How to compute $\int_0^\infty \frac{1}{(1+x^{\varphi})^{\varphi}}\,dx$?

Solution 1:

Since $\frac1\varphi=\varphi-1$, $$ \begin{align} \int_0^\infty\frac1{(1+x^\varphi)^\varphi}\,\mathrm{d}x &=(\varphi-1)\int_0^\infty\frac{x^{\varphi-2}}{(1+x)^\varphi}\,\mathrm{d}x\tag{1}\\[6pt] &=(\varphi-1)\mathrm{B}(\varphi-1,1)\tag{2}\\[6pt] &=(\varphi-1)\frac{\Gamma(\varphi-1)}{\Gamma(\varphi)}\tag{3}\\ &=\frac{\Gamma(\varphi)}{\Gamma(\varphi)}\tag{4}\\[6pt] &=1 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x^{\varphi-1}$ noting that $\varphi(\varphi-1)=1$
$(2)$: $\int_0^\infty\frac{x^{\alpha-1}}{(1+x)^\beta}\,\mathrm{d}x=\mathrm{B}(\alpha,\beta-\alpha)$
$(3)$: $\mathrm{B}(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$
$(4)$: $\alpha\,\Gamma(\alpha)=\Gamma(\alpha+1)$

Solution 2:

Hint: Make $x \mapsto \dfrac{1}{x}$ and use $\phi^2 = \phi + 1$ to further simplify. The final result should be $1$.

$$\int_0^{\infty} \frac{1}{(1+x^{\phi})^{\phi}}\,dx = \int_0^{\infty} \frac{x^{\phi^2}}{(1+x^{\phi})^{\phi}}\frac{dx}{x^2} = \int_0^{\infty} \frac{x^{\phi - 1}}{(1+x^{\phi})^{\phi}}\,dx = \, \cdots$$

Solution 3:

The antiderivative invokes an hypergeometric function $$\int \frac{dx}{(1+x^{a})^{a}}=x \, _2F_1\left(\frac{1}{a},a;1+\frac{1}{a};-x^a\right) $$ For the definite integral, as Omnomnomnom commented, the result expresses using the gamma function $$I(a)=\int_0^{\infty} \frac{dx}{(1+x^{a})^{a}} =\frac{\Gamma \left(1+\frac{1}{a}\right) \Gamma \left(a-\frac{1}{a}\right)}{\Gamma (a)}$$ and $I(\phi)=1$ since $\phi-\frac{1}{\phi}=1$ and $1+\frac{1}{\phi}=\phi$.

Amazing are $I(2)=\frac{\pi} 4$, $I(6)=\frac{124729 }{559872}\pi$ and $I(\infty)=1$.