# Prove that integral is convergent

• elabed haidar
In summary, the conversation discusses the integral of x^alpha times (lnx)^ beta from 0 to 0.5 and the question of whether it converges when alpha is greater than -1. The methods attempted include finding a bound for log(x), using substitution, integration by parts, and the theory of multiplying by x^m. The final suggestion is to apply the test that states if \int_{-\infty}^0{g} converges and \lim_{x\rightarrow-\infty}{\frac{f}{g}}=0, then \int_{-\infty}^0{f} converges.
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

## The Attempt at a Solution

So what did you try?

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.

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

i tried all methods

Try a substitution first to get rid of the logarithm.

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

i tried

So, what did you get?

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

What did you get after the substitution x=et?

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

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)

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

elabed haidar said:
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.

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 = et, 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 = tβ. 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?

Last edited:

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 $\beta$?

## 1. What does it mean for an integral to be convergent?

An integral is said to be convergent if its value approaches a finite number as the limits of integration approach a certain value.

## 2. How do you prove that an integral is convergent?

To prove that an integral is convergent, you can use various convergence tests such as the comparison test, the ratio test, or the integral test.

## 3. What are the conditions for an integral to be convergent?

For an integral to be convergent, the function being integrated must be continuous on the interval of integration, and the limits of integration must be finite.

## 4. Can an integral be convergent and divergent at the same time?

No, an integral can only be either convergent or divergent. If the integral satisfies the conditions for convergence, it is considered convergent. Otherwise, it is considered divergent.

## 5. What is the importance of proving that an integral is convergent?

Proving that an integral is convergent is important in determining the behavior of the function being integrated and in evaluating the integral accurately. It also helps in solving various mathematical problems involving integrals.

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