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

Question

How to compute the integral,

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

where, $\varphi = \dfrac{\sqrt{5}+1}{2}$ is the Golden Ratio?

Answer 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 $\varphi(\varphi-1)=1$
$(2)$: $\int_0^\infty\frac{x^{\alpha-1}}{(1+x)^\beta}\,\mathrm{d}x=\mathrm{B}(\alpha,\beta-\alpha)$
$(3)$: $\mathrm{B}(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$
$(4)$: $\alpha\,\Gamma(\alpha)=\Gamma(\alpha+1)$

Answer 2

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

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

Answer 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 $I(\infty)=1$.

Answer 4

Hint: let $x^{\varphi}=u$ then $$\dfrac{1}{\varphi}\int_{0}^{+\infty}\dfrac{u^{1/\varphi-1}}{(1+u)^{\varphi}}du=\dfrac{1}{\varphi}B(\dfrac{1}{\varphi},\varphi-\dfrac{1}{\varphi})=\dfrac{1}{\varphi}B(\dfrac{1}{\varphi},1)$$

Answer 5

For an answer avoiding the use of special functions: use $w=[x^{\varphi}(1+x^{\varphi})^{-1}]^{\varphi}$ then $$\int_0^1 \mathrm{d}w = 1$$

Answer 6

Let $y=\frac{1}{1+x^\varphi}$, then the integral is converted into $$ I=\frac{1}{\varphi} \int_0^1(1-y)^{\varphi-2} d y=\frac{1}{\varphi(\varphi-1)}=1 $$

Answer 7

Let us find the expression for a general $\ n > 0$ and then put $\ n = \varphi $

Let $\frac{1}{1+x^n} = t$. Then, $\ x = [\frac{1-t}{t}]^{\frac{1}{n}}$ and $\ dt = -n\frac{x^{n-1}dx}{(1+x^n)^2}$

Or, rearranging, we get $$\ dx = -\frac{dt}{n \cdot t^{(1+\frac{1}{n})}(1-t)^{(1-\frac{1}{n})}}$$

Notice that the lower and upper bounds of integration are 1 and 0, so to reverse it, a negative sign will appear. This negative will be cancelling the negative obtained in the expression for $\ dx$

Converting the integral in terms of $\ t$, we get:

$$\ I(n) = \frac{1}{n} \cdot \int_0^{1} t^n \cdot t^{-(1+\frac{1}{n})} \cdot (1-t)^{-(\frac{1}{n}-1)} dt $$

$$\ = \frac{1}{n} \cdot \int_0^1 t^{(n - \frac{1}{n} -1)} (1-t)^{(\frac{1}{n}-1)} dt$$

Which can be expressed in terms of the Beta function:

$$\ I(n) = \frac{1}{n} \cdot B[n-\frac{1}{n},\frac{1}{n}]$$

Now, using the fact that $\ B(x,y) = \frac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}$, we can write:

$$\ I(n) = \frac{1}{n} \cdot \frac{\Gamma(n-\frac{1}{n}) \cdot \Gamma(\frac{1}{n})}{\Gamma(n)} $$

Now, for $\ n = \varphi$, note that $\varphi - \frac{1}{\varphi} = 1$

Thus,

$$\ I(\varphi) = \frac{1}{\varphi} \cdot \frac{\Gamma(1) \cdot \Gamma(\frac{1}{\varphi})}{\Gamma(\varphi)}$$

Recall that $\Gamma(x) = (x-1)!$.

That implies, $\ x \Gamma(x) = x(x-1)! = x! = \Gamma(x+1)$.

Also recall that $ \varphi = 1 + \frac{1}{\varphi}$

Using the above facts, it follows that

$$\ I(\varphi) = \frac{\frac{1}{\varphi}\cdot\Gamma(\frac{1}{\varphi})}{\Gamma(\varphi)} = 1$$

Which is the result stated above previously.