a cool result using the gamma function

[steady sapphire]

$\forall \alpha >1$ we set $I(\alpha)=\int_{0}^{\infty} \frac{1}{1+t^{\alpha}} dt $ in that case $\forall \alpha >1$ $I(\alpha)=\frac{\frac{\pi}{\alpha}}{sin(\frac{\pi}{\alpha})}$
Proof:
For $\alpha >1$ and x>0 we set $I(\alpha,x)=\int_{0}^{\infty} \frac{e^{-x(1+t^{\alpha})}}{1+t^{\alpha}} dt$
Using the dominated convergence theorem ( not shown here but it works) $I(\alpha,\cdot): x \rightarrow I(\alpha,x)$ is a class $C^{1}$ function
And $ I=\frac{\partial I}{\partial x} (\alpha,x) =\int_{0}^{\infty} -e^{-x(1+t^{\alpha})} dt$ using the change of variables $xt^{\alpha}=u}$ we get $I=-x^{-\frac{1}{\alpha}} e^{-x} \int_{0}^{\infty} u^{\frac{1}{\alpha}-1}e^{-u}du$ Now integrating from 0 to infinity with respect to x $\int_{0}^{\infty} \frac{\partial I}{\partial x} (\alpha,x)dx$= -\int_{0}^{\infty} x^{-\frac{1}{\alpha}} e^{-x}dx \int_{0}^{\infty} u^{\frac{1}{\alpha}-1}e^{-u}du$
Which we recognize as being $-\Gamma(\frac{1}{\alpha}) \Gamma(1-\frac{1}{\alpha})$ which, using the reflection formula is $-\frac{\frac{\pi}{\alpha}}{sin(\frac{\pi}{\alpha})}$
Now using the dominated convergence theorem and the fundamental theorem of calculus the left side is equal to $-I(\alpha,0)=I(\alpha)$ thus we have the result

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[languid hamlet]

Rotor
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[steady sapphire]

Was wondering if it was correct ( this is mostly to test my Latex, but I also wanted to show a cool result)

Also I’m sorry for the lack of rigor

[radiant gazelle]

It does look pretty neat

Also maybe this command is useful for you

$I(\alpha, \cdot)$

[languid hamlet]

Rion

[steady sapphire]

Yes thanks

[elder quest]

@steady sapphire has given 1 rep to @radiant gazelle

[analog yoke]

$\int\limits_0^{ + \infty } {\frac{{dx}}{{1 + x^n }}} \underbrace = _{y = 1 + x^n }\int\limits_1^{ + \infty } {\frac{1}{y}\frac{{dy}}{{n\left( {y - 1} \right)^{\frac{{n - 1}}{n}} }}} \underbrace = _{t = y - 1}\frac{1}{n}\int\limits_0^{ + \infty } {\frac{{t^{\frac{1}{n} - 1} }}{{t + 1}}dt} = \frac{1}{n}\int\limits_0^{ + \infty } {\frac{{t^{\frac{1}{n} - 1} }}{{\left( {t + 1} \right)^{\frac{1}{n} + 1 - \frac{1}{n}} }}dt} = \frac{1}{n}B\left( {\frac{1}{n},1 - \frac{1}{n}} \right)$

$ = \frac{1}{n}\frac{{\Gamma \left( {\frac{1}{n}} \right)\Gamma \left( {1 - \frac{1}{n}} \right)}}{{\Gamma \left( 1 \right)}} = \frac{1}{n}\frac{\pi }{{\sin \left( {\frac{\pi }{n}} \right)}}$

[languid hamlet]

jamal

[analog yoke]

Let $A=\int_{0}^{\infty}\frac{1}{1+x^n}dx$.
Substitute $x=\tan^{1/n}{u}$ into $A$, then
[\begin{aligned}A&=\int_{0}^{\infty}\frac{1}{1+x^n}dx\&=\int_{0}^{\pi/2}\frac{1}{1+\tan^{2} u}\left(\frac{2}{n}\tan^{2/n-1}u \sec^{2} u du\right)=\frac{2}{n}\int_{0}^{\pi/2}\sin^{2/n-1}u \cos^{1-2/n} u du\end{aligned}]
We can use Beta function for this;
[A=\frac{2}{n}\cdot\frac{1}{2}{\rm B}\left(\frac{1}{n}, 1-\frac{1}{n}\right)=\frac{1}{n}\Gamma\left(\frac{1}{n}\right)\Gamma\left(1-\frac{1}{n}\right)]
By Euler's reflection formula,
[A=\frac{1}{n}\cdot \frac{\pi}{\sin{(\pi/n)}}=\frac{\pi}{n}\csc\frac{\pi}{n}]

[languid hamlet]

jamal