Question

Show that the integral of (1+x)/(1+x^2)dx = Pi

Answer 1

\[\int\limits_{-\infty}^{\infty}(1+x)/(1+x^2)dx=\Pi\]

Answer 2

split up the integral and it should be a snap...

Answer 3

making it \[\int\limits_{-00}^{oo}1/(1+x^2) + \int\limits_{-00}^{00}x/(1+x^2)\]

Answer 4

\[\int_{-\infty}^{\infty}\frac{1+x}{1+x^2}dx=\int_{-\infty}^{\infty}\frac1{1+x^2}dx+\int_{-\infty}^{\infty}\frac x{1+x^2}dx\]the first integrand is even and the second is odd, so this becomes\[=2\int_{0}^{\infty}\frac1{1+x^2}+0=2\tan^{-1}x|_{0}^{\infty}=2\lim_{n \rightarrow \infty}\tan^{-1}n=\pi\]badda-bing :D
any questions?

Answer 5

make sense?
yes? no? maybe?

Answer 6

heheh not yet

Answer 7

how do you identify even and odd is it the power?

Answer 8

and wheres the 2 comming from?

Answer 9

an odd function is defined as a function such that\[f(-x)=-f(x)\]and an even function is defined as a function such that\[f(-x)=f(x)\]check to see that the first integrand is even and the second is odd while I type the rest of the explanation

Answer 10

for all even functions we have the following very useful fact\[\int_{-a}^{a}f(x)=2\int_{0}^{a}f(x)\]think about the area under the graph of x^2 (which is an even function) from x=-a to x=a and this should make sense
the area is symmetric !

Answer 11

for all odd functions we have that\[\int_{-a}^{a}f(x)=0\]this should make sense if you consider the area under x^3 (an odd function) from x=-a to x=a
again the areas on each side of x=0 are the same, but they are negative on one side and positive on the other
hence the integral is zero

Answer 12

oh im seeing it

Answer 13

here are pics of what I'm saying about the areas
even:
http://www.wolframalpha.com/input/?i=integral+from+-1+to+1+of+x%5E2dx
odd:
http://www.wolframalpha.com/input/?i=integral+from+-1+to+1+of+x%5E3dx

Answer 14

the integral does not converge unless you use the Cauchy principal value

Answer 15

The who-what?
what is wrong with my analysis?

Answer 16

I did see that the integral would not yield pi unless I used the even-odd trick, but I thought I was just messing up...

Answer 17

cool yea i got it! thanks you!

Answer 18

but zarkon said something and I want to understand it :/

Answer 19

when taking the integral \[\int\limits_{-\infty}^{\infty}f(x)dx\]

you cant do this

\[\lim_{a\to\infty}\int\limits_{-a}^{a}f(x)dx\]

( which is what your odd function trick relies upon)

Answer 20

so you can't use the odd function thing at all in improper integrals?
then how do I solve this?

Answer 21

you shouldn't

Answer 22

unless you are finding the Cauchy principal value :)

Answer 23

I am looking at wikipedia for the Cauchy principle value thing...
I don't suppose you'd care to demonstrate it here, eh? I've never heard of it.

Answer 24

ex

\[\lim_{a\to\infty}\int\limits_{-a}^{a}\frac{x}{1+x^2}dx=0\]

Answer 25

the integral is actually
\[\int\limits_{-\infty}^{\infty}\frac{x}{1+x^2}dx=\lim_{a\to-\infty}\lim_{b\to\infty}\int\limits_{a}^{b}\frac{x}{1+x^2}dx\]

which does not exist

Answer 26

the Cauchy principle value is from complex analysis, or what?

Answer 27

I don't know the original context of how it was first used

Answer 28

@Lammy Zarkon is saying my proof is flawed, so don't rely on it.

Though maybe your teacher will let it slide if they are not as smart as him :P

so Zarkon, what is your advice for Lammy and I who obviously don't know this technique yet
are we without hope?

Answer 29

haaha but i liked that way

Answer 30

if he is supposed to show that the integral is pi then maybe his prof wants him to find the CPV...then your way of doing the integral is fine

Answer 31

never learn the CPV yet

Answer 32

though in reality...the integral does not converge

Answer 33

I can see I need to read about this to know what you are really saying...
the integral does not converge, but if I use this CPV then it does...and my trick becomes valid...
hm...
Until then, I suppose you will have to do with my flawed proof Lammy

Answer 34

is this for a physics class?

Answer 35

they tend to like to use the CPV without stating so.

Answer 36

no its for calulus class

Answer 37

I'm betting the professor overlooked it because he's not as good as Zarkon ;)

Answer 38

I would ask your professor about it.

Answer 39

was the problem from a book or from the prof?

Answer 40

its from the prof!

Answer 41

haha yea this prof. makes lots of mistake while lecturing

Answer 42

i will ask her on monday

Answer 43

from mathematica v8

Integrate::idiv: Integral of 1/(1+x^2)+x/(1+x^2) does not converge on {-\[Infinity],\[Infinity]}. >>


then using the CPV

In[6]:= NIntegrate[x/(1 + x^2), {x, -\[Infinity], \[Infinity]},
Method -> {"PrincipalValue"}]

Out[6]= 0.

Answer 44

so it doesn't converge to pi even with the CPV ?

Answer 45

it will if I did the entire problem

Answer 46

oh ic, you just did the second integral

Answer 47

yes...the first integral is ok...the second one is the bad one

Answer 48

yeah,because at first I didn't see that it was odd, and I got divergence when I evaluated it
then I saw the oddness and was like 'aha!', and forgot about trying to explain my earlier issue
clearly it is not to be ignored

Answer 49

just out of curiosity...is your prof a full time prof/adjunct/grad student?

Answer 50

it should not be ignored :)

Answer 51

full time prof

Answer 52

time for a shower....fun talking...later

Answer 53

thanks a lot

Answer 54

np

Answer 55

dinner for me
later y'all !