Question

\[\large\int\limits_0^{i\pi}e^x\,\mathrm dx\]

Answer 1

( nice and easy )

Answer 2

-2 ?

Answer 3

yes

Answer 4

I would have said there's no answer...

Answer 5

can we see the area or whatever that -2 represents :o

Answer 6

http://www.wolframalpha.com/input/?i=%5Cint_0%5E%7Bi%5Cpi%7De%5Ex+dx

not sure why wolfram is hesitating to plug e^(ipi) = 1

Answer 7

*e^(ipi) = -1

Answer 8

I think WA is considering \(i\) to be a variable, not the square root of -1.

Answer 9

nvm

Answer 10

\[\begin{align}
\large\int\limits_0^{i\pi}e^x\,\mathrm dx
&=e^x\Big|_0^{i\pi}\\
&=e^{i\pi}-e^0\\
&=\Big[\cos(\pi)+i\sin(\pi)\Big]-1\\
&=[-1+0]-1\\
&=-2
\end{align}\]

Answer 11

Yeah :p

Answer 12

A definite integral \(\displaystyle\int_a^b f(x)~dx\) is evaluated over an interval \((a,b)\), with \(a<b\). Is \(0<i\pi\)?

Answer 13

\(\le\)*

Answer 14

i don't know what diagram could represent the integral

i don't know if 0 ≤ iπ

Answer 15

Regarding your second doubt, it's not - the statement doesn't make sense because the complex numbers aren't ordered.

Answer 16

As for the first, you could say it's the line/contour integral over the line segment connecting the origin and (0,1) in the complex plane, but even so those limits don't make sense.

Answer 17

maybe 0 = iπ

Answer 18

Sure, if \(i\) or \(\pi\) is 0, but we both know you mean \(i=\sqrt{-1}\) and \(\pi=3.14...\), right?

Answer 19

i was thinking modulo something, but yeah i = √-1 , π ~ 3

Answer 20

\[\begin{align}
\int\limits_0^{2i\pi}e^x\,\mathrm dx
&=e^x\Big|_0^{2i\pi}\\
&=e^{2i\pi}-e^0\\
&=\Big[\cos(2\pi)+i\sin(2\pi)\Big]-1\\
&=[1+0]-1\\
&=0
\end{align}\]

\[\begin{align}
\large\int\limits_0^{3i\pi}e^x\,\mathrm dx
&=e^x\Big|_0^{3i\pi}\\
&=e^{3i\pi}-e^0\\
&=\Big[\cos(3\pi)+i\sin(3\pi)\Big]-1\\
&=[-1+0]-1\\
&=-2
\end{align}\]

Answer 21

Well we have the same problem. What does it mean to integrate over an interval complex numbers with nonzero imaginary parts?

Answer 22

I think I see what you're getting at, the pattern's there, but the notation and computation aren't valid imo

Answer 23

Then again, replacing \(i\) with \(\sqrt{-1}\) in WA gives your answer: http://www.wolframalpha.com/input/?i=Integrate%5BExp%5Bx%5D%2C%7Bx%2C0%2CSqrt%5B-1%5D*pi%7D%5D

So I don't know what to think anymore :/