Question

Integrate this with limits.

Answer 1

What needs to be integrated?

Answer 2

\[\int\limits_{0}^{\pi /4} \ln(tanx)\]

Answer 3

@myininaya @Zarkon @satellite73 @amistre64

Answer 4

im curious if a substitution would be beeficial

Answer 5

\[\int\limits \ln(1 - tanx) - \int\limits \ln(1+tanx) = I\]

Answer 6

say:

u = tan x

du = sec^2 x dx

cos^2 x du = dx

tan^-1 u = x

just a thought, migh tnot be prudent tho

Answer 7

tan = sin/cos

ln(sin/cos) = ln(sin) - ln(cos)

that might be good to work on

Answer 8

@amistre64 Hmm.
The troblem here is the ln.
How do you get rid of that with a substitution?

Answer 9

*problem

Answer 10

true, with these i have to just run through a few ideas usually before i remember things correctly :)

Answer 11

i don't think you are going to get a nice closed form for this one

Answer 12

Is this correct?
\[-\int\limits \ln(1 - tanx)dx = \int\limits_{0}^{\pi/4} \ln(1 + \tan(-x))d(-x)\]

Answer 13

http://www.wolframalpha.com/input/?i=integrate+ln%28tanx%29+dx

that does look quite nasty

Answer 14

id consider trying to see if theres a power series that you can work thru if you need to do this by hand

Answer 15

sorrt, but i cant verify sidds method. i just aint smart enough :)

Answer 16

Because.
\[I'=\int\limits_{0}^{\pi/4} \ln(1+tanx) = \int\limits_{0}^{\pi/4} \ln( 1 + \tan( \pi/4 - x)dx\]

\[I' =\int\limits_{0}^{\pi/4} \ln( 1 + (1-tanx)/(1+tanx))\]

\[I'=\int\limits_{0}^{\pi/4} (\ln2) - \int\limits_{0}^{\pi/4}\ln(1+tanx)\]

\[2I'= \int\limits_{0}^{\pi/4}\ln2dx\]

Answer 17

@mukushla Any idea how to do it?

Answer 18

emm..im thinking :)

Answer 19

your method is fine :)

Answer 20

wait u say for 0 to pi/4\[\int \ln (\tan x) \ \text{d}x=\int \ln (1-\tan x) \ \text{d}x-\int \ln (1+\tan x) \ \text{d}x\]right :)

now what?

Answer 21

and of course i agree with\[\int_0^{\pi/4} \ln(1+\tan x) \ \text{d}x=\frac{\pi}{8} \ln2 \]

Answer 22

but how about\[\int_0^{\pi/4} \ln(1-\tan x) \ \text{d}x=?\]

Answer 23

it will have ln(2tanx) in it instead of ln 2

Answer 24

1- tan (pi/4-x) = 1- ..../.... = 2tanx/(1+tan x)

Answer 25

oh yeah for the second one :\ yeah right

Answer 26

so we could find ln(tan x) right ?

Answer 27

\(\int \ln (\tan x) \ \text{d}x=\int \ln (1-\tan x) \ \text{d}x-\int \ln (1+\tan x) \ \text{d}x\)
from here

Answer 28

i think no :/

Answer 29

we have first integral, we have 2nd in terms of ln(tan x).....on right side.....
or is it that everything will cancel out?>?

Answer 30

will cancel out

Answer 31

as @satellite73 said no closed form for this one and as @amistre64 mentioned i think we must do it with series if we wanna use pen and paper only

Answer 32

letting\[t=-\ln(\tan x)\]gives\[I=\int_{0}^{\infty} -\frac{te^{-t}}{1+e^{-2t}} \ \text{d}t\]\[\frac{1}{1+e^{-2t}}=\sum_{n=0}^{\infty}(-1)^n e^{-2nt}\]so\[I=\int_{0}^{\infty} -te^{-t} \sum_{n=0}^{\infty}(-1)^n e^{-2nt}\ \text{d}t=-\sum_{n=0}^{\infty}(-1)^{n} \int_{0}^{\infty} te^{-(2n+1)t}\ \text{d}t=\]\[-\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^2}=-C=-0.9159...\]\[C : \text{Catalan's Constant}\]
http://mathworld.wolfram.com/CatalansConstant.html

Answer 33

Nice.
Guess you guys are right.

Answer 34

Catalan's constant. I havent used that before.
Thanks guys :)