Question

What is converging and diverging of integrals?

Answer 1

Please use simple examples to explain!

Answer 2

convergence of improper integral means they have some real value (not infinity)
if the value of integral comes out to be infinity or -infinity or infinity - infinity we say the integral diverges.here is an example.
\[\Large \int\limits_{1}^{\infty}\frac{1}{\sqrt{x}}\]

\[\Large 2|\sqrt{x}|_{1}^{\infty} =\infty\]
so it is divergent
now look at the following integral
\[\Large \int\limits_{2}^{\infty}\frac{1}{x^2}\]
\[\Large -1 |\frac{1}{x}|_{2}^{\infty}=-1[\frac{1}{\infty}-\frac{1}{2}]=\frac{1}{2}\]
so it is divergent because we have some number instead of just infinity.
i hope it is clear !

Answer 3

the last sentence is a typo it is convergent !!!
the second integral converges because it has some value 1/2 instead of infinity.

Answer 4

in my own words...converging - the integral exists...diverging - doesnt exist...well according to my understanding anyway =))

Answer 5

What is wrong with this question?
http://www.wolframalpha.com/input/?i=integrate+%28ln+x%29%2F%281%2B+ln+x%29^2+from+0+to+1

Answer 6

@lgbasallote yes you are right . just wanted to make it clear with example.

Answer 7

heh i never said you were wrong...i just gave my definition :)

Answer 8

@DLS integrate the above question you have mentioned then apply the limits. tell me what you get.

Answer 9

x/(1+log(x)) ?

Answer 10

yes apply limits.

Answer 11

Something strange :/

Answer 12

there is a singularity between 0 and 1