Question

Solve improper integral

$\displaystyle\int\limits^\infty_0 {\frac{x^{49}}{(1+x)^{51}} } \, dx$

Answer 1

Solution:-

There are many ways to solve this integral, I will use two common methods to solve it, out of which first method includes Euler integral of first kind.

Method 1:

We are aware about the definition of beta function.

$\boxed{\rm B(x, y) = \int_0^\infty \dfrac{u^{x-1}}{(1+u)^{x+y}}\, du}$

The given integral is,

$\displaystyle\rm I= \int_0^\infty \dfrac{x^{49}}{(1+x)^{51}} \: dx$

Can also be written as,

$\displaystyle\rm I= \int_0^\infty \dfrac{x^{50-1}}{(1+x)^{50+1}} \: dx$

On comparing this integral with definition of beta function, we get:

$\displaystyle\rm\int_0^\infty \dfrac{x^{49}}{(1+x)^{51}} \: dx = B(50,1)$

Now, we can use the Gamma Beta relation.

$\boxed{\rm B(x, y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)} }$

Using this, we get:

$\rm B(50, 1) = \dfrac{\Gamma(50) \cdot \Gamma(1)}{\Gamma(51)}$

$\rm\iff B(50, 1) = \dfrac{\Gamma(50) \cdot \Gamma(1)}{50 \cdot\Gamma(50)}$

$\rm\implies B(50, 1) = \dfrac{1}{50}$

Hence the required answer is 1/50.

$\rule{280}{1}$

Method 2:

We have to given integral,

$\displaystyle\rm I= \int_0^\infty \dfrac{x^{49}}{(1+x)^{51}} \: dx$

Can also be written as,

$\displaystyle\rm I= \int_0^\infty \dfrac{x^{49}}{ {x}^{51} (1+ \frac{1}{x} )^{51}} \: dx$

$\iff\displaystyle\rm I= \int_0^\infty \dfrac{1}{ {x}^{2} (1+ \frac{1}{x} )^{51}} \: dx$

$\sf Now\, let \: \left(1 + \frac{1}{x} \right) = t \implies - \dfrac{dx}{ {x}^{2} } = dt$

The above integral changes to,

$\displaystyle\rm I= \int_\infty^{1} - \dfrac{1}{ t^{51}} \: dt$

$\iff\displaystyle\rm I= -\dfrac{ t^{ - 50}}{ - 50}\bigg |^{1}_{ \infty}$

$\iff\displaystyle\rm I= \dfrac{ t^{ - 50}}{ 50}\bigg |^{1}_{ \infty}$

$\implies\displaystyle\rm I= \dfrac{1}{50}$

Hence the required answer is 1/50.