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There's no special initial conditions, restrictions, or anything like that right?

:) yeah i remember, for now there is not :)

this appears to be the same question that was posted earlier

@experimentX dude i could not do that :/

and today is the last day :/

which one??

they are all connected

i think i already did the first one.

you found sqrt2-i(pi)/4

i am not sure it is true or not

what's the answer supposed to be??

final answer should be -(pi)*i

for question no 1??

no they are connected each other book says only an answer and it is -(pi)*i

O ... let's see for 2

|dw:1343076384065:dw|

|dw:1343076480604:dw|

|dw:1343076552517:dw|

|dw:1343076614227:dw|

|dw:1343076688932:dw|
sorry man ... i couldn't get any better

:) okey bro thanks again

Have you had the residue theorem?

yes i guess

but i did not learn well :/

If you know the Residue Theorem you can cut your work from computing three line integrals to one.

Mimic my post above for 2 remaining lines.

hmm then will i continue with 1/4(ln4.... or i always use the i/1+iy

for the second integral
z= x -I
dz = dx
\[\int_1^{-1} \frac{1}{x-i} \,
dx=-\frac{1}{2} (i \pi )
\]

how can i understand the integral part for ex. why the second one become 1/x-i

How much is y on the second line? -1
How much is x on the seconfd line? x from 1 to -1
z= x -i

On the third line
z= -1 + i y
dz= i dy
\[
\int_{-1}^0 \frac{i}{-1+i y}
\, dy
\]

ok i understand i guess thank u :)

\[
\int_{-1}^0 \frac{i}{-1+i y}
\, dy=-\frac{1}{4} i (\pi
-2 i \log (2))
\]

log=ln

thank u so much :)

yw