Question

int from -1to1 1/x dx

Answer 1

is that a question ?

Answer 2

yes

Answer 3

what is the solution?

Answer 4

let u=1-x
du=-du
int(1-u)/udu
=-int1/u-1du
=1-x-ln(1-x)+c

Answer 5

correct

Answer 6

let's not
let's instead note that 1/x is an odd function

Answer 7

\[\int_{-a}^{a}f(x)dx=0\]if f(x) is odd

Answer 8

@TuringTest if f(x) is odd then what?

Answer 9

then the above theorem holds

Answer 10

ok

Answer 11

@TuringTest can you please show me the steps of this question?

Answer 12

wow u dont give me a meda

Answer 13

there is a theorem that\[\text{ iff }f(x)\text{ is odd, then }\int_{-a}^{a}f(x)dx=0\]do you know what the definition of an odd function is @woundedtiger40 ?

Answer 14

I never heard of it :(

Answer 15

a function f(x) is odd if\[f(-x)=f(x)\] so functions like x, x^3, and 1/x are odd, because\[f(-x)=\frac1{-x}=-\frac1x=-f(x)\]

Answer 16

but when I try to solve this question in maple it says undefined

Answer 17

because you have a singularity at x=0, oops forgot about that :/

Answer 18

I have found something about it at http://planetmath.org/encyclopedia/IntegralOfOddFunction.html but what exactly is a an odd function?

Answer 19

gotcha

Answer 20

I just defined an odd function above
when f(-x)=-f(x) the function is odd
plot a few odd functions on wolfram like x, x^3, 1/x, etc into wolfram and you will see they have a certain symmetry

Answer 21

http://www.mathsisfun.com/algebra/functions-odd-even.html

Answer 22

thanks dude

Answer 23

but because we have a singularity at x=0 this is an improper integral, so we need to convert this to\[\lim_{a\to0^-}\int_{-1}^{a}\frac{dx}x+\lim_{a\to0^+}\int_{a}^{1}\frac{dx}x\]in other words, my trick won't work here because 1/x is undefined at x=0
do yuo know anything about improper integrals yet?

Answer 24

so actually, this integral diverges unless you use the Cauchy-Principle Value, which I will not explain in detail, but basically allows you to use my odd function trick in this case
i.e.\[\int_{-1}^{1}\frac{dx}x\text{ diverges, but }PV\int_{-1}^{1}\frac{dx}x=0\]