Logarithm raised to what?

Calculus Level 5

$\large \int_0^{\pi/2} ( \ln (\cos x))^2 ( (\ln(\cos x))^2 - 6x^2) \, dx$

If the integral above is equal to $\dfrac{\pi^a}2 (\ln 2)^b - \dfrac{\pi^c}d$ , where $a,b,c$ and $d$ are positive integers, find $a+b+c+d$ .

Image Credit: Wikimedia Natural Logarithm .

The answer is 170.

1 solution

Aditya Malusare
Jan 9, 2016

This is the sort of solution that starts with something unrelated.

Consider the integral $\displaystyle\int_0^{\frac{\pi}{2}} \cos^u(x) \cos(ux)\text{d}x$ . Writing $\cos(x) = \dfrac{e^{ix} + e^{-ix}}{2}$ , the integrand becomes $\dfrac{(1+e^{2ix})^u + (1+e^{-2ix})^u}{2^{u+1}}$

Expanding using the binomial theorem the expression becomes $\displaystyle\sum_{k=0}^u\dfrac{ \displaystyle{{u}\choose{k}}(e^{2kix}+e^{-2kix}) }{2^{u+1}} =\displaystyle\sum_{k=0}^u\dfrac{ \displaystyle{{u}\choose{k}}\cos(2kx) }{2^{u}}$

$\color{}{\text{As Mark Hennings has pointed out, this is insufficient for what I am about to do. Hence, a correction:}}$

Expanding using the binomial series , we get $\displaystyle\sum_{k=0}^{\infty}\dfrac{ \displaystyle{{u}\choose{k}}(e^{2kix}+e^{-2kix}) }{2^{u+1}}$ We are allowed to do this, since $|e^{2ix}| = |e^{-2ix}| = 1$ and $u > 0$ and so the series converges.

So, $\displaystyle\sum_{k=0}^{\infty}\dfrac{ \displaystyle{{u}\choose{k}}(e^{2kix}+e^{-2kix}) }{2^{u+1}} =\displaystyle\sum_{k=0}^{\infty}\dfrac{ \displaystyle{{u}\choose{k}}\cos(2kx) }{2^{u}}$

Now, observe that for all $k > 0$ the integral is of the form $\displaystyle \int_{0}^{\frac{\pi}{2}} \cos(2kx) \text{d}x$ which is equal to zero for non-zero $k$

Hence, $\int_0^{\frac{\pi}{2}} \cos^u(x) \cos(ux)\text{d}x = \dfrac{\pi}{2^{u+1}}$

Now, seeing $\cos^u(x)$ (in the integral) and $\log(\cos(x))$ we are motivated to differentiate under the integral sign (trying 4 times).

$\dfrac{d^4}{du^4}\int_0^{\frac{\pi}{2}} \cos^u(x) \cos(ux)\text{d}x = \dfrac{d^4}{du^4}\dfrac{\pi}{2^{u+1}}$

Now, the actual differentiation is computationally intensive, so it does not seem unreasonable to give it to a computer (although it can be done manually)

Wolfram Alpha says

The messy expression reduces to $\log^4(\cos(x)) - 6x^2\log^2(\cos(x)) + x^4$ after taking $\lim_{u \to 0^{+}}$ and using DCT throughout. (See comments below)

Hence, finally, $\int_0^{\frac{\pi}{2}}(\log^4(\cos(x)) - 6x^2\log^2(\cos(x)) + x^4) \text{d}x = \dfrac{\pi \log^4(2)}{2}$

Simply taking the last term to the other side and grouping things together we get $\int_0^{\frac{\pi}{2}} \log^2(\cos(x))(\log^2(\cos(x)) - 6x^2)\text{d}x = \dfrac{\pi \log^4(2)}{2} - \dfrac{\pi^5}{160}$

This shows us that the final answer is $1+4+5+160 = \boxed{170}$

You have proved the formula $\int_0^{\frac12\pi} \cos^u x \cos(ux)\,dx \; = \; \frac{\pi}{2^{u+1}}$ for positive integer $u$ , but you then want to differentiate your result with respect to $u$ . There is something missing here. This formula is true for all $u > -1$ , but you have not proved it.

Whittaker and Watson have the identity (which they attribute to Cauchy) $\int_0^{\frac12\pi} \cos^{p+q-2}\theta \cos\big((p-q)\theta\big)\,d\theta \; = \; \frac{\pi}{2^{p+q-1}(p+q-1)B(p,q)}$ for $p+q>1$ ; putting $p=u+1$ and $q=1$ gives us what we need. The result is proved by some serious contour integration, though!

We could take this W&W integral and change parameters, obtaining $\int_0^{\frac12\pi} \cos^u\theta \cos(v \theta)\,d\theta \; = \; \frac{\pi}{2^{u+1}(u+1)B(\frac12(u+v)+1,\frac12(u-v)+1)}$ certainly valid for $-1 < u,v < 1$ , and differentiate with respect to $u$ and $v$ . In principle we could use this to obtain all integrals $\int_0^{\frac12\pi} \big[\ln(\cos \theta)\big]^m \theta^n\,d\theta \;.$

Mark Hennings - 5 years, 5 months ago

Can you please post your solution here so I can print and frame it? (Pretty please)

Relevant problem .

Pi Han Goh - 5 years, 5 months ago

My method was hard work (but I was using Mathematica to perform the algebra). I differentiated $\tfrac12B(\tfrac12(u+1),\tfrac12)$ four times, and put $u=0$ , to obtain $a \; = \; \int_0^{\frac12\pi} \big[\ln(\cos x)\big]^4\,dx\; = \; \tfrac{19}{480}\pi^5 + \tfrac14\pi^3 (\ln2)^2 + \tfrac12 \pi (\ln2)^4 + 3\pi\ln2 \zeta(3) \;.$ I then differentiated the (manipulated) W&W formula twice with respect to $u$ , putting $v=0$ , to obtain $\int_0^{\frac12\pi} \big[ \ln(\cos x)\big]^2 \cos vx\, dx \;= \; \tfrac{1}{12v^3}\left[12 + 2\pi^2v^2 - 3\pi^2v^2\mathrm{cosec}^2(\tfrac12v\pi) + 3v^2\left(H_{-\frac12v} + H_{\frac12v} + 2\ln2\right)^2\right] \sin\tfrac12v\pi\;.$ I then differentiated this twice with respect to $v$ and put $v=0$ to get the second integral. $b \; = \; \int_0^{\frac12\pi} \big[\ln(\cos x)\big]^2 x^2\,dx \; = \; \tfrac{11}{1440}\pi^5 + \tfrac{1}{24}\pi^3 (\ln 2)^2 + \tfrac12\ln 2\zeta(3) \;,$ and $\begin{array}{rcl} a - 6b & = & \left(\tfrac{19}{480}\pi^5 + \tfrac14\pi^3 (\ln2)^2 + \tfrac12 \pi (\ln2)^4 + 3\pi\ln2 \zeta(3) \right) - \left( \tfrac{11}{240}\pi^5 + \tfrac{1}{4}\pi^3 (\ln 2)^2 + 3\pi\ln 2\zeta(3) \right) \\ & = & \tfrac12\pi(\ln 2)^4 - \tfrac{1}{160}\pi^5 \;. \end{array}$

Aditya's method of using the W&W formula is much better than mine, since the integral he obtains is much easier to differentiate multiple times. I made the mistake of working out the first integral quickly, and then focused on getting the second integral on its own, rather than obtaining the two together!

My concern with Aditya's method is not that it was not a good one (it was!), but that the key formula he wanted to use needed more proof than he gave it.

Mark Hennings - 5 years, 5 months ago

@Mark Hennings – Hey, @Mark Hennings ! The problem with my method is that I have used the binomial theorem to prove the result for only positive integral values of $u$ , right? Differentiation is not allowed in that case, and it was silly of me to overlook this. But, I am now editing the solution so that I use the binomial series , which converges for the given conditions, i.e. $(1+z)^{u} = \displaystyle\sum_{k=0}^{\infty}{{u}\choose{k}}z^k$ , where $\displaystyle{{u}\choose{k}}$ is the generalized binomial coefficient. Since $|z| = 1$ and $u>-1$ the series converges for all $u$ taken, right? I hope this fixes my incomplete solution. Anyways, thanks for showing another amazing way how to do it!

Aditya Malusare - 5 years, 5 months ago

@Aditya Malusare – It's still not going to be straightforward, since the series does not converge at $x=\tfrac12\pi$ when $u < 0$ (and you want to differentiate at $0$ , so have to handle what is happening for negative $u$ ). Showing that you can integrate the series term-by-term is not automatic.

The series does converge uniformly in $x$ for all $u > 0$ , and so we can calculate the integral for all $u > 0$ by your proposed method. We can then differentiate four times, and let $u$ tend to $0$ , using continuity in $u$ of all the expressions involved, thanks to the DCT.

Mark Hennings - 5 years, 5 months ago

How will you get the last expression in terms of log ?? I think when we put u and v as zero ... We will get zero @Mark Hennings if n is odd...

Aman Rajput - 5 years, 5 months ago

Differentiate $m$ times with respect to $u$ , and $n$ times with respect to $v$ , and put $u=v=0$ . All the integrands are single-signed over $(0,\tfrac12\pi)$ , so we will never get $0$ . Don't forget we are integrating over $(0,\tfrac12\pi)$ , not over $(-\tfrac12\pi,\tfrac12\pi)$ .

Mark Hennings - 5 years, 5 months ago

@Mark Hennings – I think sir you didn't get my point :)

I was saying ... We can't convert that integral into logarithmic form as you expressed for all $n$ .

Just differentiate it with respect to u twice and differentiate with respect to v once ... And then put u=0=v You will get zero.

The form you have expressed is not valid for odd powers of theta

Aman Rajput - 5 years, 5 months ago

@Aman Rajput – Good point! This formula does not work for odd $n$ . There is a much more complicated formulae for $\int_0^{\frac12\pi} \cos^ux e^{ivx}\,dx$ which could be used (in theory) to get the formula for odd $n$ , but that takes us into the territory of Ronak's problem!

Mark Hennings - 5 years, 5 months ago

@Mark Hennings – I have solved that ronaks problem

Aman Rajput - 5 years, 5 months ago

@Aman Rajput – So have I.

Mark Hennings - 5 years, 5 months ago

My method was first to evaluate the value of integral of (log(cosx))^4 by using beta function and digamma function. Now the problem arises with the second part. I used the expansion for $(log(sinx))^2$ then evaluated the integral. Sorry since I didn't solve this problem I couldn't post my solution.

Aditya Kumar - 5 years, 5 months ago

@Pi Han Goh you could use the expansion to evaluate Ronak's problem. But the problem is to bring it to that form in which he expects the answer.

Aditya Kumar - 5 years, 5 months ago

I doubt I'm able to solve Ronak's problem, I've spent over 3 hours on that question last time, came up with nothing.

Pi Han Goh - 5 years, 5 months ago

@Pi Han Goh – Which is Ronak's problem? Can I have a shot at it?

Mark Hennings - 5 years, 5 months ago

@Mark Hennings – I've linked it in my first comment above, but here it is again !

Pi Han Goh - 5 years, 5 months ago

Logarithm raised to what?

Image Credit: Wikimedia Natural Logarithm .

The answer is 170.

1 solution

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