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Integration by parts when a limit is infinity. 
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#1
Mar1605, 05:09 PM

P: 12

I'm having a tough time trying to do integration by parts with one of my limits being infinity. My Integral looks like:
[tex] \int_0^\infty x^z e^{x} dx[/tex] with [tex] z = \frac{1}{\pi}[/tex] Now if I let [tex]u = e^{x} [/tex] and [tex]dv = x^z dx[/tex], I will have: [tex]du = e^{x} dx [/tex] and [tex] v = \frac{1}{z + 1} x^{z + 1} [/tex] and so [tex] 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[/tex] However, since one of the limit is infinity, the term [tex] uv = e^{x} \frac{1}{z + 1} x^{z + 1}[/tex] has a freakin infinity subbed in it. The answer is actually, [tex] \frac{1}{z + 1} \int_0^\infty x^{z + 1} e^{x} dx[/tex], 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
Mar1605, 05:14 PM

P: 998

What is [tex]\lim_{x\rightarrow \infty} x^ze^{x}[/tex]?



#3
Mar1605, 05:27 PM

P: 787

I think it is easier to look at the limit like this:
[tex]\lim_{x\rightarrow\infty} \frac{1}{x^{\frac{1}{\pi}}e^{x}}[/tex] 


#4
Mar1605, 05:51 PM

P: 998

Integration by parts when a limit is infinity.
[tex]x^ze^{x} = \frac{x^z}{e^x} \neq \frac{1}{x^{\frac{1}{z}}e^x}[/tex]



#5
Mar1605, 06:03 PM

P: 787

Shinjo said in his first post that [tex]z = \frac{1}{\pi}[/tex]



#6
Mar1605, 06:09 PM

P: 998

oops!
Anyways the integral actually is just equal to [tex]\Gamma \left( \frac{\pi  1}{\pi} \right)[/tex] 


#7
Mar1605, 06:51 PM

P: 12

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


#8
Mar1605, 06:54 PM

P: 998

[tex] \lim_{x \rightarrow \infty} x^ye^{x} = 0 \ \forall y \in \mathbb{R}[/tex]



#9
Mar1605, 07:06 PM

P: 12

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: [tex] \frac{1}{z + 1} \int_0^\infty x^{z + 1} e^{x} dx[/tex], but I don't know how to get rid of the [tex] uv = e^{x} \frac{1}{z + 1} x^{z + 1}[/tex] term, since one of the limits is infinity. Thank you for the help btw, it is much appreciated. 


#10
Mar1605, 07:10 PM

P: 998

[tex] \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}[/tex]
then just apply the result in my last post. 


#11
Mar1605, 07:22 PM

P: 12

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


#12
Mar1605, 07:30 PM

P: 506




#13
Mar1605, 07:30 PM

P: 787

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



#14
Mar1605, 07:36 PM

P: 12

Thank you to all once again. 


#15
Mar1605, 07:38 PM

P: 998

No nuisance. If I thought it were a nuisance to justify my math, I wouldn't be on a math forum! :)
Say [tex]y \in \mathbb{R}[/tex]. Note that if [tex]y \leq 0[/tex] the result is obvious, so assume [tex] y > 0 [/tex]. Then let [tex]\lfloor a \rfloor[/tex] and [tex] \lceil a \rceil[/tex] represent the floor and ceiling functions applied to [tex]a[/tex] respectively. Thus if [tex] m = \lfloor y \rfloor, \ M = \lceil y \rceil[/tex] we clearly have [tex]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}[/tex] (by a slightly generalized version of the squeeze theorem). Now, let [tex]a \in \mathbb{N}^+[/tex]. By l'Hopital's rule, applied [tex] a[/tex] times, we have [tex]\lim_{x \rightarrow \infty} x^ae^{x} = \lim_{x \rightarrow \infty} \frac{x^a}{e^x} = \lim_{x \rightarrow \infty} \frac{ax^{a1}}{e^x} = . \ . \ . = \lim_{x \rightarrow \infty} \frac{a!}{e^x} = a! \lim_{x \rightarrow \infty} e^{x} = 0[/tex] and since for [tex] y \ge 0[/tex] clearly [tex] M \geq m \geq 0, \ M, m \in \mathbb{N}[/tex] we get [tex] 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[/tex] as we wanted. QED. 


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