1

Can anyone help me calculate the integral using differentiation with respect to a parameter: $$I(a) = \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx, |a|<1$$

UPD: I try to solve like this: \begin{align*} I'(a)&=\int_{0}^{1} \frac{2ax^2}{\sqrt{1-x^2}(1+a^2x^2)} dx \\&=2a\left(\frac{1}{a^2}\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}} - \frac{1}{a^2}\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}(a^2x^2+1)}\right) \\&= 2a\left(\frac{1}{a^2}\arcsin(x)|_0^1-\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}(a^2x^2+1)}\right) \end{align*} Then I try to solve the second integral: I am doing a variable change like this: $$x=\sin(x) ,y=\arcsin(x),dx=\cos(y)dy$$ After that, I solve this integral and finally get: $$\frac{2ax^2\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}$$ And the problem is when substituting 1 into the result: $$\lim_{x\to 1}{\frac{2ax^2\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}$$ Nevertheless, this is a difficult way. Is there an easy way?

Quanto
  • 97,352
Michael
  • 41
  • 3

1 Answers1

1

Note

$$\lim_{x\to 1}{\frac{\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}= \frac{\arctan^{-1}(\infty)}{\sqrt{1+a^2}}= \frac{\frac\pi2}{\sqrt{1+a^2}} $$

Then, you have

$$I’(a)=\frac\pi a\left(1-\frac1{\sqrt{1+a^2}} \right) $$ and

$$ \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx = \int_0^a I’(t) dt =\pi \int_0^a \frac1t\left(1-\frac1{\sqrt{1+t^2}} \right) dt \\ = \pi\ln \frac {\sqrt{1+a^2}+1}2 $$

Quanto
  • 97,352