An amazing Integral

First substitute $x^{2017/2}=\tan z \\ => \dfrac{2017}{2} x^{2015/2} dx=\sec^2(z) dz \\ => x^{2015/2} dx=\dfrac{2}{2017}\sec^2(z)dz \\ =>dx=\dfrac{2}{2017} \dfrac{1}{x^{2015/2}} .\sec^2(z) dz \\ =>dx=\dfrac{2}{2017} \dfrac{1}{\tan(z)^{2015/2017}} .\sec^2(z) dz$

Now using this into the integral

$\displaystyle \int_{0}^{\infty} \dfrac{1}{1+x^{2017}} dx \\ =\displaystyle \int_{0}^{\pi/2} \dfrac{2}{2017} \dfrac{1}{\tan(z)^{2015/2017}} .\sec^2(z).\dfrac{1}{\sec^2(z)} dz \\ =\displaystyle \dfrac{1}{2017} \times 2 \int_{0}^{\pi/2} \sin(z)^{-2015/2017} . \cos(z)^{2015/2017} dz$

Now using definition of Beta-Function we can write

$=\dfrac{1}{2017} B(\dfrac{1}{2017},\dfrac{2016}{2017})$

Then using the property $\Gamma(z) \Gamma(1-z)=\dfrac{\pi}{\sin (\pi z)}$ we get the value

$=\dfrac{1}{2017} \Gamma(1/ 2017) \Gamma(1-1/2017) \\ =\dfrac{\pi}{2017} \dfrac{1}{\sin (\pi/2017)} \\ =\dfrac{\pi}{2017} \csc (\pi/2017)$

Rewrite there integral in terms of a double integral:

$\displaystyle\int_0^\infty\frac1{1+x^n}dx=\int_0^\infty\int_0^\infty e^{-t(1+x^n)}dxdt$

Now similar to solving a Gaussian, $u = tx^n$ , and the integral becomes

$\displaystyle\frac1{n}\int_0^\infty e^{-u}u^{\frac1{n}-1}du\int_0^\infty e^{-t}t^{\frac{-1}{n}}dt=$

By using the definition of the gamma function and Euler's Reflection Formula,

$\displaystyle\frac{\Gamma(\frac1{n})\Gamma(1-\frac1{n})}{n} = \frac{\pi}{n\sin(\frac{\pi}{n})}$

Let $I = \int_{0}^{\infty} \frac{1}{1+x^{2017}}=\int_{0}^{1} \frac{1}{1+x^{2017}}+\int_{1}^{\infty} \frac{1}{1+x^{2017}}$

Now, let $I_{1}=\int_{0}^{1} \frac{1}{1+x^{2017}} =\int_{0}^{1}(1-x^{2017}+x^{2*2017}-x^{3*2017}+...)dx$

$I_{1}=1+\sum_1^\infty \frac{(-1)^n}{2017n+1}$

Again let $I_{2}=\int_{1}^{\infty} \frac{dx}{1+x^{2017}}=\int_{1}^{\infty}\frac{dx}{x^{2017}(1+\frac{1}{x^{2017}})}=\int_{1}^{\infty}\frac{dx}{x^{2017}}(1-x^{-2017}+ x^{-2*2017}-...)$

$I_{2}=\sum_1^\infty \frac{(-1)^{n+1}}{2017n-1}$

So $I = I_{1}+I_{2} = 1+\sum_1^\infty(\frac{(-1)^{n+1}}{2017n-1}-\frac{(-1)^{n+1}}{2017n+1})=1+\sum_1^\infty(\frac{2(-1)^{n+1}}{(2017n)^{2}-1}$ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (*)

Since we know $\sin(x)=x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{(2\pi)^2})...$ taking log and differentiating with respect to x both sides we get

$\cot(x)= \frac{1}{x}-2\sum_1^\infty\frac{x}{(n\pi)^2-x}$

Using this we can show that $\sum_1^\infty\frac{2(-1)^{n+1}}{(2017n)^{2}-1}$ = $-1+\frac{\pi}{2017}(cot(\frac{\pi}{2*2017})-cot(\frac{\pi}{2017}) )$

Substituting this in (*) we get $I = \frac{\pi}{2017}(cot(\frac{\pi}{2*2017})-cot(\frac{\pi}{2017}) )$

On Simplifying this we get $I = \frac{\pi}{2017}cosec(\frac{\pi}{2017})$ so answer is $1+ 2017+2017 = 4035$

Nice 👍 application of gamma , beta function and euler reflection formula