Prove that integral is convergent

[elabed haidar]

Homework Statement

i need to know the integral of x^alpha times (lnx)^ beta from 0 to 0.5
the question is if alpha greater than -1 prove that integral convergent

Homework Equations

The Attempt at a Solution

 

[micromass]

So what did you try?

 

[henry_m]

I don't think you'll be able to actually do that integral in closed form, but you can identify when it exists. You'll need to find a bound for log(x) as x approaches 0.

 

[elabed haidar]

i don't want to know what is the answer , i just want to prove how does it converge at zero

 

[elabed haidar]

i tried all methods

 

[micromass]

Try a substitution first to get rid of the logarithm.

 

[elabed haidar]

i tried the neighbourhood at zero since the problem is at zero , but i don't think lnx has a neighbourhood at zero , so i tried change by varibale x= e^t but still didnt work
then i tried integration by parts , and that was a disaster I am stuck
and last i tried the theory which says that if you mulitply by x ^m where m is less than 1 and you reach the limit zero you will have a convergent integral but that only works with real numbers not variables so what do you think i should do? please make sure

 

[elabed haidar]

i tried

 

[micromass]

So, what did you get?

 

[elabed haidar]

i told you in the other thread its the same question they told me to write where i told you

 

[micromass]

What did you get after the substitution x=e^t?

 

[elabed haidar]

man it becomes from -infinte to zero et ^a times x ^beta but still nothing i cnt go anywhere with that

 

[micromass]

So you get

$$\int_{-\infty}^0{e^{\alpha+1}t^\beta dt}$$

right?

Now, apply partial integration to reduce the value of beta (i.e. to make beta negative)

 

[elabed haidar]

and what do i get please can you solve it i was trying all day please

 

[micromass]

and what do i get please can you solve it i was trying all day please

I'm sorry, that would be against the rules. I'm afraid you will have to solve it.

Alternatively, do you know the following test:

If $\int_{-\infty}^0{g}$ converges and $\lim_{x\rightarrow-\infty}{\frac{f}{g}}=0$, then $\int_{-\infty}^0{f}$ converges as well.

Applying this with an appropriate g could also give you the solution.

 

[Unit]

To summarize everything so far (there have been a few errors), the original integral is

$$\int_0^{0.5} x^\alpha (\ln{x})^\beta dx$$

Micromass suggested the substitution x = e^t, which leads to

$$\int_{-\infty}^{-\ln{2}} e^{t(\alpha + 1)} t^{\beta} dt$$

Note that the upper limit is -ln(2) = ln(0.5), not 0.

From here, try parts with u = t^β. The following may help:

$$\int_a^b u dv = vu \; \big|_a^b - \int_a^b v du$$

Specifically, for what values of $\alpha$ and $\beta$ will $vu \; \big|_a^b$ converge?

 

[hunt_mat]

Why not try looking at certain values of $\beta$, say for example $\beta =1$, then your integral becomes:
$$\int_{0}^{\frac{1}{2}}x^{\alpha}\ln xdx=\int_{-\infty}^{-\ln 2}ye^{(\alpha +1)y}dy$$
upon using the substitution $y=\ln x$. Now note that the interval of integration is over a negative region which means that $\alpha +1>0$ and so $\alpha >-1$, which is what we were trying to show in the first place. Can you now show this for general $\beta$?