**Basis **(n=1)

We will use integration by parts

`int u dv=uv-int v du`

`int_0^infty xe^-x dx=|[u=x,dv=e^-x dx],[du=dx,v=-e^-x]|=`

`-xe^-x|_0^infty+int_0^infty e^-x dx=(-xe^-x-e^-x)|_0^infty=`

`lim_(x to infty)(-xe^-x-e^-x)-(0-1)=`

In order to calculate the above integral we shall use L'Hospital's rule:

`lim_(x to a)(f(x))/(f(x))=lim_(x to a) (f'(x))/(g'(x))`

First we rewrite the limit so we could use L'hospital's rule.

`lim_(x to infty)-xe^-x=lim_(x to infty)-x/e^x=`

Now we differentiate.

`lim_(x to infty)-1/e^x=0`

Let us now return to calculating the integral.

`0-0-0+1=1`

As we can see the integral converges to 1.

Let us assume that integral `int_0^infty x^n e^-x dx` converges for all `n leq k.`

**Step** (n=k+1)

We will once again use integration by parts.

`int_0^infty x^(k+1)e^-x dx=|[u=x^(k+1),dv=e^-x dx],[du=(k+1)x^k dx,v=-e^-x]|=`

`-x^(k+1)e^-x|_0^infty+(k+1)int_0^infty x^k e^-x dx`

From the assumption we know that the above integral converges, therefore we only need to show that `x^(k+1)e^-x|_0^infty` also converges.

`x^(k+1)e^-x|_0^infty=lim_(x to infty)x^(k+1)e^-x-0=lim_(x to infty) x^(k+1)/e^x`

If we now apply L'Hospital's rule `k+1` times, we will get

`lim_(x to infty) ((k+1)!)/e^x=0`

Thus, we have shown that the integral converges for `n=k+1` concluding the induction.

**QED**

The image below shows graphs of the function under integral for different values of `n.` We can see that `x`-axis is asymptote for all of the graphs meaning that the function converges to zero for all `n.` The only difference is that the convergence gets a little bit slower as `n` increases and so the area under the graph increases as well. However, the area remains finite for all `n in NN,` as we have already concluded.

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