Inverse of a function in a integral

Let's find a relationship between $(1-x^{18})^\frac{1}{20}$ and $(1-x^{20})^\frac{1}{18}$ , as you may guess from the title, they are inverses of each other.

This is because if we let the following:

$y=(1-x^{18})^\frac{1}{20} \Rightarrow y^{20}=1-x^{18} \Rightarrow x=(1-y^{20})^\frac{1}{18}.$ which shows that they are inverses of each other.

Since we are integrating the function from $0$ to $1$ , we can check that the function and the inverse have the same area.

This is true if we plug in the values of $x=0,1$ into $y=(1-x^{18})^\frac{1}{20}$ we will get $y=1,0$

Then,

$\int_{0}^{1}(1-x^{18})^\frac{1}{20}-(1-x^{20})^\frac{1}{18}dx$

$=\int_{0}^{1}(1-x^{18})^\frac{1}{20}dx-\int_{0}^{1}(1-x^{20})^\frac{1}{18}dx$

$=\int_{0}^{1}(1-x^{18})^\frac{1}{20}dx-\int_{0}^{1}(1-y^{18})^\frac{1}{20}dy$

$=0.$

$\begin{aligned} I & = \int_0^1 \left((1-x^{18})^\frac 1{20} - (1-x^{20})^\frac 1{18}\right) dx \\ & = {\color{#3D99F6}\int_0^1 (1-x^{18})^\frac 1{20} dx} - \color{#D61F06} \int_0^1 (1-x^{20})^\frac 1{18} dx & \small \color{#3D99F6} \text{Let }u = x^{18} \implies du = 18 x^{17} dx \\ & = {\color{#3D99F6}\int_0^1 \frac {(1-u)^\frac 1{20}}{18u^\frac {17}{18}} du} - \color{#D61F06} \int_0^1 \frac {(1-v)^\frac 1{18}}{20v^\frac {19}{20}} dv & \small \color{#D61F06} \text{Let }v = x^{20} \implies dv = 20 x^{19} dx \\ & = \frac 1{18}\text{B}\left(\frac 1{18}, \frac {21}{20}\right) - \frac 1{20}\text{B}\left(\frac 1{20}, \frac {19}{18}\right) & \small \color{#3D99F6} \text{Beta function }\text B (m,n) = \int_0^1 t^{m-1}(1-t)^{n-1} dt \\ & = \frac {\Gamma \left(\frac 1{18}\right) \Gamma \left(\frac {21}{20}\right)}{18 \Gamma \left(\frac {199}{180}\right)} - \frac {\Gamma \left(\frac 1{20}\right) \Gamma \left(\frac {19}{18}\right)}{20 \Gamma \left(\frac {199}{180}\right)} & \small \color{#3D99F6} \text B (m,n) = \frac {\Gamma (m) \Gamma (n)}{\Gamma (m+n)} \text{, where }\Gamma(\cdot) \text{ denotes gamma function} \\ & = \frac {\Gamma \left(\frac 1{18}\right) \cdot \color{#3D99F6}\frac 1{20}\Gamma \left(\frac 1{20}\right)}{18 \Gamma \left(\frac {199}{180}\right)} - \frac {\Gamma \left(\frac 1{20}\right) \cdot \color{#3D99F6} \frac 1{18} \Gamma \left(\frac 1{18}\right)}{20 \Gamma \left(\frac {199}{180}\right)} & \small \color{#3D99F6} \text{Note that }\Gamma (1+x) = x\Gamma (x) \\ & = \boxed 0 \end{aligned}$

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References:

• Beta function
• Gamma function

Coz you know the identity;

$\large\ \displaystyle \int_{a}^{b} f(x)dx + \int_{f(a)}^{f(b)} f^{-1}(x)dx = bf(b) - af(a)$

the rest is a big ZERO.