Search found 56 matches
- Thu Feb 04, 2016 8:24 pm
- Forum: Calculus
- Topic: Integral with fractional part
- Replies: 5
- Views: 3792
Re: Integral with fractional part
That was a fun one. Kudoes fellas
- Thu Feb 04, 2016 8:41 am
- Forum: Calculus
- Topic: Integral with fractional part
- Replies: 5
- Views: 3792
Re: Integral with fractional part
Thanks for finishing, rd
- Thu Feb 04, 2016 12:29 am
- Forum: Calculus
- Topic: Integral with fractional part
- Replies: 5
- Views: 3792
Re: Integral with fractional part
Hey T. Heck, I think this one is more involved than that triple one I managed to get sometime back. Anyway, I think I have it. Though, I may have to come back to finish posting. $$\int_{0}^{1}x^{n}\left\{\frac{1}{x}\right\}^{n}dx$$ Let $x=1/u, \;\ dx=\frac{-1}{u^{2}}du$ $$\begin{align*}\int_{1}^{\in...
- Sun Jan 31, 2016 4:30 pm
- Forum: Calculus
- Topic: two wild and crazy integrals
- Replies: 2
- Views: 2512
Re: two wild and crazy integrals
Well, since no one bit, I reckon I will post a little something on the log integral. $$\int_{0}^{\frac{\pi}{2}}\log(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x))dx$$ Let $u=\tan(x), \;\ dx=\frac{1}{u^{2}+1}du$ $$\int_{0}^{\infty}\frac{\log\left(\frac{a^{2}u^{2}+b^{2}}{u^{2}+1}\right)}{u^{2}+1}du$$ $$=\int_{0}^...
- Sun Jan 31, 2016 3:50 pm
- Forum: Calculus
- Topic: Infinite product
- Replies: 3
- Views: 2969
Re: Infinite product
Hi T
No, I do not have a relation between the two. I just noticed they were the same answer and thought maybe it would be cool to show.
Yes, I saw RD's solution to the other similar one. Clever.
No, I do not have a relation between the two. I just noticed they were the same answer and thought maybe it would be cool to show.
Yes, I saw RD's solution to the other similar one. Clever.
- Sat Jan 30, 2016 7:04 am
- Forum: Calculus
- Topic: two wild and crazy integrals
- Replies: 2
- Views: 2512
Re: two wild and crazy integrals
Sorry for the typo on the first integral. I do not know why I had an 'n' there. But, maybe we can generalize. I reckon these two are not interesting or too easy?. But, for the case when n=1, a simple unit circle contour will suffice quite easily. Use the usual $$\cos(x)=\frac{z+z^{-1}}{2}, \;\ \sin(...
- Fri Jan 29, 2016 8:25 pm
- Forum: Calculus
- Topic: Infinite product
- Replies: 3
- Views: 2969
Re: Infinite product
I was kind of waiting for someone to post, but I will go ahead. Simply note that the geometric series $$1+x+x^{2}+x^{3}+x^{4}+....=\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$$ is the same as the factorization: $$(1+x)(1+x^{2})(1+x^{4})(1+x^{8})\cdot\cdot\cdot (1+x^{2^{n}})=\prod_{n=0}^{\infty}(1+x^{2^{n...
- Thu Jan 28, 2016 12:57 am
- Forum: Calculus
- Topic: two wild and crazy integrals
- Replies: 2
- Views: 2512
two wild and crazy integrals
Show that:
$$\int_{0}^{2\pi}\frac{(1-4\sin^{2}(x))\cos(2x)}{2-\cos(x)}dx=\frac{2\pi (91-52\sqrt{3})}{\sqrt{3}}$$
and
Evaluate:
$$\int_{0}^{\frac{\pi}{2}}\log\left(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x)\right)dx, \;\ 0<a<b$$
$$\int_{0}^{2\pi}\frac{(1-4\sin^{2}(x))\cos(2x)}{2-\cos(x)}dx=\frac{2\pi (91-52\sqrt{3})}{\sqrt{3}}$$
and
Evaluate:
$$\int_{0}^{\frac{\pi}{2}}\log\left(a^{2}\sin^{2}(x)+b^{2}\cos^{2}(x)\right)dx, \;\ 0<a<b$$
- Wed Jan 27, 2016 10:16 pm
- Forum: Calculus
- Topic: triple fractional part integral
- Replies: 1
- Views: 1968
Re: triple fractional part integral
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{y}{z}\right\}\left\{\frac{z}{x}\right\}\right)^{2}dxdydz$$ $$=\int_{0}^{1}\int_{0}^{1}\left(\left\{\frac{x}{y}\right\}\left\{\frac{1}{x}\right\}\right)^{2}dxdy$$ Break up integral at y: $$\int_{0}^{x}\int_{0}^{1}\left...
- Mon Jan 25, 2016 9:07 am
- Forum: Calculus
- Topic: triple fractional part integral
- Replies: 1
- Views: 1968
triple fractional part integral
check this one out: http://integralsandseries.prophpbb.com/post3814.html#p3814 I am pretty sure the curly brackets represent fractional part, thought the OP should have stated that. I have seen the double integral version, so I am pretty sure a general form can be derived for this. What I mean is so...