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$$\int_0^\infty \frac{1}{(1+x^2)(1+x^{2018})}\,dx$$

My Calculus professor asked a challenge problem to one of my friends and asked her to evaluate it. I tried partial fractions to no avail and the trig substitution $x = \tan\theta$, but that leaves me with

$$\int_0^{\pi/2} \frac{1}{(1+\tan^{2018}\theta)}\,d\theta$$

which I do not know how to evaluate. Any help would be greatly appreciated!

Asaf Karagila
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WhatsDUI
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    If it was not in the denominator, the integral would be infinite. Also, since 2018 is clearly not important here, try with $\int_0^\infty \frac{dx}{(1+x^2)(1+x^n)}$ for a few values of $n$ (e.g., on Wolfram.Alpha). You'll see that you'll get $\frac{\pi}{4}$ for every integer $n$ you try. Now, how to prove it? – Clement C. Apr 12 '18 at 04:37

2 Answers2

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You are on the right track. You have $$\tag1 I=\int_0^{\pi/2}\frac1{1+\tan^{2018}x}\,dx. $$ Do the substitution $u=\pi/2-x$. Then, after you do your algebra,
$$\tag2 I=\int_0^{\pi/2}\,\frac{\tan^{2018}u}{1+\tan^{2018}u}\,du. $$ Now combine $(1)$ and $(2)$ to get $$ 2I=\int_0^{\pi/2}\frac1{1+\tan^{2018}u}\,du+\int_0^{\pi/2}\,\frac{\tan^{2018}u}{1+\tan^{2018}u}\,du=\int_0^{\pi/2}\,1\,dx=\frac\pi2. $$ So $I=\pi/4$. Of course, the $2018$ is irrelevant, any positive number will give the same value.

Martin Argerami
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Let $$I=\int_0^{\infty} \frac {dx}{(1+x^n)(1+x^2)}$$

Now substitute $x=\frac 1y$ to get $$I=\int_0^{\infty} \frac {y^n dy}{(1+y^n)(1+y^2)}=\int_0^{\infty} \frac {x^n dx}{(1+x^n)(1+x^2)}$$

Now since $$I=\int_0^{\infty} \frac {dx}{(1+x^n)(1+x^2)}=\int_0^{\infty} \frac {(1+x^n-x^n)dx}{(1+x^n)(1+x^2)}=\int_0^{\infty} \frac {(1+x^n)dx}{(1+x^n)(1+x^2)}-\int_0^{\infty} \frac {x^n dx}{(1+x^n)(1+x^2)}=\arctan x\big\vert_0^{\infty}-I$$

Hence $$2I=\frac {\pi}{2}\Rightarrow I=\frac {\pi}{4}$$

Rohan Shinde
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