The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

The Proof is Trivial!

Scroll to see replies

The solution I have only uses A level knowledge. But it requires a rather insane observation. I would discount it as a possibility, to be honest!
Original post by FireGarden
The solution I have only uses A level knowledge. But it requires a rather insane observation. I would discount it as a possibility, to be honest!


I always love a challenge :tongue:
Original post by FireGarden
Now I'm going to be a dick head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

Problem 266***

Evaluate ∫0π211+tan⁡(x)2 dx \displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx


∫0π211+tan⁡(x)2 dx=∫0π2tan⁡(x)21+tan⁡(x)2 dx    (x→π2−x) \displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx =\int_{0}^{\frac{\pi}{2}} \dfrac{ \tan(x)^{\sqrt{2}}}{1+ \tan(x)^{\sqrt{2}}} \ dx\;\;(x\to \tfrac{\pi}{2}-x)

Add both, result is π4\frac{\pi}{4}
...That's just ridiculous. And outstandingly beautiful. I shall be back.
Original post by Lord of the Flies
∫0π211+tan⁡(x)2 dx=∫0π2tan⁡(x)21+tan⁡(x)2 dx    (x→π2−x) \displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx =\int_{0}^{\frac{\pi}{2}} \dfrac{ \tan(x)^{\sqrt{2}}}{1+ \tan(x)^{\sqrt{2}}} \ dx\;\;(x\to \tfrac{\pi}{2}-x)

Add both, result is π4\frac{\pi}{4}


Is there anything you can't do?
Original post by MathsNerd1
Is there anything you can't do?


It's annoying isn't it :smile:
Original post by bananarama2
It's annoying isn't it :smile:


I thought after learning a few more tricks I'd be able to answer these questions, sadly I'm not of this calibre :redface:
Original post by FireGarden
Now I'm going to be a dick head and post something I think not even LotF will evaluate without a lot of head scratching. The solution I have is only using knowledge from A level.. but I doubt it'll be used. It has a nice answer, too.

Problem 266***

Evaluate ∫0π211+tan⁡(x)2 dx \displaystyle\int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan(x)^{\sqrt{2}}} \ dx


Is it pi/4? I want to check it's right before posting my working.

EDIT: Dam, beaten to it. I used the same method as LOTF. Is there a way to do it with differentiation under the integral?
(edited 10 years ago)
Original post by MathsNerd1
I thought after learning a few more tricks I'd be able to answer these questions, sadly I'm not of this calibre :redface:


Makes me wish I'd seen the light at a younger age.

To be fair he posted this trick earlier :tongue:
Original post by bananarama2
Makes me wish I'd seen the light at a younger age.

To be fair he posted this trick earlier :tongue:


Yeah, I only focused at the start of this year really, far too late really and yeah I've seen it a few times before but just forgot about it when first looking at the integral, oh well :redface: I'd love a good Laplace transform question as I've recently covered them :tongue:
Next try. Taking an integral of not too much difficulty, and making it ugly. Though the "ugliness" can be thrown out the window, given the use of the right theorem, which i believe is not widely known. Let's see :colone:

Problem 267**

Evaluate ∫0πxsin⁡(x)1+cos⁡2(x) dx \displaystyle\int_{0}^{\pi} \dfrac{x\sin(x)}{1+\cos^2(x)} \ dx
Solution 267

∫0πxsin⁡x1+cos⁡2x dx=∫0π(π−x)sin⁡x1+cos⁡2xdx(x→π−x)\displaystyle\int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)

Hence I=π2∫0πsin⁡x1+cos⁡2x dx=π2∫−11dt1+t2=π24I=\displaystyle\frac{\pi}{2}\int_0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx=\frac{\pi}{2}\int_{-1}^{1}\frac{dt}{1+t^2}=\frac{\pi^2}{4}
Original post by james22
Is there a way to do it with differentiation under the integral?


Sure.

f(s)=∫0π2dx1+tan⁡sx⇒f′(s)=∫0π2−ln⁡tan⁡xtan⁡sx(1+tan⁡sx)2=0f(s)=\displaystyle\int_0^{\frac{\pi}{2}} \frac{dx}{1+\tan^s x}\Rightarrow f'(s)=\int_0^{\frac{\pi}{2}} -\frac{\ln \tan x \tan^s x}{(1+\tan^s x)^2}=0

((let x↦x+π4x\mapsto x+\frac{\pi}{4}, use tan⁡(x+π4)=cot⁡(π4−x)\tan (x+\frac{\pi}{4})=\cot (\frac{\pi}{4}-x) to show that the integrand is odd))

Hence ff is constant, set s=0s=0 and we have f(s)=∫0π2dx2=π4\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \frac{dx}{2}=\frac{\pi}{4}
Exactly how much have you practiced integration? It's just silly how quickly you do basically all of them!
Original post by Lord of the Flies
Sure.

f(s)=∫0π2dx1+tan⁡sx⇒f′(s)=∫0π2−ln⁡tan⁡xtan⁡sx(1+tan⁡sx)2=0f(s)=\displaystyle\int_0^{\frac{\pi}{2}} \frac{dx}{1+\tan^s x}\Rightarrow f'(s)=\int_0^{\frac{\pi}{2}} -\frac{\ln \tan x \tan^s x}{(1+\tan^s x)^2}=0

((let x↦x+π4x\mapsto x+\frac{\pi}{4}, use tan⁡(x+π4)=cot⁡(π4−x)\tan (x+\frac{\pi}{4})=\cot (\frac{\pi}{4}-x) to show that the integrand is odd))

Hence ff is constant, set s=0s=0 and we have f(s)=∫0π2dx2=π4\displaystyle f(s)=\int_0^{\frac{\pi}{2}} \frac{dx}{2}=\frac{\pi}{4}


There's a little puzzle that's been niggling at me - thought you could shed some light on it :P
Are there any non-integer zeros of the Riemann zeta function with real part not equal to 1/2?
Reply 1915
Avatar for Jkn
Jkn
11
Problem 268***

Evaluate ∫0∞sin⁡(2tan⁡−1(t))(1+t2)(eπt+1) dt\displaystyle \int_0^{\infty} \frac{\sin( 2 \tan^{-1} (t) )}{(1+t^2)(e^{\pi t}+1)} \ dt

Edit: Btw, had another look at the dreaded 12(tanxcos(tanx))\frac{1}{2} (tanx cos (tanx)) integral and have made a few breakthroughs. I feel as though I might be nearing a solution but I'm not quite there yet! Has anyone else done anything substantial? :smile:
(edited 10 years ago)
I am currently learning this stuff, so forgive me if I have made mistakes or chosen an odd contour.

Solution 268

Consider f(z)=2z(1+z2)2(eπz+1)\displaystyle f(z)=\frac{2z}{(1+z^2)^2(e^{\pi z}+1)}, which has a pole of order 3 at ii and simple poles at (2k+1)i(2k+1)i for k≥1k\geq 1.

Let ζ\zeta be the positively oriented square formed by the lines λ,ω,h,h′\lambda,\omega,h,h' indicated below.



We find that:

Res(f,i)=i(18π+π24)\displaystyle\text{Res}(f,i)=i \left(\frac{1}{8\pi}+\frac{\pi}{24}\right) and Res(1eπz+1,(2k+1)i)=−1π\displaystyle\text{Res}\left( \frac{1}{e^{\pi z}+1},(2k+1)i\right)=-\frac{1}{\pi} so Res(f,(2k+1)i)=−(4k+2)iπ(4k2+4k)2\displaystyle\text{Res}(f,(2k+1)i)=-\frac{(4k+2)i}{\pi(4k^2+4k)^2}

Therefore, as n→∞:n\to\infty:

Unparseable latex formula:

\begin{aligned}\displaystyle\int_{\zeta} f(z)\,dz=2\pi i\left[i \left(\frac{1}{8\pi}+\frac{\pi}{24}\right)-\sum_{k\geq 1}\frac{(4k+2)i}{\pi(4k^2+4k)^2} \right]=2\pi i\left[i \left(\frac{1}{8\pi}+\frac{\pi}{24}\right)-\frac{i}{8\pi}\left]=-\frac{\pi^2}{12}



Furthermore,

Unparseable latex formula:

\begin{aligned}\displaystyle \left|\int_{h}f(z)\,dz \right|\leq \int_{0}^{1} 2n\left| f\big(n(1+2ti)\big) \right|\,dt\leq \sqrt{5}e^{-n\pi}\to 0

and similarly
Unparseable latex formula:

\displaystyle \int_{h'} f(z)\dz\to 0

and
Unparseable latex formula:

\displaystyle \int_{w} f(z)\dz\to 0



Thus,

∫λf(z) dz→∫−∞∞2x dx(1+x2)2(eπx+1)=−π212\displaystyle \int_{\lambda} f(z)\,dz\to \int_{-\infty}^{\infty} \frac{2x\,dx}{(1+x^2)^2(e^{\pi x}+1)}=-\frac{\pi^2}{12}. Last, observe that if p=∫0∞f(x) dxp=\displaystyle\int_0^{\infty} f(x)\,dx and q=∫−∞0f(x) dxq=\displaystyle\int_{-\infty}^0 f(x)\,dx then:

p−q=∫0∞2x dx(1+x2)2=1p-q=\displaystyle \int_0^{\infty} \frac{2x\,dx}{(1+x^2)^2}=1, which together with p+q=−π212p+q=-\dfrac{\pi^2}{12} yields:

∫0∞2x dx(1+x2)2(eπx+1)=12−π224\displaystyle\int_{0}^{\infty} \frac{2x\,dx}{(1+x^2)^2(e^{\pi x}+1)}=\frac{1}{2}-\frac{\pi^2}{24}
Original post by Smaug123
There's a little puzzle that's been niggling at me - thought you could shed some light on it :P
Are there any non-integer zeros of the Riemann zeta function with real part not equal to 1/2?


:lol:

Original post by FireGarden
Exactly how much have you practiced integration? It's just silly how quickly you do basically all of them!


Quite a bit, and thank you.
Reply 1918
Avatar for Jkn
Jkn
11
Original post by Lord of the Flies
x

Mind-blowing stuff as usual! How do you make your diagrams?!

Problem 269***

Using the previous problem as a hint,

Express ζ\zeta in the form ζ(s)=f(s)+∫0∞g(s,t) dt\displaystyle \zeta(s)=f(s)+\int_0^{\infty} g(s,t) \ dt where f and g are combinations of 'elementary' functions and s∈Cs \in \mathbb{C} \ {1} \{1 \}.
(edited 10 years ago)
Original post by Jkn
Problem 268***

Evaluate ∫0∞sin⁡(2tan⁡−1(t))(1+t2)(eπt+1) dt\displaystyle \int_0^{\infty} \frac{\sin( 2 \tan^{-1} (t) )}{(1+t^2)(e^{\pi t}+1)} \ dt


Is this possible without contour integration? Contour integration is horrible :frown:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you