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Does there exists a closed form of$$ \displaystyle \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { x } } \log { (\cos { x } ) }\ dx }$$

If exists can someone find a way to tackle this integral and provide a closed-form of it. Many similar integrals have closed form and I believe this one, too.

Quanto
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Ronak Agarwal
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  • $I\approx-5.8716450278229730501607669875771\ldots$ – Lucian Mar 08 '15 at 06:11
  • But does there exists a closed form, wolfram alpha also returns me this numerical value. – Ronak Agarwal Mar 08 '15 at 06:38
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    The x complicates things. If it would have been absent, then all integrals of the form $~\displaystyle\int_0^\tfrac\pi2\sin^ax~\cos^bx~\ln^k(\sin x)~dx~$ and $~\displaystyle\int_0^\tfrac\pi2\sin^ax~\cos^bx~\ln^k(\cos x)~dx~$ possess a closed form for $k\in$ N. See Wallis' integrals, beta function and polygamma function for more information. – Lucian Mar 08 '15 at 06:45
  • Yes that's the main problem for me too. I have solved the integral without x. It was asked on a different site and the asker ensures that it has a closed form , I don't know whether he is lying or not. – Ronak Agarwal Mar 08 '15 at 07:02
  • But it exist a closed form of a integral $ \displaystyle \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { (x) } } \log { (\tan { (x) } ) } dx }=\frac{\pi\sqrt{2}}{48}(3{\pi^2}+48G-6{\pi}{ln{2}}) $ Where G is the Catalan's constant. – user178256 Mar 09 '15 at 17:20
  • @RonakAgarwal Did you consider posting a note in B ? And is this question from B ? –  Mar 13 '15 at 17:56
  • Yes here is the answer to your problem, Find it's closed form, you see user178256 has provided an answer. @AzhaghuRoopeshM – Ronak Agarwal Mar 14 '15 at 06:17
  • @RonakAgarwal well it's your call, it'd be nice if you were to post the solution to the question on B rather than me doing it . What say ? After all , I didn't create this question , a friend of mine did it . –  Mar 14 '15 at 13:26

2 Answers2

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Substitute $t = \sqrt{\tan x}$

\begin{align} I=\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\cos x})\ dx=J(1)\\ \end{align} where $J(a)= -\int_0^\infty \frac{t^2 \ln(1+t^4) \tan^{-1}(a^2t^2)}{1+t^4}dt$ \begin{align} J’(a)=&-\int_0^\infty \frac{2at^2 \ln(1+t^4)}{(1+t^4)(1+a^4t^4)}dt\\ =& \ \frac{2a}{1-a^4}\int_0^\infty \left(\frac{\ln(1+t^4)}{1+t^4}- \frac{\ln(1+t^4)}{1+a^4t^4} \right)dt \end{align} Let $K(a,b)=\int_0^\infty \frac{\ln(1+b^4t^4)}{1+a^4t^4}dt$

\begin{align} K_b’(a,b)=&\int_0^\infty \frac{4b^3 t^4}{(1+a^4t^4)(1+b^4t^4)}dt =\frac{\sqrt2\pi b^2}{a(a+b)(a^2+b^2)} \end{align}

Then \begin{align} K(a,b)= &\int_0^b K_b’(a,s)ds = \frac\pi{\sqrt2 a}\left(\frac12\ln\frac{a^2+b^2}{a^2} +\ln\frac{a+b}a-\tan^{-1}\frac ba\right)\\ J’(a)=&\ \frac{2a}{1-a^4}\left[K(1,1)-K(a,1) \right]\\ =& \frac{\sqrt2\pi}{1-a^4} \left[a\left(\frac32\ln2-\frac\pi4 \right)-\frac12 \ln\frac{1+a^2}{a^2} -\ln\frac{1+a}{a}+\cot^{-1}a \right] \end{align} and

\begin{align} I=&\ J(1)=\int_0^1 J’(a)da\\ =& -\sqrt2\pi \bigg[ \left(\frac32\ln2-\frac\pi4 \right)\int_0^1 \frac{1-a}{1-a^4}da +\frac12\int_0^1 \frac{\ln\frac{1+a^2}{2a^2}}{1-a^4}da\\ &\>\>\>\>\>\>\>\>\>\>\>\>\>\>\> +\int_0^1 \frac{\ln\frac{1+a}{2a}}{1-a^4}da +\int_0^1 \frac{\frac\pi4-\cot^{-1}a}{1-a^4}da\bigg]\\ =& -\sqrt2\pi \bigg[ \left(\frac32\ln2-\frac\pi4 \right)\left(\frac\pi8+\frac14\ln2\right) +\frac12\left(\frac12G+\frac{3\pi^2}{32} +\frac\pi{8}\ln2 \right)\\ &\>\>\>\>\>\>\>\>\>\>\>\>\>\>\> +\left(\frac12G+\frac{\pi^2}{24} -\frac\pi{16}\ln2\right) +\left(-\frac14G-\frac{\pi^2}{64}\right)\bigg]\\ =&-\frac\pi{\sqrt2}\left(G+\frac{\pi^2}{12}+\frac\pi4\ln2+\frac34\ln^22 \right) \end{align}

Quanto
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  • Can this method be used for the same integral with squared log? I stumbled across and linked an old question asking about it. – user170231 Jan 06 '23 at 20:50
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$ \displaystyle \int _{ 0 }^{ \pi /2 }{ x\sqrt { \tan { (x) } } \log { (\cos { (x) } ) } dx }=-\frac{\pi\sqrt{2}}{8}(\frac{\pi^2}{3}+4G+3\ln^22+{\pi}{\ln{2}}) $

dustin
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user178256
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    In order to avoid further downvotes, it is recommended to add at least the sketch of a proof which might justify the above result. – Lucian Mar 09 '15 at 22:43
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    Whoa that's great, please give some hints for me on how to solve the integral, I thought all the time that a closed form for this does not exist. Great. – Ronak Agarwal Mar 10 '15 at 06:40
  • Use this relations: A=∫_0^(π/2)▒〖〖cos〗^(u-n-1) (x) 〖sin〗^(n-1) (x) cos⁡(ux)dx=Γ(n)Γ(u-n)/Γ(u) cos⁡(nπ/2) 〗 B=∫_0^(π/2)▒〖〖cos〗^(u-n-1) (x) 〖sin〗^(n-1) (x) sin⁡(ux)dx=Γ(n)Γ(u-n)/Γ(u) sin⁡(nπ/2) 〗 u≥1, 0<n<1 View the book "Advanced Calculus" by G.A.Gibson page 466. Calculate (∂^2 A)/(∂^2 u) and (∂^2 B)/(∂^2 u),then do u=1,n=1/2 – user178256 Mar 10 '15 at 14:12
  • Is also used I= ∫_0^(π/2)▒〖〖sin〗^(2α-1) x〖cos〗^(2β-1) xdx=1/2 Γ(α)Γ(β)/Γ(α+β) 〗 Calculate (∂^2 I)/(∂^2 α) and (∂^2 I)/(∂^2 β) for α=1/4,β=3/4 and α=3/4, β=1/4 By addition and subtraction des relationships found,we have: ∫_0^(π/2)▒〖((1+tan⁡x ))/√(tan⁡x ) Log cos⁡x dx=-π√2〗 (π/4 Log2+G) ∫_0^(π/2)▒〖((1+tan⁡x ))/√(tan⁡x ) Log cos⁡x dx=(π√2)/4 (π^2/3+3〖Log〗^2 2) 〗 – user178256 Mar 10 '15 at 14:13
  • ∫_0^(π/2)▒〖x(1+tan⁡x )/√(tan⁡x ) Log cos⁡x dx=-π√2〗 (π/4 Log2+G) ∫_0^(π/2)▒〖((1+tan⁡x ))/√(tan⁡x ) 〖Log〗^2 cos⁡x dx=(π√2)/4 (π^2/3+3〖Log〗^2 2) 〗 – user178256 Mar 12 '15 at 00:22
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    Kindly latex your mathematics, please I can't understand what you have written. – Ronak Agarwal Mar 12 '15 at 15:11