I hope someone helps me with this. Utilizing the parametric derivative, calculate the improper integral

$$I(\alpha)=\int _0^1\frac{\ln(1-\alpha^2x^2)}{x^2\sqrt{1-x^2}}\,\mathrm dx$$

The farthest I could get is differentiating w.r.t. $\alpha$, which results

$$I'(\alpha)=-\int_0^1\frac{2\alpha}{(1-\alpha^2x^2)\sqrt{1-x^2}}\,\mathrm dx$$

I don't know how to integrate this perhaps I differentiated wrong. Can someone help out?


1 Answer 1


For $a\in [0,1] $, let $$I(a) = \int_0^1 \frac{\ln(1-a^2x^2)}{x^2\sqrt{1-x^2}}\, \mathrm dx $$

Differentiating both sides w.r.t. $a$, we get $$I'(a) = -2a\int_0^1 \frac{\mathrm dx}{(1-a^2x^2)\sqrt{1-x^2}}$$

Now substituting $x=\sin u$, we get

$$\begin{align} I'(a) &= -2a \int_0^{\pi/2}\frac{\mathrm du}{1-a^2\sin^2u} \\ &= -2a \int_0^{\pi/2}\frac{\sec^2u}{\sec^2u-a^2\tan^2u}\,\mathrm du \\ &\overset{ \tan u =t }{=} -2a \int_0^\infty \frac{\mathrm dt}{1+(1-a^2)t^2} \\ I'(a) &= -\frac{\pi a}{\sqrt{1-a^2}}\end{align}$$

Now keeping in mind that $I(0)=0$, we integrate both sides w.r.t. $a$ from $0$ to $a$.

$$\begin{align}\int_0^a I'(a)\,\mathrm da &=-\pi \int_0^a\frac a{\sqrt{1-a^2}}\,\mathrm da \\ I(a) &= \pi\left[ \sqrt{1-a^2}\right]_0^a \\ I(a) &= \pi(\sqrt{1-a^2}-1)\end{align}$$

  • 1
    $\begingroup$ nice (+1). We could also use contour integration to find $I'(a)$ $\endgroup$
    – Svyatoslav
    Oct 22, 2021 at 12:04
  • $\begingroup$ @Svyatoslav Yep. $\endgroup$ Oct 22, 2021 at 13:27

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