2

Over the past week I've come across several 'Natural Logarithm Integrals': $$\int_0^{\pi/2}\ln(\sin(x))\tan(x)dx, \int_0^{\pi/2}\ln(\sin(x))\ln(\tan(x))dx$$ and so on so forth. This lead to me creating a question:

Evaluate $$I=\int_0^{\pi/2} \ln^3(\cot(x))\ln^3(\sec(x)) dx$$

To solve this, I tried the following approach:$$$$ Consider $$\beta\bigg (\dfrac{1-a}{2},\dfrac{a-b+1}{2}\bigg ) =2\int_0^{\pi/2} \sin^{-a}(x)\cos^{a}(x)\cos^{-b}(x)dx$$ $$$$ $$=\dfrac{\Gamma \bigg (\dfrac{1-a}{2}\bigg )\Gamma \bigg (\dfrac{a-b+1}{2}\bigg )}{\Gamma \bigg (\dfrac{2-b}{2}\bigg )}$$ $$$$ $$\Rightarrow \dfrac{1}{2} \dfrac{\partial^6}{\partial a^3 \partial b^3}\beta\bigg (\dfrac{1-a}{2},\dfrac{a-b+1}{2}\bigg ) \bigg|_{a=0 , b=0} = I$$ $$$$ $$=\dfrac{1}{2} \dfrac{\partial^6}{\partial a^3 \partial b^3}\dfrac{\Gamma \bigg (\dfrac{1-a}{2}\bigg )\Gamma \bigg (\dfrac{a-b+1}{2}\bigg )}{\Gamma \bigg (\dfrac{2-b}{2}\bigg )} \bigg |_{a=0 , b=0}$$ $$$$ Unfortunately, this method would become really, really messy and tedious. I was thus hoping for a better, neater method to find the closed form of the Integral. $$$$ I would be extremely grateful if somebody could please help me solve this problem. Many, many thanks in advance for your time and consideration!

User1234
  • 3,958
  • 1
    From the result (containing logarithms multiplied with both $\zeta(3)$ (even squared) and $\zeta(5)$), I doubt there will be anything "neat" here... – mickep Jun 15 '15 at 08:38
  • This integrals become really tedious if you have take more then a few derivatives (let's say two or so) – tired Jun 15 '15 at 09:02
  • Is there any way in which additional terms could be introduced to the Integrand so that upon Differentiating, certain terms get cancelled out leading to easier calculations? I confess I don't see anything myself. – User1234 Jun 15 '15 at 09:14

1 Answers1

2

After many calculations, but without difficulties Ifind $$I=-(\frac{15}{64}{\pi^5}\ln^22+\frac{63}{16}{\pi^3}\zeta{(3)}\ln2+\frac{279}{8}{\pi}\zeta{(5)}\ln2+\frac{441}{32}{\pi}\zeta{(3)}^2+\frac{75}{512}{\pi^7})$$ Using functions psi.

user178256
  • 5,467
  • Mind sharing the calculations? – Alice Ryhl Jun 16 '15 at 15:24
  • Thanks very much Sir! But would you please share how you got to this result? Also, which method did you use (hopefully not my horrible method) ?Many many thanks! – User1234 Jun 16 '15 at 16:21
  • I suggest you start with $$\dfrac{1}{2} \dfrac{\partial}{\partial a }\dfrac{\Gamma \bigg (\dfrac{1-a}{2}\bigg )\Gamma \bigg (\dfrac{a-b+1}{2}\bigg )}{\Gamma \bigg (\dfrac{2-b}{2}\bigg )} $$

    And so on,...

    $$I=\int_0^{\pi/2} \ln^2(\cot(x))\ln^2(\sec(x)) dx=\frac{11}{192}{\pi^5}+\frac{1}{8}{\pi^3}\ln^2+\frac{7}{4}{\pi}\zeta{(3)}\ln2$$

     – user178256 Jun 17 '15 at 10:41
  • I suggest you start with $$\dfrac{1}{2} \dfrac{\partial}{\partial a }\dfrac{\Gamma \bigg (\dfrac{1-a}{2}\bigg )\Gamma \bigg (\dfrac{a-b+1}{2}\bigg )}{\Gamma \bigg (\dfrac{2-b}{2}\bigg )} $$

    And so on,...

    $$I=\int_0^{\pi/2} \ln^2(\cot(x))\ln^2(\sec(x)) dx=\frac{11}{192}{\pi^5}+\frac{1}{8}{\pi^3}\ln^2+\frac{7}{4}{\pi}\zeta{(3)}$$

     – user178256 Jun 17 '15 at 10:44
  • @user178256 Sir, I don't know how I did not notice this comment till now. Sir, please could you rewrite the closed form for $$I=\int_0^{\pi/2} \ln^2(\cot(x))\ln^2(\sec(x)) dx?$$ The $\ln^2$ term did not have any argument. – User1234 Jul 30 '15 at 15:45
  • @Better Word, ok, sorry.$$\frac{\pi^3}{8}ln^22$$ – user178256 Jul 31 '15 at 18:48
  • Sir, so the final answer is $$\int_0^{\pi/2} \ln^2(\cot(x))\ln^2(\sec(x)) dx$$$$=\frac{11}{192}{\pi^5}+\frac{1}{8}{\pi^3}\ln^2(2)+\frac{7}{4}{\pi}\zeta{(3)}$$ – User1234 Aug 01 '15 at 16:09
  • @Better Word,I am really sorry.Correction:$$\frac{7\pi}{4}\zeta{(3)}\ln{2}$$ I=23.975279345464785702... – user178256 Aug 01 '15 at 18:24
  • @user178256 Sir, it's OK, there's no need to be sorry::) Sir, please could you type the entire closed form once again just so I can understand all the changes that have to be made? – User1234 Aug 02 '15 at 13:28
  • @Better Word. $$I=\int_0^{\pi/2}\ln^2(\cot(x))\ln^2(\sec(x))dx=\frac{11}{192}{\pi^5}+\frac{\pi^3}{8}\ln^2{2}+\frac{7}{4}{\pi}\zeta{(3)}\ln{2}$$ I=23.97527934546478702... – user178256 Aug 02 '15 at 21:45