Plugging in Infinite for Improper Integrals?

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needhelpwithcalculus

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when solving improper integrals from 0 to infinite, i know you're supposed to plug in infinite for X and 0 for X after solving for the integrals. but when you're solving for the infinite portion, does infinite act as 0? for example:

integral from 1 to infinite of: ln(x) / (x^2)dx :
the integral is: -(lnx + 1 / x) and when I plug in ? for x,
why does - ln(A)+1 / 1 = 0 and not infinite?

if you add an undefined number + 1 and divide it by 0, wouldn't it be DNE?
the answer to the whole problem is 1.

sorry if this sounds confusing, thanks in advance!
 
needhelpwithcalculus said:
when solving improper integrals from 0 to infinite, i know you're supposed to plug in infinite for X and 0 for X after solving for the integrals. but when you're solving for the infinite portion, does infinite act as 0? for example:

integral from 1 to infinite of: ln(x) / (x^2)dx :
the integral is: -(lnx + 1 / x) and when I plug in ? for x,
why does - ln(A)+1 / 1 = 0 and not infinite?

if you add an undefined number + 1 and divide it by 0, wouldn't it be DNE?
the answer to the whole problem is 1.

sorry if this sounds confusing, thanks in advance!
Click to expand...

You are supposed to calculate the limit of the integrand as x ? infinity.

When simply replace 'x' by "infinity' - you get the indeterminate form (infinity)/(infinity)

Apply L'Hospital's Theorem.
 
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