Integral of ln(1+tanx) from 0 to pi/4 | Homework Equations

[XtremePhysX]

Homework Statement

Evaluate the following integral:

Homework Equations

$$\int_{0}^{\frac{\pi }{4}}ln(1+tanx)dx$$

The Attempt at a Solution

I tried to split is into ln(1) times ln(tanx) but didnt work.

 

[Millennial]

Because you can't split it up like that perhaps?

Use this identity of integrals: Substituting x=a-t, dx=-dt:

$$\int_{0}^{a}f(a-t)\,dt=-\int_{a}^{0}f(x)\,dx=\int_{0}^{a}f(x)\,dx$$

 

[XtremePhysX]

can u please show me how to do it, i tried it this way but i messed it up.

 

[clamtrox]

That is a pretty difficult integral. If you really wrote it down correctly, the best approach is likely to evaluate it numerically.

 

[Millennial]

I'll give you a headstart, but you should solve it yourself.
Applying the result I gave above to your problem, we get that your integral is equal to
$$\int_{0}^{\pi/4}\log(1+\tan(\pi/4-x))dx$$
Now, note that $\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$ so that
$$\tan(\pi/4-x)=\frac{\sin(\pi/4-x)}{\cos(\pi/4-x)}=\frac{\sin(\pi/4)\cos(x)-\cos(\pi/4)\sin(x)}{\cos(\pi/4)\cos(x)+\sin(\pi/4)\sin(x)}=\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}$$
This is easily seen to be equal to
$$\frac{1-\tan(x)}{1+\tan(x)}$$
Now, please take it from there.

 

[Millennial]

That is a pretty difficult integral. If you really wrote it down correctly, the best approach is likely to evaluate it numerically.

The indefinite integral is in terms of the logarithmic integral function but the definite integral has a solution.

 

[XtremePhysX]

Thank u a lot.

great forum.

 

[Millennial]

What does that mean? Do you have the answer??

 

[XtremePhysX]

I don't have the answer but thanks to u, i solved it, the answer is pi/4 ln(2)

 

[XtremePhysX]

guys, I am learning harder integration so can anyone give me an example of an extremely difficult integral?

 

[Millennial]

$$\int_{0}^{\infty}\frac{\cos(x)\,dx}{1+x^2}$$
Good luck!

 

[XtremePhysX]

wow that's a hard one
do you have so really big, difficult and intimidating integrals that looks almost impossible to solve

 

[Millennial]

Discussion here is not really appropriate, I shall send some to you via PM.

 

[XtremePhysX]

thank you
im new here so excuse me please :)

 

[gabbagabbahey]

I don't have the answer but thanks to u, i solved it, the answer is pi/4 ln(2)

Is it? You might want to sketch the integrand and see if your result is reasonable for the area under the curve ;0).

 

[srl17]

Your really close, you probably made a simple arithmetic mistake in your work. The graph of that is very close to a triangle and the area you found wouldn't fit in the bounds of a triangle. That was fun integral though!

 

[Ray Vickson]

I don't have the answer but thanks to u, i solved it, the answer is pi/4 ln(2)

Wrong. If we let $J = \int_0^{\pi/4} \, \ln(1 + \tan(x))\, dx,$ your claim is that $J = \pi/4\ln(2).$ If you mean $J = (\pi/4) \ln(2) \doteq 0.5443965,$ that is wrong. If you mean $J = \pi/(4 \ln(2)) \doteq 1.1330900,$ that is also wrong. The correct answer is
$$J = \frac{i}{2}\text{dilog}\left(\frac{1+i}{2}\right) - \frac{i}{2}\text{dilog}\left(\frac{1-i}{2}\right) + \frac{\pi \ln(2)}{4} - C,$$
where C is Catalan's constant
$$C = \sum_{i=0}^{\infty} \frac{(-1)^i}{(2i+1)^2}$$
and
$$\text{dilog}(w) = \int_1^w \frac{\ln(t)}{1-t} \, dt$$
and $i = \sqrt{-1}.$ The numerical value of the correct answer is
$J \doteq 0.27219826.$ (All this courtesy of Maple11).


RGV

 

[srl17]

J≐0.27219826 looks like $$\frac{\pi}{8}\ln{2}$$ which is why I think it likely he made a simple arithmetic mistake.

 

[Bohrok]

He probably made the same mistake I did when I worked it out and got the same answer as him but I did eventually get$$\frac{\pi}{8}\ln2$$