Find : $ \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx $

Question

Find:

$$ \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx $$

My initial thoughts were to use a trigonometric substitution of $x=\tan(\theta)$ which implies that $ dx= \sec^2(\theta) d\theta$ this transforms our integral into

$$ \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx = \int^\frac{\pi}{2}_0 \frac{1}{1+(\tan(\theta))^{2015}} d\theta $$

But I'm having trouble evaluating it after.

I also tried a substitution of $y=\frac{1}{x}$ but that did not get me anywhere either.

Thanks for any help in advance :)

Aforest · Answer 1 · 2016-12-21T08:11:58.670

2

Hint. Try again $y=\frac{1}{x}$.

\begin{align} \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx&=\int^1_0 \frac{1}{(1+x^{2015})(1+x^2)}dx+\int^\infty_1 \frac{1}{(1+x^{2015})(1+x^2)}dx\\ &=\int^1_0 \frac{1}{(1+x^{2015})(1+x^2)}dx+\int^1_0 \frac{y^{2015}}{(1+y^{2015})(1+y^2)}dy\\ &=? \end{align}

edited Dec 21 '16 at 08:11

answered Dec 21 '16 at 08:09

Aforest

2,665

Very bad to give the same hint as that which was answered in the duplicate. – Dec 21 '16 at 08:10
1

@Rohan To be fair the duplicate was suggested after the answer was posted. – Winther Dec 21 '16 at 08:12
Yeah. I didn't know it's duplicate... – Aforest Dec 21 '16 at 08:15

score 2 · Answer 2 · answered Dec 21 '16 at 08:10

You can use $$\int_a^bf(x) \, dx = \int_a^bf(a+b-x) \, dx$$ to get $$I=\int^\frac{\pi}{2}_0 \frac{1}{1+(\tan(\theta))^{2015}} d\theta=\int^\frac{\pi}{2}_0 \frac{1}{1+(\cot(\theta))^{2015}} d\theta$$ Then \begin{align*} 2I & = \int^\frac{\pi}{2}_0 \frac{1}{1+(\tan(\theta))^{2015}} d\theta+\int^\frac{\pi}{2}_0 \frac{1}{1+(\cot(\theta))^{2015}} d\theta\\ 2I & = \int_{0}^{\frac{\pi}{2}}1 \, d \theta \end{align*}

Find : $ \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx $

2 Answers2