $0<\alpha <\beta <\frac{\pi }{2}$. Then

$$\begin{array}{rl}I(r,s)& ={\int }_{-1}^1\frac{1+x}{x\sqrt{1-x^2}}\log \left(\frac{1+2rsx+(r^2+s^2-1)x^2}{1-2rsx+(r^2+s^2-1)x^2}\right)dx\\ & ={\int }_0^1\frac{2}{x\sqrt{1-x^2}}\log \left(\frac{1+2rsx+(r^2+s^2-1)x^2}{1-2rsx+(r^2+s^2-1)x^2}\right)dx(\because \text{ parity})\\ & ={\int }_1^{\mathrm{\infty }}\frac{2}{\sqrt{x^2-1}}\log \left(\frac{x^2+2rsx+(r^2+s^2-1)}{x^2-2rsx+(r^2+s^2-1)}\right)dx(x\mapsto x^{-1})\\ & ={\int }_0^1\frac{2}{t}\log \left(\frac{{(t+t^{-1})}^2+4rs(t+t^{-1})+4(r^2+s^2-1)}{{(t+t^{-1})}^2-4rs(t+t^{-1})+4(r^2+s^2-1)}\right)dt,\end{array}$$

where in the last line we utilized the substitution $x=\frac{1}{2}(t+t^{-1})$. If we introduce the quartic polynomial

$$\begin{array}{r}p(t)=t^4+4rs{t}^3+(4r^2+4s^2-2)t^2+4rst+1,\end{array}$$

then by the property $p(1/t)=t^{-4}p(t)$, we can simplify

$$\begin{array}{rl}I(r,s)& =2{\int }_0^1\frac{\log p(t)-\log p(-t)}{t}dt={\int }_0^{\mathrm{\infty }}\frac{\log p(t)-\log p(-t)}{t}dt\\ & =-{\int }_0^{\mathrm{\infty }}(\frac{p^{\prime }(t)}{p(t)}+\frac{p^{\prime }(-t)}{p(-t)})\log tdt=-\frac{1}{2}\mathrm{\Re }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{p^{\prime }(z)}{p(z)}+\frac{p^{\prime }(-z)}{p(-z)})\log zdz,\end{array}$$

where we choose the branch cut of $\log$ in such a way that it avoids the upper-half plane
$$\mathbb{H}=\{z\in \mathbb{C}:\mathrm{\Im }z>0\}.$$

Step 3. (Residue calculation)

Now let $Z_{+}$ be the set of zeros of $p(z)$ in $\mathbb{H}$ and $Z_{-}$ be the set of zeros of $p(z)$ in $-\mathbb{H}$. Since
$$f(z):=(\frac{p^{\prime }(z)}{p(z)}+\frac{p^{\prime }(-z)}{p(-z)})\log z=O\left(\frac{\log z}{z^2}\right)\text{as }z\to \mathrm{\infty },$$
by replacing the contour of integration by a semicircle of sufficiently large radius, it follows that

$$\begin{array}{r}I(r,s)=-\frac{1}{2}\mathrm{\Re }[2\pi i\sum _{z_0\in Z_{+}}{\mathrm{Res}}_{z=z_0}f(z)+2\pi i\sum _{z_0\in -Z_{-}}{\mathrm{Res}}_{z=z_0}f(z)].\end{array}$$

But by a simple calculation, together with the condition $0<\alpha <\beta <\frac{\pi }{2}$, we easily notice that the zeros of $p(z)$ are exactly

$$e^{\pm i(\alpha +\beta )}\text{and}-e^{\pm i(\alpha -\beta )}.$$

So we have

$$Z_{+}=\{e^{i(\beta +\alpha )},-e^{-i(\beta -\alpha )}\}\text{and}{Z}_{-}=\{e^{-i(\beta +\alpha )},-e^{i(\beta -\alpha )}\}$$

and therefore

$$\begin{array}{rl}I(r,s)& =-\frac{1}{2}\mathrm{\Re }2\pi i[\sum _{z_0\in Z_{+}}\log z_0+\sum _{z_0\in -Z_{-}}\log z_0]\\ & =-\frac{1}{2}\mathrm{\Re }2\pi i\{\log e^{i(\beta +\alpha )}+\log e^{i(\pi -\beta +\alpha )}-\log e^{i(\pi -\beta -\alpha )}-\log e^{i(\beta -\alpha )}\}\\ & =-\frac{1}{2}\mathrm{\Re }2\pi i\{i(\beta +\alpha )+i(\pi -\beta +\alpha )-i(\pi -\beta -\alpha )-i(\beta -\alpha )\}\\ & =\pi \{(\beta +\alpha )+(\pi -\beta +\alpha )-(\pi -\beta -\alpha )-(\beta -\alpha )\}\\ & =4\pi \alpha =4\pi \arcsin r.\end{array}$$

This completes the proof. $◻$

참고문헌

1. Laila Podlesny, Integral ${\int }_{-1}^1\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln \left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)dx$ - Math StackExchange