The Student Room Group

The Proof is Trivial!

Scroll to see replies

Reply 2480
Original post by Felix Felicis
Well, arctantanx=x\arctan{\tan{x}} = x as you're integrating over [0,π2]\left[0,\frac{\pi}{2}\right] but you've done the substitution incorrectly, you've forgotten to make an appropriate substitution for dxdx. :wink: Your integral is actually transformed to:

0π2lnsecxtan2x dx=u=tanx120ln(u2+1)u2(u2+1) du\displaystyle \int_0^{\frac{\pi}{2}} \frac{\ln{\sec{x}}}{\tan^2 x} \ dx \overset{u=\tan{x}}= \frac{1}{2}\int_0^{\infty} \frac{\ln{(u^2 + 1)}}{u^2 (u^2 + 1)} \ du.


It may be of use to note

log(1+xn)1+xn=k1Hk(x)nk\displaystyle \frac{\log(1+x^n)}{1+x^n}=-\sum_{k \ge 1}H_k (-x)^{nk}

which comes in handy in many places, for example page 7 here http://129.81.170.14/~vhm/papers_html/rmt-final.pdf . You may be able to use Ramanujan's Master theorem in those questions, although I haven't tried it and you may end up taking Γ(n),nN0\Gamma(-n), n \in \mathbb{N}_0.
Original post by henpen
It may be of use to note

log(1+xn)1+xn=k1Hk(x)nk\displaystyle \frac{\log(1+x^n)}{1+x^n}=-\sum_{k \ge 1}H_k (-x)^{nk}

which comes in handy in many places, for example page 7 here http://129.81.170.14/~vhm/papers_html/rmt-final.pdf . You may be able to use Ramanujan's Master theorem in those questions, although I haven't tried it and you may end up taking Γ(n),nN0\Gamma(-n), n \in \mathbb{N}_0.

Well now you've lost me but I look forward to seeing solution (google will have to be my friend for the strange stuff though)
Reply 2482
Problem 400 **

Prove
ex+e1x1+cosh(x)=n0xnn!{n!e}, \displaystyle \frac{e^{-x}+e^{-1}}{x-1}+\cosh(x)=\sum_{n \ge 0}\frac{x^ n}{n!}\left \{ \frac{n!}{e} \right \},
where the squiggly brackets pertain to the fractional part function.
(edited 10 years ago)
Reply 2483
Original post by Mladenov
Sorry for that regrettably late reply; an elementary example of function with this property is the real logarithm.
Neutrices were introduced, as far as I know, (I am not quite knowledgeable in the area) to help statisticians/theoretical physicists deal with asymptotic expressions, i.e. they generalize OO and oo notions. A neutrix is an abelian (additively written) group NN, consisting of functions SGS \to G, where GG is an abelian group, such that the only constant function in NN is the zero function. The functions in NN are called negligible.
As an example consider S=ZS = \mathbb{Z}, G=RG = \mathbb{R}, and the negligible functions in NN being finite (linear) sums of nxlnk1nn^{x} \ln^{k-1}n and lnkn\ln^{k}n, where x>0x>0, kk - positive integer, and all functions which tend to zero as nn \to \infty.
This article is relevant.

Spoiler






The result in the hint is attainable using log(1u)1u=k1Hkuk\frac{\log(1-u)}{1-u}=- \sum_{k \ge 1}H_k u^k. I've had no luck simplifying the sum, though.
Original post by henpen
The result in the hint is attainable using log(1u)1u=k1Hkuk\frac{\log(1-u)}{1-u}=- \sum_{k \ge 1}H_k u^k. I've had no luck simplifying the sum, though.

Yep, I just wanted to give an example.

To evaluate this sum, we can proceed as follows:

Note Hn=011xn1xdx\displaystyle H_{n} = \int_{0}^{1} \frac{1-x^{n}}{1-x}dx. Absolute convergence implies that we can interchange the order of integration and summation, and doing so we get

Unparseable latex formula:

\displaystyle \sum_{1}^{\infty} H_{n}\frac{1}{(n+1)^{3}} &= \int_{0}^{1} \frac{1}{1-x} \sum_{1}^{\infty} \frac{1-x^{n}}{(n+1)^{3}}dx

.

Now, integration by parts will do the job. For more general identities, which I am not keen to work out, look at this blog post.

By the way, there is a better way to evaluate the integral; consider the function 0xln(1t)tdt\displaystyle \int_{0}^{x} \frac{\ln (1-t)}{t}dt.
posted in the wrong thread sorry
(edited 10 years ago)
Original post by Mladenov
Problem 83**

Let xx, aa, and bb be positive integers such that xa+b=abbx^{a+b}=a^{b}b. Then a=xa=x and b=xxb=x^{x}.


Not sure what the problem is here:

x(a+b)=xaxbx^{(a + b)} = x^ax^b

xaxb=abbx^ax^b = a^bb

Easy to see it holds that a=xa = x and b=xxb = x^x

Is there anything else to do?
Original post by bananarama2

Hint for my problem



It lands on the tower, how far down depends on how far out you hold the ball.
Original post by JamesyB
Not sure what the problem is here:

x(a+b)=xaxbx^{(a + b)} = x^ax^b

xaxb=abbx^ax^b = a^bb

Easy to see it holds that a=xa = x and b=xxb = x^x

Is there anything else to do?


You have to prove that there are no other possibilities, not only to check trivially that this is a solution.
Reply 2489
Found it difficult to put my problem into words, if you want me to clarify something because I've explained it poorly please don't hesitate to quote/inbox me :smile: Really wanted to try posting something a bit different

Problem 401**/***

iii^i, where ii is the complex unit, is a multi-valued function that quite surprisingly always takes on real values. Each value of the function can be expressed in terms of k-k, where kk can be any integer.

Find the infinite sum of the values of iii^i for which kk is a positive integer.
(edited 10 years ago)
Original post by Flauta
Found it difficult to put my problem into words, if you want me to clarify something because I've explained it poorly please don't hesitate to quote/inbox me :smile: Really wanted to try posting something a bit different

Problem 401**/***

iii^i, where ii is the complex unit, is a multi-valued function that quite surprisingly always takes on real values. Each value of the function can be expressed in terms of k-k, where kk can be any integer.

Find the infinite sum of the values of iii^i for which kk is a positive integer.


It's quite late, so I might well have messed something up…

Spoiler

(edited 10 years ago)
Here are some nice and easy questions for A-level students who may find themselves lost on this thread due to us mostly considering higher level problems :smile:

Problem 402*

Find, for ψ,η0\psi, \eta \neq 0

I=eηxsin(ψx)dxI = \displaystyle \int e^{\eta x} \sin(\psi x) dx

and give a condition on xx such that I>0I>0

Problem 403*

Find ψ\psi and η\eta such that

ln(ψ+η)=ln(ψ)+ln(η)\ln(\psi + \eta) = \ln(\psi) + \ln(\eta)
(edited 10 years ago)
Can't help but feel that this is a bit verbose and there should be a more direct way to evaluate this but here goes...

Solution 394

01lnxln2(1x)x dx=x1x01ln2xln(1x)1x dx=01ln(1x)ln2xn=0xn dx=n=001ln(1x)ln2xxn dx=n=001ln2xxnm=1xmm dx=n=0m=101xmmln2xxn dx=I.B.P.2n=1m=11m(m+n)3\displaystyle \begin{aligned} \int_0^1 \frac{\ln x \ln^2 (1-x)}{x} \ dx & \overset{x \mapsto 1-x}= \int_0^1 \frac{\ln^2 x \ln (1-x)}{1-x} \ dx \\ & = \int_0^1 \ln (1-x) \ln^2 x \sum_{n=0}^{\infty} x^n \ dx \\ & = \sum_{n=0}^{\infty} \int_0^1 \ln (1-x) \ln^2 x \cdot x^n \ dx \\ & = -\sum_{n=0}^{\infty} \int_0^1 \ln^2 x \cdot x^n \cdot \sum_{m=1}^{\infty} \frac{x^m}{m} \ dx \\ & = - \sum_{n=0}^{\infty} \sum_{m=1}^{\infty} \int_0^1 \frac{x^m}{m} \cdot \ln^2 x \cdot x^n \ dx \\ & \overset{\text{I.B.P.}}= - 2 \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(m+n)^3} \end{aligned}


Interchanging the summation bases and adding yields:

2n=1m=11m(m+n)3=n=1m=11mn(m+n)2=2k=2Hk1k3=2(k=1Hkk3ζ(4))\displaystyle\begin{aligned} -2\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m(m+n)^3} & = - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{mn(m+n)^2} \\ & = -2 \sum_{k=2}^{\infty} \frac{H_{k-1}}{k^3} \\ & = -2 \left( \sum_{k=1}^{\infty} \frac{H_k}{k^3} - \zeta (4)\right) \end{aligned}


So it remains to evaluate k=1Hkk3\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{k^3}.

Using the relation Hk=011xk1x dx:\displaystyle H_k = \int_0^1 \frac{1-x^k}{1-x} \ dx:

k=1Hkk3=0111xk=11xkk3 dx=01ζ(3)k=1xkk31x dx=I.B.P.ln(1x)(ζ(3)k=1xkk3)01=001ln(1x)k=1xk1k2 dx=01k=1xkkk=1xk1k2 dx=01k=1xk1kk=1xkk2 dx=I.B.P.12(k=1xkk2)201=12ζ2(2)\displaystyle \begin{aligned} \sum_{k=1}^{\infty} \frac{H_k}{k^3} & = \int_0^1 \frac{1}{1-x} \sum_{k=1}^{\infty} \frac{1-x^k}{k^3} \ dx \\ & = \int_0^1 \frac{\zeta (3) - \sum_{k=1}^{\infty} \frac{x^k}{k^3}}{1-x} \ dx \\ & \overset{\text{I.B.P.}}= \underbrace{-\ln (1-x) \cdot \left(\zeta(3) - \sum_{k=1}^{\infty} \frac{x^k}{k^3}\right)\bigg|_0^1}_{=0} - \int_0^1 \ln (1-x) \cdot \sum_{k=1}^{\infty} \frac{x^{k-1}}{k^2} \ dx \\ & = \int_0^1 \sum_{k=1}^{\infty} \frac{x^k}{k} \cdot \sum_{k=1}^{\infty} \frac{x^{k-1}}{k^2} \ dx \\ & = \int_0^1 \sum_{k=1}^{\infty} \frac{x^{k-1}}{k} \cdot \sum_{k=1}^{\infty} \frac{x^k}{k^2} \ dx \\ & \overset{\text{I.B.P.}}= \frac{1}{2} \left( \sum_{k=1}^{\infty} \frac{x^k}{k^2} \right)^2 \bigg|_0^1 \\ & = \tfrac{1}{2} \zeta^2 (2) \end{aligned}


Putting everything together, we have:

01lnxln2(1x)x dx=2ζ(4)ζ2(2)=12ζ(4)\displaystyle \int_0^1 \frac{\ln x \ln^2 (1-x)}{x} \ dx = 2 \zeta(4) - \zeta^2 (2) = \boxed{-\tfrac{1}{2} \zeta (4)}
(edited 10 years ago)
Original post by Felix Felicis

Spoiler



Well done.

Problem 404***

Evaluate 0xlnx1+x2dx\displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \ln x}{1+x^{2}}dx.
Original post by Indeterminate
Problem 402*

Find, for ψ,η0\psi, \eta \neq 0

I=eηxsin(ψx)dxI = \displaystyle \int e^{\eta x} \sin(\psi x) dx

and give a condition solely dependent on xx such that I>0I>0


Solution-ish 402* :ninja:

Spoiler

(edited 10 years ago)
Reply 2495
Original post by Mladenov
Well done.

Problem 404***

Evaluate 0xlnx1+x2dx\displaystyle \int_{0}^{\infty} \frac{\sqrt{x} \ln x}{1+x^{2}}dx.


Substitute x=u\sqrt{x}=u, and using that the integrand is now even,

I=ln(u2)1+u4du\displaystyle I= \int_{-\infty}^{\infty} \frac{ \ln (u^2)}{1+u^{4}}du.

The integrand has simple poles at the fourth roots of unity, [ζ4k]1k4[\zeta^{k}_4]_{1\le k \le 4}. Make a semicircular contour, integrating anticlockwise, and noting that the curvy bit tends to 00 as the radius of the semicircle \rightarrow \infty. Thus

ln(u2)1+u4du=2πi(Res[ln(z2)1+z4,z=ζ41]+Res[ln(z2)1+z4,z=ζ42])\displaystyle \int_{-\infty}^{\infty} \frac{ \ln (u^2)}{1+u^{4}}du=2\pi i (Res[\frac{\ln(z^2)}{1+z^4}, z=\zeta^1_4]+Res[\frac{\ln(z^2)}{1+z^4}, z=\zeta^2_4])

=2πi(2ln(ζ41)(ζ41ζ42)(ζ41ζ43)(ζ41ζ44)+2ln(ζ42)(ζ42ζ41)(ζ42ζ43)(ζ42ζ44))\displaystyle =2\pi i \left( \frac{2\ln(\zeta^1_4)}{(\zeta^1_4-\zeta^2_4)(\zeta^1_4-\zeta^3_4)(\zeta^1_4-\zeta^4_4)}+ \frac{2\ln(\zeta^2_4)}{(\zeta^2_4-\zeta^1_4)(\zeta^2_4-\zeta^3_4)(\zeta^2_4-\zeta^4_4)} \right)

=2πi(2iπ4(2)(2(1+i))(2i)+2i3π4(2)(2i)(2(1+i)))\displaystyle =2\pi i \left( \frac{2\frac{i\pi}{4}}{(\sqrt{2})(\sqrt{2}(1+i))(\sqrt{2}i)}+ \frac{2\frac{i 3\pi}{4}}{(-\sqrt{2})(\sqrt{2}i)(-\sqrt{2}(-1+i))} \right)

=2πi(π(1i)82+π(1i)82)=π222.\displaystyle =2\pi i \left( \frac{\pi(1-i)}{8 \sqrt{2}}+\frac{\pi(-1-i)}{8 \sqrt{2}}\right)=\frac{\pi^2}{2 \sqrt{2}}.

What's incorrect with this method?
Solution 2: A real method.

I=4J=202ln(u)1+u4du.\displaystyle I=4J=2 \int_0^\infty \frac{2 \ln(u)}{1+u^4}du.

J=01ln(u)1+u4du+1ln(u)1+u4du\displaystyle J = \int_0^1 \frac{ \ln(u)}{1+u^4}du+ \int_1^\infty \frac{ \ln(u)}{1+u^4}du

=01ln(u)k0(1)ku4kdu+1ln(u)k0(1)ku4(k+1)du\displaystyle = \int_0^1 \ln(u) \sum_{k \ge 0} (-1)^ku^{4k}du+\int_1^\infty \ln(u) \sum_{k \ge 0}(-1)^ku^{-4(k+1)}du

=k0(1)k(4k+1)2+k0(1)k(4k+3)2=G\displaystyle =- \sum_{k \ge 0} \frac{(-1)^k}{(4k+1)^2}+\sum_{k \ge0} \frac{(-1)^k}{(4k+3)^2}=G

Where G is Catalan's constant. This answer is incorrect, but I cannot see what's incorrect in the method.
(edited 10 years ago)
Reply 2496
Problem 405**

Evaluate

0πxcos(x)1+sin2(x)dx.\displaystyle \int_0^\pi \frac{x \cos(x)}{1+\sin^2(x)}dx.

Problem 406**

Find a closed form (i.e. no series) for

k1ζ(2k+1)1k+1.\displaystyle \sum_{k \ge 1} \frac{\zeta(2k+1)-1}{k+1}.
(edited 10 years ago)
Original post by henpen

Spoiler



Alternate solution to 404 (still can't do contour integration :nopity:)

We wish to evaluate:

0xlnx1+x2 dx\displaystyle\int_0^{\infty} \frac{\sqrt{x} \ln x}{1+x^2} \ dx


Start by letting xtanx:x \mapsto \tan x :

0xlnx1+x2 dx=0π2tanxlntanx dx=0π2tanxlnsinx dxI0π2tanxlncosx dxJ\displaystyle\begin{aligned} \int_0^{\infty} \frac{\sqrt{x} \ln x}{1+x^2} \ dx & = \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \ln \tan x \ dx \\ & = \underbrace{\int_0^{\frac{\pi}{2}} \sqrt{\tan x} \ln \sin x \ dx}_{I} - \underbrace{\int_0^{\frac{\pi}{2}} \sqrt{\tan x} \ln \cos x \ dx}_{J} \end{aligned}


I\bullet \quad I: For α>0\alpha > 0, define

K:=B(1α4,14)=01xα4(1x)34 dx=xsin2x20π2sinα+12xtanx dx    Kαα=1=I\displaystyle\begin{aligned}K := \text{B}\left(1 - \tfrac{\alpha}{4}, \tfrac{1}{4}\right) & = \int_0^1 x^{-\frac{\alpha}{4}} (1-x)^{-\frac{3}{4}} \ dx \\ & \overset{x \mapsto \sin^2 x}= 2 \int_0^{\frac{\pi}{2}} \sin^{\frac{-\alpha + 1}{2}} x \sqrt{\tan x} \ dx \\ \implies \frac{\partial K}{\partial \alpha} \bigg|_{\alpha = 1} & = - I\end{aligned}

Kαα=1=2π4(3ln2π2)    I=2π4(π23ln2)\displaystyle\frac{\partial K}{\partial \alpha} \bigg|_{\alpha = 1} = \frac{\sqrt{2}\pi}{4} \left(3 \ln 2 - \frac{\pi}{2}\right) \implies I = \frac{\sqrt{2}\pi}{4} \left(\frac{\pi}{2} - 3 \ln 2\right)


J\bullet \quad J: Now, for β>0\beta > 0, define

L:=B(34,13β4)=01x14(1x)3β4 dx=xsin2x20π2tanxcos32(1β)x dx    Lββ=1=3J\displaystyle\begin{aligned} L := \text{B} \left(\tfrac{3}{4}, 1 - \tfrac{3\beta}{4}\right) & = \int_0^1 x^{-\frac{1}{4}} (1-x)^{-\frac{3\beta}{4}} \ dx \\ & \overset{x \mapsto \sin^2 x}= 2 \int_0^{\frac{\pi}{2}} \sqrt{\tan x} \cos^{\frac{3}{2} (1 - \beta)} x \ dx \\ \implies \frac{\partial L}{\partial \beta}\bigg|_{\beta = 1} & = -3J\end{aligned}

Lββ=1=32π4(π2+3ln2)    J=2π4(π2+3ln2)\displaystyle \begin{aligned} \frac{\partial L}{\partial \beta} \bigg|_{\beta = 1} = \frac{3\sqrt{2}\pi}{4}\left( \frac{\pi}{2} + 3 \ln 2\right) \implies J = - \frac{\sqrt{2}\pi}{4} \left( \frac{\pi}{2} + 3 \ln 2\right)\end{aligned}


Bringing everything together:

0xlnx1+x2 dx=IJ=2π24\displaystyle \int_0^{\infty} \frac{\sqrt{x} \ln x}{1+x^2} \ dx = I - J = \frac{\sqrt{2}\pi^2}{4}
(edited 10 years ago)
Looks like my system of differential equations was a bit beyond you guys I guess. I'll post the solution in the very near future.

Posted from TSR Mobile

Quick Reply

Latest