MIT Integration Bee $21$

Question

Question:

$$\int_0^\infty {{1}\over{(1+x^{2015})({1+x^2}})}dx$$

How to solve these integrals. I have no idea.

Answer 1

Use the substitution $x\mapsto 1/x$:

\begin{align*} I &= \int_0^\infty {{1}\over{(1+x^{2015})({1+x^2}})}dx \\ &= \int_0^\infty {{1}\over{(1+x^{-2015})({1+x^{-2}}})} \frac{1}{x^2}dx\\ &= \int_0^\infty {{x^{2015}}\over{(1+x^{2015})({1+x^{2}}})} dx \end{align*}

Then,

\begin{align*} 2I &= \int_0^\infty {{1}\over{(1+x^{2015})({1+x^2}})}dx+\int_0^\infty {{x^{2015}}\over{(1+x^{2015})({1+x^2}})}dx\\ &=\int_0^\infty {{1+x^{2015}}\over{(1+x^{2015})({1+x^2}})}dx\\ &=\int_0^\infty \frac{1}{1+x^2}dx = \frac{\pi}{2} \end{align*}

So $I=\pi/4$.