# Does this integral converge or diverge?

• Dell
In summary, the given integral \int\frac{dx}{ln(x)} (from 0 to 1) is divergent. This can be seen by transforming the integral using the substitution t=ln|x| and evaluating the limits at negative infinity and zero, which show that the integral of exp(t)/t converges at negative infinity and diverges at zero. Other methods were also attempted, such as finding a similar function with known convergence or divergence properties, but ultimately the conclusion remained the same. The area between 0.4 and 0.6 was not taken into account since the function is constant there.

#### Dell

the given integral is:

$$\int$$$$\frac{dx}{ln(x)}$$ (from 0 to 1)

now i have a problem at both ends of the integral, because i have ln(0) is undefinde and ln(1) is 0 and 1/0 is undefined, so i can integrate from 0 to 0.5 and 0.5 to 1, but i tried to integrate this and just couldnt, i tried substitution
t=ln(x)
x=et
dt=dx/x

and then tried integration in parts but this just didnt work,

so then i thought maybe i need to find a similar function which i know more about or can easily find more about, but again i have problems at both limits, so i think i need 2 separate functions, since i know that 1/ln(x) is negative in the limits 0-1, i will take the abs value of 1/ln(x) and look for positive functions...
i am looking for an integral that is either
*bigger than my original integral and converges
*smaller than my original integral and diverges

what i found was
$$\int$$dx/$$\sqrt{x}$$ (from 0 to 0.4) which converges and is bigger than $$\int$$dx/|ln(x)| (from 0 to 0.4 )

and my second integral is $$\int$$1/x (from 0.6 to 1) which diverges and is smaller than $$\int$$dx/|ln(x)| (from 0.6 to 1 )

therefore i know that the integral $$\int$$$$\frac{dx}{ln(x)}$$ (from 0 to 1) diverges near 1, and converges near 0, so all in all it diverges.

is this okay? is it alright that i didnt take into account the area between 0.4 and 0.6 since the function is constant there anyway??

can you see any better way to solve this??

Dell said:
is this okay? is it alright that i didnt take into account the area between 0.4 and 0.6 since the function is constant there anyway??

can you see any better way to solve this??

I came to a similar conclusion using your first substitution t=ln|x|. I found this easier to visualize. The lower limit of zero transforms to negative infinity and it is clear that the integral of exp(t)/t converges at negative infinity. However the upper limit of 1 transforms to zero and it is clear that the integral of exp(t)/t diverges at zero, since the numerator goes to unity and the denominator is a divergent 1/t form.