# About the integral $\int_{0}^{2\pi} \ln(s^2x^2 + 1) \, ds$

I've encountered an integral that has piqued my interest and am seeking insights or methods for its evaluation. The integral in question is:

$$\int_{0}^{2\pi} \ln(s^2x^2 + 1) \, ds$$

where ($$x$$) is treated as a constant parameter. This integral arises in a specific context I'm studying, and I'm particularly interested in understanding how its value changes with different values of ($$x$$).

Despite my efforts, I've found that this integral does not yield easily to standard techniques or elementary functions for its antiderivative. Numerical methods and special functions seem to be viable paths, but I'm curious about any analytic approaches or insights the community might have.

## Figure Description

Attached is a plot showing how the value of the integral changes as a function of ($$x$$) over the range from -2 to 2. The plot illustrates the integral's sensitivity to changes in ($$x$$) and provides a visual representation of the problem at hand.

I'm looking for any of the following:

• Analytic approaches to tackle this integral.
• Insights into the behavior of this integral with respect to ($$x$$).
• References to similar integrals or relevant mathematical literature.

Integrate by parts

\begin{align} &\int_{0}^{2\pi} \ln(s^2x^2 + 1)ds\\ =& \ 2\pi\ln(4\pi^2 x^2+1) - \int_{0}^{2\pi} \frac{2s^2 x^2}{s^2 x^2 + 1}ds\\ =&\ 2\pi\ln(4\pi^2 x^2+1) +\frac{2\tan^{-1}(2\pi x)}x-4\pi \end{align}

This is likely unnecessary to the OP and his specific problem as @Quanto's solution is far more efficient. Still, it is another solution for those who stumble upon this question in the future.

You can also expand the logarithm as a series. \begin{align*} \int_0^{2\pi}\ln(s^2x^2+1)\;{\rm d}s&=\int_0^{2\pi}\Bigg[-\sum_{n=1}^{\infty}\frac{(-1)^n(xs)^{2n}}{n}\Bigg]\;{\rm d}s\\ &=-\sum_{n=1}^{\infty}\frac{(-1)^n(x)^{2n}}{n}\int_0^{2\pi}s^{2n}\;{\rm d}s\\ (\text{partial fraction decomp.})&=\sum_{n=1}^{\infty}(-1)^{n+1}(x)^{2n}(2\pi)^{2n+1}\Bigg[\frac1n-\frac{2}{2n+1}\Bigg]\\\text{(distribute)}&=\color{blue}{2\pi\sum_{n=1}^{\infty}\frac{(-4\pi^2x)^n}{n}}\color{red}{-4\pi}\color{green}{+\frac2x\sum_{n=0}^{\infty}\frac{(-1)^n(2\pi x)^{2n+1}}{2n+1}}\\&=\color{blue}{2\pi\ln(4\pi^2x^2+1)}\color{red}{-4\pi}\color{green}{+\frac{2}{x}\tan^{-1}(2\pi x)} \end{align*} Making sure the indices don't mess us up, $$4\pi$$ is subtracted because the series after the third equal sign starts from $$n=1$$, whereas the arctangent series (at least of the form that was used) begins at $$n=0$$. Applying the series expansion formula for the logarithm, but now we use it to contract the expression, we arrive at $$\int_0^{2\pi}\ln(s^2x^2+1)\;{\rm d}s=\boxed{2\pi\ln(4\pi^2x^2+1)+\frac2x\tan^{-1}(2\pi x)-4\pi}$$ Which is exactly what @Quanto got.

Do you know Feynman's trick? It's an interesting trick .

we have : $$I(x)=\int^{2\pi}_{0}\ln(1+s^2x^2)ds$$

We differentiate both sides with respect to the variable $$x$$

$$\frac{dI(x)}{dx}=\int^{2\pi}_{0}\frac{2xs^2}{1+s^2x^2}ds=\int^{2\pi}_{0}\frac{2}{x} ds -\frac{2}{x^2}\int^{2\pi}_{0}\frac{x}{1+s^2x^2}ds$$

Therfore: $$\frac{dI(x)}{dx}=\frac{4\pi}{x}-\frac{2}{x^2}\arctan(2\pi x)$$

So $$I(x)=4\pi\ln(x)+\frac{2}{x}\arctan(2\pi x)-\int \frac{4\pi}{x(1+4\pi^2x^2)}dx=4\pi\ln(x)+\frac{2}{x}\arctan(2\pi x)-\int \frac{4\pi}{x}-\frac{16\pi^2 x}{1+4\pi^2x^2}dx$$ Finally:

$$I(x)=\frac{2}{x}\arctan(2\pi x)+2\pi\ln(1+4\pi^2x^2)+c$$ We have $$I(0)=0$$ which implies $$c=-4\pi$$.

So: $$\int^{2\pi}_{0}\ln(1+s^2x^2)ds=\frac{2}{x}\arctan(2\pi x)+2\pi\ln(1+4\pi^2x^2)-4\pi$$