Integrate $\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}$

Question

Challenge question: solve $$\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}$$

It's a generalization of a recent Math.SE question, but how would one normally approach it?

i have only found a numerical value of your integral $$0.2383627095847316199842751304939977634136721384754172600021618182991202279207338258104065665273398798$$ — Dr. Sonnhard Graubner, Dec 30 '16 at 17:32
@user190080 : This one by JohnChessant. He has a function $I(a)$, and he is asking for $I(2)$. I simply posed $I(4)$ (this question). — THA, Dec 31 '16 at 00:34
@THA I have been summoned. :D Yes, $I(4)$ is equivalent to the integral in this question, and by my formula the answer is $\frac{1}{16}\pi\left(4−\pi\right)\sqrt{2}$ (talk of funny answers ;) ). — Ant, Dec 31 '16 at 01:08

score 10 · Accepted Answer · edited Apr 13 '17 at 12:21

10

Hint. By the change of variable $$ \sqrt{\tan x}=t,\quad x=\arctan (t^2),\quad dx=\frac{2t\:dt}{1+t^4}, $$ one gets

$$ \int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}=\int_0^\infty \frac{4t^4\ln t}{\left(1+t^4\right)^2}\:dt=\frac{\pi \sqrt{2}}{4}-\frac{\pi ^2\sqrt{2}}{16}. \tag1 $$

Addendum. One may recall the identity $$ \int_0^\infty\frac{x^\alpha}{1+x}\:dx=-\frac{\pi}{\sin \alpha \pi}, \quad -1<\alpha<0, \tag2 $$ giving, with an integration by parts, $$ \int_0^\infty\frac{x^\alpha}{(1+x)^2}\:dx=\frac{\alpha \pi}{\sin \alpha \pi}, \quad -1<\alpha<1 \tag3 $$ differentiating $(3)$ yields $$ \int_0^\infty\frac{x^\alpha\ln x}{(1+x)^2}\:dx=\frac{\pi}{\sin \alpha \pi}-\frac{\pi^2\alpha \cos \alpha \pi}{\sin^2 \alpha \pi}, \quad -1<\alpha<1, \tag4 $$ then putting $x=t^4$, $\alpha=\frac14$ gives $(1)$.

edited Apr 13 '17 at 12:21

Community

1

answered Dec 30 '16 at 17:43

Olivier Oloa

120,989

1

What is the method for evaluating the last integral? – Rene Schipperus Dec 30 '16 at 18:00
Did you use integration by parts in the end? If yes, Could you please explain how did u evaluate $\int{\frac{4t^4}{(1+t^4)^2}dt}$ – Serenity Dec 30 '16 at 18:06
@ShreyasS did you forget the log in your comment ? – Rene Schipperus Dec 30 '16 at 18:09
@OlivierOloa What is IPP ?Am I dumb ? – Rene Schipperus Dec 30 '16 at 18:11
@ReneSchipperus I assumed he has integrated by parts, taking $lnt$ as the first function – Serenity Dec 30 '16 at 18:12
@ReneSchipperus I've edited my hint. Thanks. – Olivier Oloa Dec 30 '16 at 18:33
@ReneSchipperus Thank you very much. Happy new year to you. – Olivier Oloa Dec 30 '16 at 18:54
+1 Happy new year all! And it appears there is no other way to do it; I asked the question that OP referenced; you can check that my formula was essentially equivalent to that of @OlivierOloa. – Ant Dec 31 '16 at 03:50
@John Chessant Thank you very much. Happy new year to all! – Olivier Oloa Dec 31 '16 at 07:57

score 0 · Answer 2 · answered Dec 30 '16 at 17:33

0

Write $\ln\tan x=\ln\sin x-\ln\cos x$, replace $\sqrt{\tan x}$ also, and do the integral of the two summands separately. For the first summand put $u=\sin x$ and integrate by parts. For the second summand put $u=\cos x$.

answered Dec 30 '16 at 17:33

user2747939

188

score 0 · Answer 3 · edited Dec 30 '16 at 18:21

0

Using $$\tan(x)\cos(x)\sin(x)=\sin^2(x)=\frac{\tan^2(x)}{1+\tan^2(x)}$$ and putting $t=\tan(x)$, we get

$$I=\int_0^{+\infty}t^{\frac{3}{2}}dt=+\infty.$$

edited Dec 30 '16 at 18:21

amWhy

209,954

answered Dec 30 '16 at 17:50

hamam_Abdallah

62,951

There was nothing modified that could have modified the question's meaning. In fact, the only edit was to include \int in the dollar signs. – amWhy Dec 30 '16 at 18:21
If you want to delete your answer, you are free to do so. The question now is, as all long, not changed. – amWhy Dec 30 '16 at 18:22

Integrate $\int_0^{\pi/2} {\sin x \cos x \sqrt{\tan x} \ln{\tan x} \,dx}$

3 Answers3