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How do you evaluate the integral #int 1/x# from [0,1]?

Answer 1

It's undefined.

The integral #int_0^1 1/xdx# is equivalent to #int_0^1x^-1dx#. All integrals of the form #int_0^t x^ndx#, where #t# is a real parameter, are equal to #t^(n+1)/(n+1)#, so if #t = 1# and #n=-1#, #int_0^1 1/xdx# = #1^0/0#, which is undefined.

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Answer 2

Christopher Allers

Diverging...

#int_0 ^1 1/x dx = [ln(x) ]_0 ^1 #

#=> ln1 - ln0 = ln 0 #

We know #ln0# is undefined, but #lim_(n to 0^+ ) ln n = -oo#

So hence the integral is diverging...

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Answer 3

Emma Aldridge

The integral ( \int_{0}^{1} \frac{1}{x} , dx ) is improper because it has a singularity at ( x = 0 ). To evaluate it, we need to use the limit definition of the integral:

[ \int_{0}^{1} \frac{1}{x} , dx = \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} , dx ]

This becomes:

[ \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} , dx = \lim_{a \to 0^+} \left[ \ln|x| \right]_{a}^{1} ]

[ = \lim_{a \to 0^+} \left( \ln|1| - \ln|a| \right) ]

[ = \lim_{a \to 0^+} \left( 0 - \ln|a| \right) ]

[ = \lim_{a \to 0^+} -\ln|a| ]

[ = -\infty ]

Therefore, the integral ( \int_{0}^{1} \frac{1}{x} , dx ) diverges to negative infinity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.