Question

The integral (3.1) is improper because of infinite upper limit and it is also improper for 0 < p < 1 because x^p-1becomes infinite at the lower limit. However, the integral is convergent for any p>0. Prove this.

Short answer

Hence, the given statement is proved.

Step-by-step solution

Given Information

The given integral is${\int }_0^{\infty }{x}^{p-1}{e}^{-x}dx$

Definition of improper integral

Any integral is said to be improper if has either one or both the limits infinite or integrand tending to infinity at one or more points within limits of integration.

Improper Integral because of infinite upper limit

Substitute by infinity, the integral will become improper.

For $0<p<1$, the integral is also improper because $x^{p-1}$becomes infinite at the lower limit .

As $p<1$, substitute by p in $x^{p-1}$all the values of power will become negative which implies that x will be in denominator.

Substitute by 0 in $x^{p-1}$the function will be infinite as $\frac{n}{0}=\infty :n$: is any number.

Convergence of given integral

For $p>0$, two cases are given below.

Case1:$p\ge 1$

At the lower limit zero, it is obvious that x will be zero and integral will have the value.

In the upper limit, the first function $x^{p-1}$will tend to infinity and the second function $e^{-x}$tends to zero.

So $0\times \infty$has the value.

Case 2:$0<p<1$

In this case all the numbers are fraction less than 1 but it is also convergent.

So by using the relations mentioned below the values of the integral will be finite.

$$\begin{gathered} \tau \left(\frac{1}{2}\right)=\sqrt{\pi } \\ \tau \left(p\right)\tau \left(1-p\right)=\frac{\pi }{\sin \pi p} \end{gathered}$$

The integral is convergent.

The integral ${\int }_0^{\infty }{x}^{p-1}{e}^{-x}dx$is improper because of infinite upper limit and it is also improper for$0<p<1$ because$x^{p-1}$ becomes infinite at the lower limit. However, the integral is convergent for any$p>0$ is proved.