0

Integrate: $$\int_0^1 \frac{x^2}{\ln(1+x^2)}\;dx$$

How do I calculate this?

creative
  • 3,539
MrJam
  • 1
  • @MrJam Integrate how? It is an improper integral? How do you know that it exists, for example? – AnyAD Oct 11 '18 at 08:23
  • Since $\log(1+x^2)\approx \sqrt{1+x^2}-\frac{1}{\sqrt{1+x^2}}$, your integral is close to $\frac{1}{2}\left[\sqrt{2}+\log(1+\sqrt{2})\right]$. – Jack D'Aurizio Oct 11 '18 at 18:15

1 Answers1

1

I do not think that a closed form would exist (even using special functions).

What you could do is to use Taylor series expansions $$\log(1+x^2)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\frac{x^{10}}{5}-\frac{x^{12}}{6}+O\left(x^{14}\right)$$ $$\frac{x^2}{\log \left(x^2+1\right)}=\frac{x^2}{x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\frac{x^{10}}{5}-\frac{x^{12}}{6}+O\left(x^{14}\right) }$$ Now, long division to make $$\frac{x^2}{\log \left(x^2+1\right)}=1+\frac{x^2}{2}-\frac{x^4}{12}+\frac{x^6}{24}-\frac{19 x^8}{720}+\frac{3 x^{10}}{160}+O\left(x^{12}\right)$$ Integrate termwise and use the bounds.

Claude Leibovici
  • 260,315