Math Help - Collecting integral problems

1. Collecting integral problems

Yes, post an interesting integral or a problem related to integration (like a sum, double sum, etc which can be computed applyin' integration), 'cause I want to collect problems and their solutions to make a PDF.

I'm not gonna start yet, may be in few hours, but if someone wants to start, no problem.

(I hope we can share lots of interesting integrals.)

2. $\int \frac {sin\;xcos\;x}{sin^4x+cos^4x}\;dx$

too easy? too boring?

3. can i get a copy of the PDF when your done?

4. Originally Posted by polymerase
$\int \frac {sin\;xcos\;x}{sin^4x+cos^4x}\;dx$

too easy? too boring?
Cool. . (Set $t=\sin^2 x$ ).

5. Originally Posted by polymerase
$\int \frac {sin\;xcos\;x}{sin^4x+cos^4x}\;dx$

too easy? too boring?
Sorry to say, but too easy and too boring. (Read above, it's about interesting integrals.)

Originally Posted by polymerase
can i get a copy of the PDF when your done?
Yes. I hope people post problems.

---

Okay, here's my problem: Evaluate $\int_0^1 {\int_0^1 {\frac{{x - 1}}
{{(1 + xy)\ln (xy)}}\,dx} \,dy} .$

Solution:

For the inner integral, substitute $u=xy.$ After reversing integration order we have

$\int_0^1 {\int_0^1 {\frac{{x - 1}}
{{(1 + xy)\ln (xy)}}\,dx} \,dy}= \underbrace {\int_0^1 {\frac{{1 - u}}
{{(1 + u)\ln u}}\,du} }_\ell + \ln 2.$

Now the remaining challenge is to compute $\ell.$ Expanding as a sum we get

$\ell = \sum\limits_{k = 0}^\infty {( - 1)^k \ln \frac{{k + 1}}
{{k + 2}}} = \ln \frac{{1 \cdot 3 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n - 1)(2n + 1)}}
{{2 \cdot 2 \cdot 4 \cdot 4 \cdot \cdots \cdot (2n)(2n)}}.$

Inside logarithm we have the Wallis Product and its limit is $\frac2\pi.$ The rest follows.

6. If $x>0$ show that: $\int_0^{\infty} \frac{\cos (xt) + t \sin (xt)}{1+t^2} dt = \pi e^{-x}$.

7. Hey K:

Here's one you're probably familiar with, but if not then it's a good one:

$\int_{-\infty}^{\infty}\frac{1}{\sqrt{(x^{2}+a^{2})(x^{2} +b^{2})}}dx=\frac{\pi}{agm(a,b)}$

I think this was one of Gauss's babies.

8. Okay guys, but I dunno if you read well above, you need to post problem + solution. (To add those problems into a PDF.)

Here's another problem: Evaluate $\sum\limits_{k = 0}^\infty {\frac{1}
{{(k + 2)k!}}} .$

Start with series expansion for $e^x$ and multiply by $x,$

$e^x = \sum\limits_{k = 0}^\infty {\frac{{x^k }}
{{k!}}} \implies xe^x = \sum\limits_{k = 0}^\infty {\frac{{x^{k + 1} }}
{{k!}}} .$

Integrate $\int_0^x {ue^u \,du} = \sum\limits_{k = 0}^\infty {\frac{{x^{k + 2} }}
{{(k + 2)k!}}} .$

The integral can be evaluated easily via integration by parts, this yields

$xe^x - e^x + 1 = \sum\limits_{k = 0}^\infty {\frac{{x^{k + 2} }}
{{(k + 2)k!}}} .$
By setting $x=1$ we have the desired answer.

9. Is this interesting by your standards?

$\lim_{x\to\infty}\left(\frac{n}{1+3n^2}+\frac{n}{4 +3n^2}+\frac{n}{9+3n^2}+...+\frac{n}{n^2+3n^2}\rig ht)=?$

Ill post the answer if it is.

10. Originally Posted by polymerase
$\lim_{n\to\infty}\left(\frac{n}{1+3n^2}+\frac{n}{4 +3n^2}+\frac{n}{9+3n^2}+...+\frac{n}{n^2+3n^2}\rig ht)=?$
$\sum_{k=1}^n \frac{n}{k^2+3n^2} = \sum_{k=1}^n \frac{1}{\left( \frac{k}{n} \right)^2 + 3} \cdot \frac{1}{n}$

This is the Riemann sum for $\int_0^1 \frac{dx}{x^2+3}$ with equal sub-intervals of length $1/n$ using right-endpoints.

Thus, (by fundamental theorem) this limit converges to $\sqrt{3} \tan^{-1} \frac{x}{\sqrt{3}} \bigg|_0^1 = \sqrt{3} \tan^{-1} \frac{1}{\sqrt{3}} = \frac{\pi \sqrt{3}}{6}$

11. Yeah, we had an obviously Riemann sum there.

Riemann sums are routine problems, unless you to post some interesting limit.

12. Originally Posted by ThePerfectHacker
$\sum_{k=1}^n \frac{n}{k^2+3n^2} = \sum_{k=1}^n \frac{1}{\left( \frac{k}{n} \right)^2 + 3} \cdot \frac{1}{n}$

This is the Riemann sum for $\int_0^1 \frac{dx}{x^2+3}$ with equal sub-intervals of length $1/n$ using right-endpoints.

Thus, (by fundamental theorem) this limit converges to $\sqrt{3} \tan^{-1} \frac{x}{\sqrt{3}} \bigg|_0^1 = \sqrt{3} \tan^{-1} \frac{1}{\sqrt{3}} = \frac{\pi \sqrt{3}}{6}$
One tiny problem, it converges to $\frac{1}{\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\r ight)$ Answer should be $\frac{\pi}{6\sqrt{3}}$

13. Originally Posted by Krizalid
Yes, post an interesting integral or a problem related to integration (like a sum, double sum, etc which can be computed applyin' integration), 'cause I want to collect problems and their solutions to make a PDF.

I'm not gonna start yet, may be in few hours, but if someone wants to start, no problem.

(I hope we can share lots of interesting integrals.)
Well, this one is probably more beautiful than interesting. You've no doubt met it. If it nevertheless meets your criteria for interesting, I'll take the time to post a solution.

$\int_0^\infty \frac{u^{x-1}}{e^{u} - 1} \, du = \zeta(x) \Gamma(x),$

where $\zeta (x)$ and $\Gamma (x)$ are, respectively, the famous zeta and Gamma functions.

14. Originally Posted by Krizalid
Yes, post an interesting integral or a problem related to integration (like a sum, double sum, etc which can be computed applyin' integration), 'cause I want to collect problems and their solutions to make a PDF.

I'm not gonna start yet, may be in few hours, but if someone wants to start, no problem.

(I hope we can share lots of interesting integrals.)
Ditto to my previous post:

$\int_0^1 \frac{1}{x^x} \, dx = \frac{1}{1^1} + \frac{1}{2^2} + \frac{1}{3^3} + ....$

15. Originally Posted by mr fantastic
Well, this one is probably more beautiful than interesting. You've no doubt met it. If it nevertheless meets your criteria for interesting, I'll take the time to post a solution.

$\int_0^\infty \frac{u^{x-1}}{e^{u} - 1} \, du = \zeta(x) \Gamma(x),$

where $\zeta (x)$ and $\Gamma (x)$ are, respectively, the famous zeta and Gamma functions.
That is nice . But that is not so hard to prove. It should be worthy to be such a problem.

This is Mine 86th Post!!!

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