- #1

theperthvan

- 184

- 0

The reason I was wondering is because integrating 1/(1+x^2) WRT x.

I know the answer is arctan(x), but how about breaking it into partial fractions by doing 1/(1-ix)(1+ix)?

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- Thread starter theperthvan
- Start date

- #1

theperthvan

- 184

- 0

The reason I was wondering is because integrating 1/(1+x^2) WRT x.

I know the answer is arctan(x), but how about breaking it into partial fractions by doing 1/(1-ix)(1+ix)?

- #2

KTC

- 90

- 0

Yes, though it's a mutivalued function so it's not as simple as with natural number.

- #3

quasar987

Science Advisor

Homework Helper

Gold Member

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So, log(i), whatever it is, should be a number such that exp(log(i))=i. Let's say log(i)=z=x+iy... then z is such that exp(z)=exp(x+iy)=exp(x)exp(iy)=exp(x)[cos(y)+isin(y)]=i. Well, x=0, y=pi/2 does the trick. But so does x=0, y=pi/2+2npi for any integer n. So log(i) is actually a

- #4

Alkatran

Science Advisor

Homework Helper

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Given a complex number x = a + bi we want to find its logarithm

[tex]c + di = ln(a + bi)[/tex]

Take the exponential of both sides

[tex]e^{c + di} = e^{ln(a + bi)}[/tex]

[tex]e^c e^{di} = a + bi[/tex]

[tex]e^c (cos(\frac{d}{\pi}) + i sin(\frac{d}{\pi})) = a + bi[/tex]

Separating the imaginary and real terms:

[tex]e^c cos(\frac{d}{\pi}) = a[/tex]

[tex]e^c sin(\frac{d}{\pi}) = b[/tex]

To get d, divide one by the other:

[tex]\frac{e^c}{e^c} \frac{sin(\frac{d}{\pi})}{cos(\frac{d}{\pi})} = \frac{b}{a}[/tex]

[tex]tan(\frac{d}{\pi})} = \frac{b}{a}[/tex]

[tex]d = tan^{-1}(\frac{b}{a}) \pi[/tex]

To get c, use the absolute values and Pythagoras:

[tex]e^c e^{di} = a + bi[/tex]

[tex]|e^c e^{di}|^2 = |a + bi|^2[/tex]

[tex]|e^c|^2 = a^2 + b^2[/tex]

[tex]e^c = \sqrt{a^2 + b^2}[/tex]

[tex]c = ln(\sqrt{a^2 + b^2})[/tex]

Note that you can add/subtract 2pi to d to get more solutions, which is a typical side effect of using arctan.

[tex]c + di = ln(a + bi)[/tex]

Take the exponential of both sides

[tex]e^{c + di} = e^{ln(a + bi)}[/tex]

[tex]e^c e^{di} = a + bi[/tex]

[tex]e^c (cos(\frac{d}{\pi}) + i sin(\frac{d}{\pi})) = a + bi[/tex]

Separating the imaginary and real terms:

[tex]e^c cos(\frac{d}{\pi}) = a[/tex]

[tex]e^c sin(\frac{d}{\pi}) = b[/tex]

To get d, divide one by the other:

[tex]\frac{e^c}{e^c} \frac{sin(\frac{d}{\pi})}{cos(\frac{d}{\pi})} = \frac{b}{a}[/tex]

[tex]tan(\frac{d}{\pi})} = \frac{b}{a}[/tex]

[tex]d = tan^{-1}(\frac{b}{a}) \pi[/tex]

To get c, use the absolute values and Pythagoras:

[tex]e^c e^{di} = a + bi[/tex]

[tex]|e^c e^{di}|^2 = |a + bi|^2[/tex]

[tex]|e^c|^2 = a^2 + b^2[/tex]

[tex]e^c = \sqrt{a^2 + b^2}[/tex]

[tex]c = ln(\sqrt{a^2 + b^2})[/tex]

Note that you can add/subtract 2pi to d to get more solutions, which is a typical side effect of using arctan.

Last edited:

- #5

MarkUS ViFe

- 1

- 0

but I had an Problem too:

i^0 = 1

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

...

It's going on like that,

but if you make de lg of the rationaly numbers...

lg(i^0)=lg(1) ____ lg(i^2)=lg(-1) ____ lg(i^4)=lg(1) ...

then there will be...

0*lg(i)=lg(1) ____ 2*lg(i)=lg(-1) _____ 4*lg(i)=lg(1) ...

____ 0=lg(1) ______ lg(i)=lg(-1)/2 _____ lg(i)=lg(1)/4 ...

NOW IS THAT POSSIBLE?

lg(1)=0 ... That's True...

BUT:

lg(-1)=?

My first Problem...

Just forget it for the moment...

SO: lg(i)=0/4 => lg(i)=0 => lg(i)=lg(1) => i=1 !

Now where was I wrong...?

or is it just that you can't work that way...

as an other example...

i^2= -1

i = -/-1]

i*i = -/ -1 ]*-/ -1 ]

So...

i*i = -/ -1*-1 ]

i*i = -/ 1 ]

i*i = 1 ??

Both time's i=1...

- #6

BloodyFrozen

- 353

- 1

I haven't read everything you wrote jet...

but I had an Problem too:

i^0 = 1

i^1 = i

i^2 = -1

i^3 = -i

i^4 = 1

...

It's going on like that,

but if you make de lg of the rationaly numbers...

lg(i^0)=lg(1) ____ lg(i^2)=lg(-1) ____ lg(i^4)=lg(1) ...

then there will be...

0*lg(i)=lg(1) ____ 2*lg(i)=lg(-1) _____ 4*lg(i)=lg(1) ...

____ 0=lg(1) ______ lg(i)=lg(-1)/2 _____ lg(i)=lg(1)/4 ...

NOW IS THAT POSSIBLE?

lg(1)=0 ... That's True...

BUT:

lg(-1)=? -> log(-1)=[π/ln(10)]i and in case your wondering ln(-1)=πi

My first Problem...

Just forget it for the moment...

SO: lg(i)=0/4 => lg(i)=0 => lg(i)=lg(1) => i=1 !

Now where was I wrong...?

or is it just that you can't work that way...

as an other example...

i^2= -1

i = -/-1]

i*i = -/ -1 ]*-/ -1 ]

So...

i*i = -/ -1*-1 ]

i*i = -/ 1 ]

i*i = 1 ??

Both time's i=1...

The fallacy is that √(xy)=√x√y is generally valid only if atleast one of the two numbers x or y is positive. (From Wiki)

- #7

Dickfore

- 2,988

- 5

[tex]

\log{(i)} = i \, \left(\frac{\pi}{2} + 2 \, k \, \pi\right), \; k \in \mathbb{Z}

[/tex]

\log{(i)} = i \, \left(\frac{\pi}{2} + 2 \, k \, \pi\right), \; k \in \mathbb{Z}

[/tex]

- #8

awkward

- 365

- 0

You don't actually need to find ln(i) in order to perform the integration using partial fractions, but I think you will need the following formula:

The reason I was wondering is because integrating 1/(1+x^2) WRT x.

I know the answer is arctan(x), but how about breaking it into partial fractions by doing 1/(1-ix)(1+ix)?

[tex]\tan^{-1}(z) = \frac{1}{2i} \ln \frac{1-iz}{1+iz}[/tex]

You might want to see if you can derive it (it's not hard).

- #9

michael urbano

- 2

- 0

log (i) = x

convert imaginary to exponential form...

thus;

(r)(exp^i) => i = (1)exp^(∏/2)i

then substitute to equation...

log (1exp^(∏/2)i)

basic logarithmic law...

log (ab) = log a + log b

therefore...

log (1exp^(∏/2)i) = log 1 + log exp^((∏/2)i)

log 1 = 0

log exp^((∏/2)i) = i log exp^(∏/2) = 0.6822 i

therefore...

log (i) = 0.6822i

- #10

michael urbano

- 2

- 0

log (i) = ?

let log (i) = x

therefore...

10^x = i

take the natural logarithm...

ln 10^x = ln i

ln 10^x = x ln 10

ln (i) = ln (1 exp^((∏/2)i) = ln 1 + ln exp^(∏/2)i = (∏/2)i

x ln 10 = (∏/2)i

x = (∏/2)i / ln 10

x = 0.6822 i

same as the answer of my first post...

different solution but one answer...

- #11

pwsnafu

Science Advisor

- 1,082

- 85

same as the answer of my first post...

different solution but one answer...

If you want a unique answer you need to pick a branch cut first.

- #12

thelema418

- 132

- 4

While the complex numbers contains all the real numbers, you may need to justify why you can get a solution of a real variable using a method with complex numbers.

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