Integration by parts when a limit is infinity.

[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}<br /> and so
uv - \int_0^\infty v du<br /> = 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?

 

[Data]

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

 

[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}}

 

[Data]

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

 

[Jameson]

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

 

[Data]

oops!

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

 

[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}<br /> and so
uv - \int_0^\infty v du<br /> = 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?

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.

 

[Data]

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

 

[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.

 

[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.

 

[Shinjo]

\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.

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?

 

[hypermorphism]

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?

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) = x^z+1/e^x. 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)!/e^x which approaches zero as x increases without bound.

 

[Jameson]

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

 

[Shinjo]

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) = x^z+1/e^x. 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)!/e^x.

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.

 

[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 &gt; 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.