Impossible integral (1 Viewer)

Physicklad · Sep 26, 2015

could someone please destroy this integral it deprives my sleep ty

integral95 · Sep 26, 2015

sorry mate, not even wolfram alpha can give you a simple answer.

Mathsisfun15 · Sep 26, 2015

i.e. beyond the scope of the extension 2 course

calamebe · Sep 26, 2015

Mathsisfun15 said:
i.e. beyond the scope of the extension 2 course

Nah that seems pretty simple

Ekman · Sep 26, 2015

I was able to manipulate it to:

$\int -x\cot{x} dx$

Which I believe cant be integrated, via elementary methods.

Mathsisfun15 · Sep 26, 2015

Mathsisfun15 said:
i.e. beyond the scope of the extension 2 course

although i'm curious to why the 3rd line is there
surely
-log(x^2+1)+log(x-i)+log(x+i)
=-log(x^2+1)+log(x^2+1)
=0

integral95 · Sep 26, 2015

Mathsisfun15 said:
although i'm curious to why the 3rd line is there
surely
-log(x^2+1)+log(x-i)+log(x+i)
=-log(x^2+1)+log(x^2+1)
=0

The log law doesn't apply to complex numbers

jonathan_lam · Sep 26, 2015

There is a similar question in a Sydney Grammar textbook: $\int_0^1\frac{\log(1+x)}{1+x^2}$ . It also appeared in a Putnam a long time ago.

Use the substitution $x=\tan\theta$ so $I= \int_0^{\frac{\pi}{4}} \log(1+\tan\theta) \, d\theta$

Use the fact that $\int_0^a f(x) =\int_0^a f(a-x)$ so

$I = \int_0^{\frac{\pi}{4}} \log\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \, d\theta = \int_0^{\frac{\pi}{4}} \log\biggl(\frac{2}{1+\tan\theta} \biggr) \, d\theta$

Add the two equations together and we get

$2I = \int_0^{\frac{\pi}{4}} \log(2) \, d\theta$

Ekman · Sep 26, 2015

jonathan_lam said:
The question was in a Sydney Grammar textbook.

Use the substitution $x=\tan\theta$ so $I= \int_0^{\pi/4} \log(1+\tan\theta) \, d\theta$

Use the fact that $\int_0^a f(x) =\int_0^a f(a-x)$ so

$I = \int_0^{\pi/4} \log\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \, d\theta = \int_0^{\pi/4} \log\biggl(\frac{2}{1+\tan\theta} \biggr) \, d\theta$

Add the two equations together and we get

$2I = \int_0^{\pi/4} \log(2) \, d\theta$

Correction* Your integral should look like this after the substitution:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+tan^{2}\theta)} d\theta$

Ekman · Sep 26, 2015

jonathan_lam said:
After you simplify it you get what I wrote.

Can you please show your working out, because I don't see how else you can get:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+\tan^{2}\theta)} d\theta$

to equal:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+\tan\theta)} d\theta$

and get to the answer you suggested

Drsoccerball · Sep 26, 2015

Ekman said:
Can you please show your working out, because I don't see how else you can get:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+\tan^{2}\theta)} d\theta$

to equal:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+\tan\theta)} d\theta$

and get to the answer you suggested

i got what you got

jonathan_lam · Sep 26, 2015

Ekman said:
Correction* Your integral should look like this after the substitution:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+tan^{2}\theta)} d\theta$

I just checked the textbook and I typed the question wrong. There should be a square in the denominator but not inside the log. This might be what OP is asking?

psyc1011 · Sep 26, 2015

Ekman said:
Can you please show your working out, because I don't see how else you can get:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+\tan^{2}\theta)} d\theta$

to equal:

$\int_{0}^{\frac{\pi}{4}} \ln{(1+\tan\theta)} d\theta$

and get to the answer you suggested

His integral is 1+x instead of 1+x^2 for denominator

Ekman · Sep 26, 2015

psyc1011 said:
His integral is 1+x instead of 1+x^2 for denominator

You mean inside the log, not the denominator?

Impossible integral (1 Viewer)

Physicklad

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