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I need help showing that this integral converges $$\int_0^1 \frac{\ln(1-\alpha ^2x^2)}{\sqrt{1-x^2}}dx$$ for $|\alpha| \le 1$

I thought of using comparison but I don't know what to use.

I would really appreciate some help

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  • $\begingroup$ Have you tried splitting up the domain of integration and checking for convergence around $x = 1$, which seems to be the only point of issue. $\endgroup$
    – mattos
    Nov 22 '19 at 1:07
  • $\begingroup$ Near $x=1$, you might want to compare the integrand with $1/(1-x^2)^{3/4}$. $\endgroup$
    – irchans
    Nov 22 '19 at 1:24
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    $\begingroup$ This might be of interest. $\endgroup$
    – Mnifldz
    Nov 22 '19 at 1:49
  • $\begingroup$ May be of help to note that $$\log(1-\alpha^2x^2)=\log(1-\alpha x)+\log(1+\alpha x).$$ $\endgroup$
    – Allawonder
    Nov 22 '19 at 4:50
  • $\begingroup$ It is an application of Newton 2nd theorem of derivatives of integral function and solutions of intigral from that. $\endgroup$ Nov 22 '19 at 5:56
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Looking at the integrand, use Taylor series to show that, close to $0$ $$\frac{\ln(1-\alpha ^2x^2)}{\sqrt{1-x^2}}=-\alpha ^2 x^2+O\left(x^4\right)$$ and, close to $1$, $$\frac{\ln(1-\alpha ^2x^2)}{\sqrt{1-x^2}}=-\frac{\log \left(1-\alpha ^2\right)}{\sqrt{2(1- x)}}+O((1-x)^{\frac 12})$$ So, no problem at the bounds and the integral is finite.

In fact $$\int_0^1 \frac{\ln(1-\alpha ^2x^2)}{\sqrt{1-x^2}}\,dx=\pi \log \left(\frac{1+\sqrt{1-\alpha ^2}}{2} \right)$$ as long as $\Im(\alpha )\neq 0\lor -1<\Re(\alpha )<1$.

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Here is a proof of convergence by direct comparison.

The integrand is continuous for $0 \leqslant x \leqslant 1-\epsilon < 1$ and

$$\tag{*}\int_0^{1-\epsilon}\frac{-\ln(1-\alpha^2x^2)}{\sqrt{1-x^2}}\, dx = \int_0^{1-\epsilon}\frac{-\ln(1-|\alpha| x)}{\sqrt{1-x^2}}\, dx - \int_0^{1-\epsilon}\frac{\ln(1+|\alpha| x)}{\sqrt{1-x^2}}\, dx$$

The second integral on the RHS of (*) converges as $\epsilon \to 0$ since

$$\frac{\ln(1+|\alpha| x)}{\sqrt{1-x^2}} = \frac{\ln(1+|\alpha| x)}{\sqrt{1-x}\sqrt{1+x}} \leqslant \frac{\ln(2)}{\sqrt{1-x}},$$

and $\frac{1}{\sqrt{1-x}}$ is improperly integrable over $[0,1]$.

For the first integral on the RHS of (*), since $|\alpha| \leqslant 1$ we have

$$|\alpha|x \leqslant x \implies 1- |\alpha|x \geqslant 1- x\implies -\ln(1-|\alpha|x) \leqslant - \ln(1-x),$$

and

$$\frac{-\ln(1-|\alpha| x)}{\sqrt{1-x^2}} \leqslant \frac{-\ln(1-x)}{\sqrt{1-x}}$$

If we can show that $\frac{-\ln(1-x)}{\sqrt{1-x}}$ is improperly integrable over $[0,1]$ then the first integral on the RHS of (*) also converges as $\epsilon \to 0$ by the comparison test.

Since $-\ln(y) \leqslant \frac{1-y}{y}$ for $0 < y < 1$ we have

$$\tag{**}\int_0^{1-\epsilon}\frac{-\ln(1-x)}{\sqrt{1-x}} \, dx = \int_\epsilon^1 \frac{-\ln(x)}{\sqrt{x}} \, dx = \int_\epsilon^1 \frac{-4\ln(x^{1/4})}{\sqrt{x}} \, dx \\ \leqslant \int_\epsilon^1 \frac{4(1-x^{1/4})}{x^{1/4}}\frac{1}{\sqrt{x}} \, dx \leqslant \int_\epsilon^1 \frac{4}{x^{3/4}} \, dx$$

Since the integral on the RHS converges as $\epsilon \to 0$, we have proved convergence of both integrals on the RHS of (*) and convergence of

$$\int_0^{1}\frac{\ln(1-\alpha^2x^2)}{\sqrt{1-x^2}}\, dx $$

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With $t\in[0,1]$, the integral is equivalent to

$$I(t)=\int_0^1 \frac{\ln(1-tx^2)}{\sqrt{1-x^2}}dx$$

$$|I(t)|\le|I(1)|=\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}}dx=-2\int_0^{\pi/2}\ln\cos u\>du=\pi\ln 2$$

where the substitution $u=\sin x$ is used.

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