Integration Question (1 Viewer)

[KFunk]

I was doing a past 3/4 yearly paper from my school (2003) and the last question really stumped me. Basically it was:

Evaluate ∫ ln(1 +tanx) dx {between 0 and π/4}

(*note) in the part just before that you had to show that:
∫ ln(1 +tanx) dx {between 0 and π/4} = ∫ ln[2/(1+tanx)] dx { 0 and π/4}


I was having a go doing the test in the time limit. I got everything else done but spent the rest of the time on this question. I punched it into mathematica to see if it could help me out and it gave me this as the answer:

1/4( i Log[-1 - i]² - i Log[-1 +i]² - πLog[2] )

¿qué?
What is that doing in our test? ... in other words, what did I do wrong??

 

[Trev]

Putting both y=ln[(1+tan[x]) and y=ln(2/(1+tan[x])) into Graphmatica, they only intersect when x= π/8. (Around that anyway, don't have a calculator around to check).

Maybe it was a typo in the exam that they didn't bother correcting on the paper you were given?

 

[Templar]

1/4( i Log[-1 - i]² - i Log[-1 +i]² - πLog[2] )

1/8 Pi Log[2]

 

[Slidey]

1/4( i Log[-1 - i]^2 - i Log[-1 +i]^2 - πLog[2] )

Recognise that e^(i@)=cis(@)
Now -1-i=sqrt(2)cis(5pi/4)
e^(5ipi/4)=-1-i
e^(3ipi/4)=-1+i
1/4( i Log[-1 - i]^2 - i Log[-1 +i]^2 - πLog[2] )
=1/4( i*(5i*pi/4)^2 - i*(3i*pi/4)^2 - πLog[2] )
=pi/4.ln2-25ipi/16+9ipi/16
=pi/4*ln2-ipi

how did you get your answer, Templar?

 

[Slidey]

1/4( i(Log[-1 - i]^2 - Log[-1 +i]^2) - πLog[2] )
Imaginary part equals:
(ln[-1-i]+ln[-1+i])(ln[-1-i]-ln[-1+i])
=ln2(ln[(-1-i)/(-1+i)])
=ln2.ln([-1-i]^2/2)=ln2(2ln[-1-i]-1)
=ln2.(5ip/2-1)
So the whole thing equals:
1/4(-[5pi/2]ln2-i.ln2 - pi.ln2)
=(ln2)/4(-i-7pi/2)

Now I'm confused.

 

[KFunk]

Answers are cheap (but impressive )... Anyone know how to connect the dots?

 

[Trev]

How do you integrate complex numbers?

 

[dawso]

a complex number is a constant hence treat it like any other, ie -

int i = ix + c

 

[Slidey]

CAAAAAREFUL, Matt. That's a whole can of worms better left closed for now.

 

[dawso]

but steele m8, i already opened it, and a can is made of aluminium, as such the skill required to weld it back shut is greater than wat i possess.....

seeing as though we dont need 2 know it, i say we move on, and in regards to ur question kfunk, no idea why it was in ur test??

 

[Trev]

a complex number is a constant hence treat it like any other, ie -

int i = ix + c

I'm an idiot. lol.

 

[Slidey]

No, Trev, it's a valid question. Complex Analysis is a huge area of maths at uni.

 

[Will Hunting]

How do you integrate complex numbers?

There should never, this year, arise a necessity to integrate an expression involving the complex number, i. Although a logical scrutiny will suggest that this is possible, no Extension 2 student should, by any means, feel penitent over an absence of knowledge thereof.

KFunk, the integrand is of the form ln[f(x)]

Let u = ln(tanx + 1), u' = sec²x/(tanx + 1)
Also, v' = 1, v = x

Now, by parts,

I = xln(tanx + 1) - ∫[(xsec²x)/(tanx + 1)]dx = xln(tanx + 1) - I₂

(From the limit 0 to π/4)

Let y = tanx, dy = sec²xdx
Also, x = tan^-1y
And, y = 0, 1 when x = 0, pi/4

Now,

I₂ = ∫[(tan^-1y).(1/(y + 1))]dy

Let u₂ = tan^-1y, u₂' = 1/(y² + 1)
Also, v₂' = 1/(y + 1), v₂ = ln(y + 1)

Now, by parts,

I₂ = tan^-1y.ln(y + 1) - ∫[cos²x.ln(tanx + 1)] = tan^-1y.ln(y + 1) - I₃

(From [ln(y + 1)]/(y² + 1) = [ln(tanx + 1)]/sec²x = cos²x.ln(tanx + 1))

Let u₃ = ln(tanx + 1), u₃' = sec²x/(tanx + 1)
Also, v₃' = cos²x, v₃ = x/2 + sin2x/4

Then,

I₃ = 1/2[xln(tanx + 1)] + 1/4[sin2xln(tanx + 1)] - 1/2I₂ - 1/4∫[(sin2xsec²x)/(tanx + 1)]dx

So, I = 3/2xln(tanx + 1) - tan^-1(tan x).ln(tanx + 1) +1/4[sin2xln(tanx + 1)] - 1/2I₂ - 1/4∫[(sin2xsec²x)/(tanx + 1)]dx

(NB. tan^-1(tan x).ln(tanx + 1) has been transformed from y into x for consistency, but could safely be left in y, provided the limits, 0 and 1, in y are used)

Ya, a mouthful, to say the least, but my outlook on this is still positive. I feel that that the emergence of the final integral, and the occurence, therein, of sin2x do provide some light at the end of the tunnel. A successful dissection of that integral would also enable further treatment of 1/2I₂ and, possibly, some closure on this saga.

 

[KFunk]

Thanks for taking a crack at it will. I just about got as far as you did but I wasn't nearly so optimistic. This question itself is only out of 2 or 3 marks so I'm doubtful about it being as epic as that. My three main problems are:

-Techniques I've tried (namely int. by parts) are long winded and don't seem to have a quick solution... at all.
-Mathematica evaluates it as a complex integral.
-It's placed on a highschool paper in such a manner as to suggest it isn't too difficult a question and that it has a real solution.

grr.

 

[ngai]

∫ ln(1 +tanx) dx {between 0 and π/4}

given
∫ ln(1 +tanx) dx {between 0 and π/4} = ∫ ln[2/(1+tanx)] dx { 0 and π/4}


ok surely the exam setters arent that evil...they (usually) dont give u nonsense

let I = ∫ ln(1 +tanx) dx {between 0 and π/4}
ln(1+tanx) = -ln(1/(1+tanx) = -[ln(2/2(1+tanx))]
which gives ln(1+tanx) = -ln(2/(1+tanx)) + ln2 (log rules)
so integrate both sides from 0 to pi/4
I = -I + integral of ln2 from 0 to pi/4
2I = 1/2 int of ln2
= 1/2 ln2 (pi/4 - 0)
= pi/8 ln2

 

[martin]

let I = ∫ ln(1 +tanx) dx {between 0 and π/4}
ln(1+tanx) = -ln(1/(1+tanx) = -[ln(2/2(1+tanx))]
which gives ln(1+tanx) = -ln(2/(1+tanx)) + ln2 (log rules)
so integrate both sides from 0 to pi/4
I = -I + integral of ln2 from 0 to pi/4
2I = 1/2 int of ln2
= 1/2 ln2 (pi/4 - 0)
= pi/8 ln2

Yeah, nice one ngai.

To show that Mathematica's answer is equivalent we need to use the complex logarithm:
if z=r*e^(it)
Log(z) = ln(r) + it
(Note that this has most of the features of the real log, like exp(log(z)) = z, but is defined for all complex numbers).

Then -1-i = sqrt(2)*e^(-i3pi/4) and -1+i=sqrt(2)*e^(i3pi/4)
so Log(-1-i)=ln(sqrt(2)) + -i3pi/4 and Log(-1+i)=ln(sqrt(2)) + i3pi/4

as (a-b)^2 - (a+b)^2 = -4ab
[Log(-1-i)]^2 - [Log(-1+i)]^2 = -4*ln(sqrt(2))*i3pi/4 = -i3pi*ln(sqrt(2)) = -i3pi*ln(2)/2

so 1/4[i(log(-1-i))^2 - i(log(-1+i))^2 - pi*log2]
= 1/4[(3/2)pi*ln2 - pi*ln(2)]
= 1/4[(1/2)pi*ln2]
= (1/8)pi*ln2

So whats the moral of the story? Computers can do all integrations that are possible but they don't always give a simple answer. It might be easier to just do the integration numerically than symbolically. For instance in Matlab, int(log(1+tanx),0,pi/4) gives an even more complicated answer, but quad(log(1+tanx),0,pi/4) = 0.2722 (using something like simpsons rule) which is approximately (1/8)pi*ln2

 

[KFunk]

∫ ln(1 +tanx) dx {between 0 and π/4}

given
∫ ln(1 +tanx) dx {between 0 and π/4} = ∫ ln[2/(1+tanx)] dx { 0 and π/4}


ok surely the exam setters arent that evil...they (usually) dont give u nonsense

let I = ∫ ln(1 +tanx) dx {between 0 and π/4}
ln(1+tanx) = -ln(1/(1+tanx) = -[ln(2/2(1+tanx))]
which gives ln(1+tanx) = -ln(2/(1+tanx)) + ln2 (log rules)
so integrate both sides from 0 to pi/4
I = -I + integral of ln2 from 0 to pi/4
2I = 1/2 int of ln2
= 1/2 ln2 (pi/4 - 0)
= pi/8 ln2

Ahh, genius. Thank you very much Ngai, you have put my mind at ease. I'll have to remember that kind of manipulation for the future.

 

[lfc_reds2003]

he should not only be thge master of the complex planes n conic sections but manipulation n integration ... lol

Pristine mathematics Andy boi LOL

 

[KFunk]

Perhaps we should send a letter to the editor correcting them?

 

[ngai]

lol complex and conics...they just asked me to name some things that are in the 4u course..so i listed some of the topics...and i suppose they chose those 2