# Integration by parts when a limit is infinity.

1. Mar 16, 2005

### Shinjo

I'm having a tough time trying to do integration by parts with one of my limits being infinity. My Integral looks like:

$$\int_0^\infty x^z e^{-x} dx$$ with $$z = \frac{-1}{\pi}$$

Now if I let $$u = e^{-x}$$ and $$dv = x^z dx$$,
I will have: $$du = -e^{-x} dx$$ and $$v = \frac{1}{z + 1} x^{z + 1}$$ and so
$$uv - \int_0^\infty v du = e^{-x} \frac{1}{z + 1} x^{z + 1} - \int_0^\infty \frac{1}{z + 1} x^{z + 1} -e^{-x} dx$$

However, since one of the limit is infinity, the term
$$uv = e^{-x} \frac{1}{z + 1} x^{z + 1}$$ has a freakin infinity subbed in it. The answer is actually,
$$\frac{1}{z + 1} \int_0^\infty x^{z + 1} e^{-x} dx$$, which is what my integral part looks like. What am I doing wrong? Why do I get a term with infinity at the front?

2. Mar 16, 2005

### Data

What is $$\lim_{x\rightarrow \infty} x^ze^{-x}$$?

3. Mar 16, 2005

### Jameson

I think it is easier to look at the limit like this:

$$\lim_{x\rightarrow\infty} \frac{1}{x^{\frac{1}{\pi}}e^{x}}$$

4. Mar 16, 2005

### Data

$$x^ze^{-x} = \frac{x^z}{e^x} \neq \frac{1}{x^{\frac{1}{z}}e^x}$$

Last edited: Mar 16, 2005
5. Mar 16, 2005

### Jameson

Shinjo said in his first post that $$z = -\frac{1}{\pi}$$

6. Mar 16, 2005

### Data

oops!

Anyways the integral actually is just equal to $$\Gamma \left( \frac{\pi - 1}{\pi} \right)$$

7. Mar 16, 2005

### Shinjo

But the way to calculate the gamma function is to do that actual integral...

anywayz, do you see my problem though? when my $$v$$ becomes $$v = \frac{1}{z + 1} x^{z + 1},$$my $$x^{z + 1}$$ term gets moved to the top and turns into infinity.

8. Mar 16, 2005

### Data

$$\lim_{x \rightarrow \infty} x^ye^{-x} = 0 \ \forall y \in \mathbb{R}$$

9. Mar 16, 2005

### Shinjo

Hmm...maybe I should have mentioned this earlier, but I am not trying to solve the integral. I have to get it in a specific form so I can use Gaussian Quadrature. In order to do that I have to integrate by parts.

The answer is supposed to be: $$\frac{1}{z + 1} \int_0^\infty x^{z + 1} e^{-x} dx$$, but I don't know how to get rid of the $$uv = e^{-x} \frac{1}{z + 1} x^{z + 1}$$ term, since one of the limits is infinity.

Thank you for the help btw, it is much appreciated.

10. Mar 16, 2005

### Data

$$\left[e^{-x}\frac{1}{z+1}x^{z+1}\right]_0^\infty = \lim_{x\rightarrow \infty} \left(\frac{e^{-x}x^{z+1}}{z+1}\right) - 0 = \frac{1}{z+1}\lim_{x\rightarrow \infty} e^{-x}x^{z+1}$$

then just apply the result in my last post.

11. Mar 16, 2005

### Shinjo

Ahh...I see what you're saying. Thank you.

I don't mean to question your method, I know I might be becoming a nuisance by now, but can you tell me why $$\lim_{x \rightarrow \infty} x^ye^{-x} = 0 \ \forall y \in \mathbb{R}$$? I mean, doesn't $$\lim_{x \rightarrow \infty} x^y = \infty$$?

12. Mar 16, 2005

### hypermorphism

That's not really relevant when you have a denominator that is also a function of x. For example, the trivial f(x) = x/x or g(x) = x/(x2 + 1), neither of which diverge as x increases without bound. Now refer to your problem f(x) = xz+1/ex. Note that both numerator and denominator diverge as x increases without bound, so we may apply L'Hospital's rule to examine the equivalent limit (z+1)!/ex which approaches zero as x increases without bound.

Last edited: Mar 16, 2005
13. Mar 16, 2005

### Jameson

I think he meant $$x^z$$ This gives you $$\lim_{x\rightarrow\infty} \frac{1}{x^{\frac{1}{\pi}}e^{x}} = 0$$

14. Mar 16, 2005

### Shinjo

Ah, Thank you. I just noticed, when the x is an exponent, it increases the term much faster than when x is a base. So I can see how x^y / e^x will = 0 for all y.

Thank you to all once again.

15. Mar 16, 2005

### Data

No nuisance. If I thought it were a nuisance to justify my math, I wouldn't be on a math forum! :)

Say $$y \in \mathbb{R}$$. Note that if $$y \leq 0$$ the result is obvious, so assume $$y > 0$$. Then let $$\lfloor a \rfloor$$ and $$\lceil a \rceil$$ represent the floor and ceiling functions applied to $$a$$ respectively. Thus if $$m = \lfloor y \rfloor, \ M = \lceil y \rceil$$ we clearly have $$x^me^{-x} \leq x^ye^{-x} \leq x^Me^{-x} \ \forall x \geq 1 \Longrightarrow \lim_{x\rightarrow \infty}x^me^{-x} \leq \lim_{x \rightarrow \infty} x^ye^{-x} \leq \lim_{x \rightarrow \infty} x^Me^{-x}$$ (by a slightly generalized version of the squeeze theorem).

Now, let $$a \in \mathbb{N}^+$$. By l'Hopital's rule, applied $$a$$ times, we have

$$\lim_{x \rightarrow \infty} x^ae^{-x} = \lim_{x \rightarrow \infty} \frac{x^a}{e^x} = \lim_{x \rightarrow \infty} \frac{ax^{a-1}}{e^x} = . \ . \ . = \lim_{x \rightarrow \infty} \frac{a!}{e^x} = a! \lim_{x \rightarrow \infty} e^{-x} = 0$$

and since for $$y \ge 0$$ clearly $$M \geq m \geq 0, \ M, m \in \mathbb{N}$$ we get $$0 = \lim_{x \rightarrow \infty} x^m e^{-x} \leq \lim_{x\rightarrow \infty} x^ye^{-x} \leq \lim_{x \rightarrow \infty} x^M e^{-x} = 0 \Longrightarrow \lim_{x \rightarrow \infty} x^y e^{-x} = 0$$ as we wanted. QED.

Last edited: Mar 16, 2005