Stuck on solving improper integral $\int_{0}^{1} \frac{\log(1+3x^2)}{x^2} dx$

Question

Hello I have to find the convergence and solve this improper integral: $$\int_{0}^{1} \frac{\log(1+3x^2)}{x^2} dx$$

I did the convergence part.Now, for the solving I have some problems.

I did the following so far(used parts formula): $$\int_{0}^{1} \frac{\log(1+3x^2)}{x^2} dx = $$ $$\int_{0}^{1} \log(1+3x^2)\left(\frac{1}{x}\right) dx = $$ $$ -\frac{\log(1+3x^2)}{x}\bigg\rvert^1_0 + \int_{0}^{1} \frac 1x *\frac {6x}{1+3x^2} dx = $$

I'm stuck at this point. Can you give me some help? Thank you.

Answer 1

After using integration by parts, you have this:

$\displaystyle\int_{0}^{1}\dfrac{\log(1+3x^2)}{x^2}\,dx$ $= -\left[\dfrac{\log(1+3x^2)}{x}\right]_{0}^{1} + \displaystyle\int_{0}^{1}\dfrac{1}{x} \cdot \dfrac{6x}{1+3x^2}\,dx$.

Clearly, at $x = 1$, we have $\dfrac{\log(1+3x^2)}{x} = \log 4$.

Also, for $y \approx 0$, we have $\log(1+y) = y + O(y^2)$. Hence, for $x \approx 0$, we have $\log(1+3x^2) = 3x^2 + O(x^4)$, and thus, $\dfrac{\log(1+3x^2)}{x} = 3x + O(x^3)$.

Thus, $\displaystyle \lim_{x \to 0^+}\dfrac{\log(1+3x^2)}{x} = 0$, and so, $\left[\dfrac{\log(1+3x^2)}{x}\right]_{0}^{1} = \log 4 - 0 = 2\log 2$

(Sidenote: If you don't like using big-O notation here, you can use L'Hopital's Rule instead).

Finally, $\displaystyle\int_{0}^{1}\dfrac{1}{x} \cdot \dfrac{6x}{1+3x^2}\,dx$ $= \displaystyle\int_{0}^{1}\dfrac{6}{1+3x^2}\,dx$ $= \displaystyle\int_{0}^{1}\dfrac{2\sqrt{3}}{1+(x\sqrt{3})^2}\sqrt{3}\,dx$

$= \displaystyle\int_{0}^{\sqrt{3}}\dfrac{2\sqrt{3}}{1+u^2}\,du$ $= \displaystyle \left[2\sqrt{3}\arctan u\right]_{0}^{\sqrt{3}}$ $= \dfrac{2\pi\sqrt{3}}{3}$.

Therefore, $\displaystyle\int_{0}^{1}\dfrac{\log(1+3x^2)}{x^2}\,dx = \dfrac{2\pi\sqrt{3}}{3} - 2\log 2$.

Answer 2

$$\int^1_0\dfrac{1}{x}\times\dfrac{6x}{1+3x^2}dx=\int^1_0\times\dfrac{6}{1+3x^2}dx$$ Now, use the fact that $$\int^1_0\dfrac{dx}{1+x^2}=\left.\arctan(x)\right|^1_0$$ With a suitable choice of $u$-substitution.

Answer 3

You can use another way to do calculation. It is much easier. Let $$ I(\alpha)=\int_0^1\frac{\log(1+\alpha x^2)}{x^2}dx. $$ Then $$ I'(\alpha)=\int_0^1\frac{1}{1+\alpha x^2}dx=\frac{1}{\sqrt{\alpha}}\arctan\sqrt{\alpha}. $$ So \begin{eqnarray*} I(3)&=&\int_0^3\frac{1}{\sqrt{\alpha}}\arctan\sqrt{\alpha}d\alpha\\ &=&2\sqrt{\alpha}\arctan\sqrt{\alpha}-\log(1+\alpha)|_0^3\\ &=&2\sqrt{3}\arctan\sqrt{3}-2\log 2\\ &=&\frac{2\pi\sqrt{3}}{3}-2\log 2. \end{eqnarray*}