1. ## Doing Fun Integrals

I'd like to initiate an $\displaystyle \int\text{Marathon}$

The rules are simple
• The person who solves an integral, proposes the another one, but before proposing it, he must have the approval (if the answer is correct or not) of the person who proposed the previous integral.
• Only propose integrals “of one variable”; this is because all do not know multiple integration (I include myself). The level of the proposed ones must be middle-upper.
• Indefinite, Definite and Improper integrals are allowed.
• Use LaTeX
I start, here's the first one:

$\displaystyle \int\frac1{\sin(x-1)\sin(x-2)}\,dx$

Greetings

2. $\displaystyle \int \frac{1}{\sin(x-1)\sin(x-2)} dx$

We can write,

$\displaystyle \sin(x-1)\sin(x-2) = \frac{1}{2}[ \cos 1 - \cos (2x-3) ]$

Thus, if you use $\displaystyle t=2x-3$:

$\displaystyle \frac{1}{4}\int \frac{1}{\cos 1 - \cos t} dt$

Now, there is a way to get,
$\displaystyle \int \frac{1}{\cos 1 - \cos t} dt$
It is just really ugly.
~~~

We can use the formula that if $\displaystyle A\not = 0 \mbox{ and }|A|<1$ then:
$\displaystyle \int \frac{1}{A - \cos t} dt = - \frac{2}{\sqrt{1-A^2}}\tan^{-1} \left(\frac{(A+1)\tan \frac{t}{2}}{\sqrt{1-A^2}} \right)+C$

This formula can of course be painfully derived by the substitution $\displaystyle u=\tan \frac{t}{2}$.

3. Ufff, sorry Hacker for not asked you the permission.

I mistook of integral, sorry.

Here's the "original"

Compute $\displaystyle \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx$

(this is not hard)

4. Originally Posted by Krizalid
$\displaystyle \int_0^1\frac{\ln(x+1)}{x^2+1}\,dx$
(I do substitutions in a strange way).

Write,
$\displaystyle \int_0^1 \frac{(x+1)}{x^2+1} \cdot \frac{1}{x+1} \cdot \ln (x+1) dx$

Let, $\displaystyle t=\ln (x+1)$ then $\displaystyle t' = \frac{1}{x+1}$

That means, $\displaystyle x+1 = e^t \mbox{ and }x^2+1 = (e^t-1)^2+1$

Thus, (by substitution theorem)
$\displaystyle \int_0^{\ln 2} \frac{e^t}{(e^t-1)^2+1} dt$
Let $\displaystyle u=e^t-1 \mbox{ thus }u'=e^t$.
Use substitution theorem again,
$\displaystyle \int_0^1 \frac{1}{u^2+1} du = \tan^{-1} u\big|_0^1 = \tan^{-1} (1) - \tan^{-1} (0) = \boxed{\frac{\pi}{4}}$.

5. Sorry, but I'm afraid the answer is not correct.

Try again.

6. Originally Posted by Krizalid
Sorry, but I'm afraid the answer is not correct.
Indeed! I forgot the a factor of $\displaystyle t$ in my substitution.

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Here is an indirect way of doing this integral.
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$\displaystyle \zeta = \int_0^1 \frac{\ln (x+1)}{x^2+1} dx$

Use the same substitution $\displaystyle t=\ln(x+1)$ as above to get correctly this time,

$\displaystyle \int_0^1 \frac{e^t\cdot t}{(e^t-1)^2+1} dt$

Let $\displaystyle u=t \mbox{ and }v'=\frac{e^t}{(e^t-1)^2+1} \mbox{ then }u'=1 \mbox{ and }v = \tan^{-1} (e^t-1)$

Integration by parts,

$\displaystyle t\cdot \tan^{-1} (e^t-1) \big|_0^{\ln 2} - \int_0^{\ln 2} \tan^{-1}(e^t-1) dt = \frac{\pi}{4}\cdot \ln 2 - \int_0^{\ln 2} \tan^{-1}(e^t-1) dt$

Now consider the integral,
$\displaystyle \int_0^{\ln 2} \tan^{-1}(e^t-1) dt$

Use substitution $\displaystyle s=e^t-1$ to get,

$\displaystyle \int_0^{1} \frac{\tan^{-1} s}{s+1} ds$

Let $\displaystyle a=\tan^{-1} s \mbox{ and }b=\frac{1}{s+1}$. This integration by parts gives that:

$\displaystyle \int_0^1 \frac{\tan^{-1} s}{s+1} dx = \int_0^1 \frac{\ln (x+1)}{x^2+1)} dx = \zeta$

Thus,

$\displaystyle \zeta = \frac{\pi}{4}\cdot \ln 2 - \zeta$

$\displaystyle 2\zeta = \frac{\pi}{4}\cdot \ln 2$

$\displaystyle \boxed{\zeta = \frac{\pi}{8}\ln 2}$

7. Now is correct.

8. Originally Posted by Krizalid
Now is correct.

I saw this integral more than a year ago. I loved it. So I written it down in my notebook.

Let $\displaystyle a>0$:

$\displaystyle \int_0^{\infty} \ln (1+e^{-ax}) dx$

9. Delete this thread, it didn't have much relevance that we say...

Since in any case, interesting integrals are exposed that we want to share without the necessity to do a marathon.

10. Originally Posted by ThePerfectHacker
I saw this integral more than a year ago. I loved it. So I written it down in my notebook.

Let $\displaystyle a>0$:

$\displaystyle \int_0^{\infty} \ln (1+e^{-ax}) dx$
I know I've seen this thing somewhere. I just can't find it!

Yes, it's a useless post to others, but at least I get to express my frustration.

-Dan

11. i think it was a great idea, and a good opportunity to see some interesting integrals. of course it also defeats the purpose if we get stuck on one integral too long. maybe we can modify the rules a bit to keep things moving. like put a time limit on when the integral should be solved--if not solved within that limit, the person who suggested the limit in the first place should give the solution and choose another integral. and of course, there should be some reward for coming up with an integral that no one can solve (within reason)

12. Solution:

$\displaystyle \int_0^{\infty} \ln (1+e^{-ax} ) dx$ for $\displaystyle a>0$.

We use the formula: $\displaystyle \ln (1+x) = x - \frac{x^2}{2}+\frac{x^3}{3} - .... \mbox{ for }-1<x\leq 1$.

Since $\displaystyle 0<e^{-ax}\leq 1 \mbox{ on }[0,\infty)$:

$\displaystyle \int_0^{\infty} e^{-ax} - \frac{e^{-2ax}}{2} + \frac{e^{-3ax}}{3} - \frac{e^{-4ax}}{4}+ ... dx$

Integrate term-by-term*

$\displaystyle -\frac{e^{-ax}}{a} + \frac{e^{-2ax}}{2^2a} - \frac{e^{-3ax}}{3^2a} + ... \big|_0^{\infty}$

$\displaystyle \frac{1}{a} - \frac{1}{2^2a}+\frac{1}{3^2a}-\frac{1}{4a} + ... = \frac{1}{a} \left( 1 - \frac{1}{2^2}+\frac{1}{3^2} - \frac{1}{4^2}+... \right) = \frac{\pi^2}{12a}$

*)This is legal because of something called uniform convergence.

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Note, this integral was taken from here.

13. Originally Posted by ThePerfectHacker
Solution:

$\displaystyle \int_0^{\infty} \ln (1+e^{-ax} ) dx$ for $\displaystyle a>0$.

We use the formula: $\displaystyle \ln (1+x) = x - \frac{x^2}{2}+\frac{x^3}{3} - .... \mbox{ for }-1<x\leq 1$.

Since $\displaystyle 0<e^{-ax}\leq 1 \mbox{ on }[0,\infty)$:

$\displaystyle \int_0^{\infty} e^{-ax} - \frac{e^{-2ax}}{2} + \frac{e^{-3ax}}{3} - \frac{e^{-4ax}}{4}+ ... dx$

Integrate term-by-term*

$\displaystyle -\frac{e^{-ax}}{a} + \frac{e^{-2ax}}{2^2a} - \frac{e^{-3ax}}{3^2a} + ... \big|_0^{\infty}$

$\displaystyle \frac{1}{a} - \frac{1}{2^2a}+\frac{1}{3^2a}-\frac{1}{4a} + ... = \frac{1}{a} \left( 1 - \frac{1}{2^2}+\frac{1}{3^2} - \frac{1}{4^2}+... \right) = \frac{\pi^2}{12a}$

*)This is legal because of something called uniform convergence.

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Note, this integral was taken from here.
that's nice. i never thought of using power series, i was trying all these weird substitutions and integrations by parts and getting nowhere. what hints to you that such an appraoch is necessary--or even applicable in the first place?

14. Originally Posted by Jhevon
--or even applicable in the first place?
Experience. The two most useful techiques in integration are:
1)Power Series
2)Contour Integration

#2 is usually the best. But I do not understand it yet. One of the things I am learning over the summer is in order to under approach #2. If you look in the link I gave, TD!, computes the first integral using that approach.

15. i think it was a great idea, and a good opportunity to see some interesting integrals. of course it also defeats the purpose if we get stuck on one integral too long. maybe we can modify the rules a bit to keep things moving. like put a time limit on when the integral should be solved--if not solved within that limit, the person who suggested the limit in the first place should give the solution and choose another integral. and of course, there should be some reward for coming up with an integral that no one can solve (within reason)
I think it is a good idea too. But we should try to follow the rules and if no response let the original poster answer the question. I think it will give a chance to let people see some hard/interesting integrals.

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$\displaystyle \int_0^{\pi/2} \frac{dx}{1+(\tan x)^{\sqrt{2}}}$