Solving the Improper Integral of ( x^2 e^{-x} ) from 0 to Infinity Using Integration by Parts

The integral in question is the improper integral ( I int_0^{infty} x^2 e^{-x} ,dx ). To solve this, we can apply the method of integration by parts iteratively. Let's walk through the steps.

Stepwise Solution

Step 1: Choosing u and dv

We start by choosing appropriate values for ( u ) and ( dv ) in the integration by parts formula ( int u , dv uv - int v , du ).

u x^2, which implies ( du 2x , dx ).

dv e^{-x} , dx, which implies ( v -e^{-x} ).

Step 2: Applying Integration by Parts

Using the integration by parts formula:

[ I left[ -x^2 e^{-x} right]_0^{infty} - int_0^{infty} 2x e^{-x} , dx ]

Step 3: Evaluating the Boundary Term

Let's evaluate the boundary term:

[ lim_{b to infty} left[ -x^2 e^{-x} right]_0^b lim_{b to infty} left( -b^2 e^{-b} - 0 right) ]

Since the exponential function ( e^{-b} ) grows faster than any polynomial ( b^2 ), it follows that ( b^2 e^{-b} ) approaches 0 as ( b to infty ). Therefore, the term evaluates to 0.

Step 4: Calculating the Remaining Integral

We now need to evaluate the integral:

[ int_0^{infty} 2x e^{-x} , dx ]

Again, let's use integration by parts:

u 2x, which implies ( du 2 , dx ).

dv e^{-x} , dx, which implies ( v -e^{-x} ).

Applying the integration by parts formula again:

[ int 2x e^{-x} , dx left[ -2x e^{-x} right]_0^{infty} - int_0^{infty} 2 e^{-x} , dx ]

Step 5: Evaluating the Boundary Term Again

Evaluating the boundary term:

[ lim_{b to infty} left[ -2x e^{-x} right]_0^b lim_{b to infty} left( -2b e^{-b} - 0 right) 0 ]

As before, the exponential term dominates the polynomial term.

Step 6: Calculating the Remaining Simple Integral

Now we compute the remaining integral:

[ int_0^{infty} 2 e^{-x} , dx 2 left[ -e^{-x} right]_0^{infty} 2(0 - (-1)) 2 ]

Step 7: Combining the Results

Putting everything together:

[ I 0 - 2 2 ]

Conclusion

The value of the improper integral is:

[ int_0^{infty} x^2 e^{-x} , dx 2 ]

The Importance of the Exponential Function

The formula in the statement of this question is a little unclear. Certainly, the correct expression of the function was: [ f(x) x^2 e^{-x} frac{x^2}{e^x} ]

The function is everywhere continuous and integrable over the real axis ( mathbb{R} ), including the interval ([0, infty)). The integral ( int_0^{infty} f(x) , dx ) should not simply be solved but rather calculated, as it is an improper integral. In such cases, it's often necessary to study its convergence or divergence.

Application and Convergence

For positive functions on ([0, infty)), if ( F(x) ) is its indefinite integral or antiderivative, we determine:

[ int_0^{t} f(x) , dx F(t) - F(0) ]

And thus:

[ int_0^{infty} f(x) , dx lim_{t to infty} left( F(t) - F(0) right) ]

This integral is typical for the application of integration by parts, and this method will need to be applied twice.

Integration by Parts Example

Let's consider the function again:

[ int x^2 e^{-x} , dx ]

We let:

u x^2, which implies ( du 2x , dx ).

dv e^{-x} , dx, which implies ( v -e^{-x} ).

Performing the first integration by parts:

[ F(x) -x^2 e^{-x} 2 int x e^{-x} , dx ]

For the second integral, we let:

u x, which implies ( du dx ).

dv e^{-x} , dx, which implies ( v -e^{-x} ).

And applying integration by parts again:

[ int x e^{-x} , dx -x e^{-x} int e^{-x} , dx -x e^{-x} - e^{-x} ]

Plugging this back into the first result:

[ F(x) -x^2 e^{-x} 2 left( -x e^{-x} - e^{-x} right) -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} ]

Evaluating at the boundaries and taking the limit as ( t to infty ), we find:

[ F(infty) 0, quad F(0) -2 ]

Thus:

[ int_0^{infty} x^2 e^{-x} , dx F(infty) - F(0) 0 - (-2) 2 ]

Final Conclusion

Therefore, the value of the improper integral is:

[ int_0^{infty} x^2 e^{-x} , dx 2 ]