Integral 111

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Integral 111

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1309 posts

#1 h Dec 29, 2018, 3:19 AM • 1 Y

Y by Adventure10

Greetings, calculate: $\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x} dx$ Bonus: $\int_0^\frac{\pi}{2} x^2 \sqrt{\cot x} dx$

This post has been edited 1 time. Last edited by Yimself, Dec 29, 2018, 7:29 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

809 posts

#2 h Dec 29, 2018, 4:44 AM • 2 Y

Y by Adventure10, Mango247

Partial progress..

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3465 posts

#3 h Dec 29, 2018, 5:47 AM • 2 Y

Y by Adventure10, Mango247

~~redacted~~

This post has been edited 2 times. Last edited by integrated_JRC, Jan 1, 2019, 10:58 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

809 posts

#4 h Dec 29, 2018, 6:39 AM • 2 Y

Y by Adventure10, Mango247

integrated_JRC wrote:

We put $\tan x=z^2\quad$ ; $~dx=\frac{2z}{z^4+1}\, dz$
$\begin{align*} \mathfrak{I}&=\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x} dx\\ &=2\int_0^{\infty}\frac{\Big(\tan^{-1}z^2\Big)^2z}{z^4+1}\, dz\\ \end{align*}$

It should be $\mathfrak{I}=2\int_0^{\infty}\frac{\Big(\tan^{-1}z^2\Big)^2z^2}{z^4+1}\, dz$

This post has been edited 2 times. Last edited by Maths1998, Dec 29, 2018, 6:40 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1668 posts

#5 h Dec 30, 2018, 9:59 AM • 5 Y

Y by Yimself, mrtaurho, MassiveMonster, Adventure10, Mango247

Yimself wrote:

Greetings, calculate: $\int_0^\frac{\pi}{2} x^2 \sqrt{\tan x} dx$

According to my very long calculations, the answer is $\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96},$ which is surprisingly nice.

The idea is similar to the one I used to find $\int_0^{\pi/2}x^2\sqrt{\cos x} \ dx,$ only the solution is a little more involved. By the way, I said in that post that $\int_0^{\pi/2}x^2\sqrt{\sin x} \ dx$ doesn't seem to be as nice as $\int_0^{\pi/2}x^2\sqrt{\cos x} \ dx$ but I really never tried to find it; maybe I should now.

OK, I'm going to skip many details to keep the solution short. Start with the substitution $\sin x = \sqrt{t}$ to get
$I:=\int_0^{\pi/2} x^2\sqrt{\tan x} \ dx = \frac{1}{2}\int_0^1 (\sin^{-1}\sqrt{t})^2t^{-1/4}(1-t)^{-3/4}dt$ and then the Maclaurin series of $(\sin^{-1}\sqrt{t})^2$ gives
$I=\sum_{n \ge 1}\frac{4^{n-1}}{n^2\binom{2n}{n}}\int_0^1 t^{n-1/4}(1-t)^{-3/4}dt=\sum_{n \ge 1} \frac{4^{n-1}}{n^2\binom{2n}{n}} \cdot \frac{\Gamma(n+3/4)\Gamma(1/4)}{\Gamma(n+1)}$ $=\frac{\sqrt{\pi}\Gamma(1/4)}{4}\sum_{n\ge 1} \frac{\Gamma(n+3/4)}{n^2\Gamma(n+1/2)}=\frac{\sqrt{\pi}\Gamma(1/4)}{4\Gamma(3/4)}\sum_{n\ge 1}\frac{n+1/2}{n^2}B(n+3/4, 3/4)$ $=\frac{\sqrt{\pi}\Gamma(1/4)}{4\Gamma(3/4)}\sum_{n \ge 1}\left(\frac{1}{n}+\frac{1}{2n^2}\right)\int_0^1t^{n-1/4}(1-t)^{-1/4} dt$ $=\frac{\sqrt{\pi}\Gamma(1/4)}{4\Gamma(3/4)}\int_0^1t^{-1/4}(1-t)^{-1/4}\left(-\ln(1-t)+\frac{1}{2}\text{Li}_2(t)\right)dt.$ Let
$J:=\int_0^1t^{-1/4}(1-t)^{-1/4}\ln(1-t) \ dt, \ \ \ K:=\int_0^1t^{-1/4}(1-t)^{-1/4}\text{Li}_2(t) \ dt.$ So
$I=\frac{\sqrt{\pi}\Gamma(1/4)}{4\Gamma(3/4)}\left(-J+ \frac{1}{2}K\right). \ \ \ \ \ \ (*)$ Let's find $J$ first. Let $\psi$ be the digamma function. We have
$J=\int_0^1t^{-1/4}(1-t)^{-1/4}\ln t \ dt=\frac{\partial}{\partial x}B(x,y) \Big\vert_{x=y=3/4}=B(3/4,3/4)(\psi(3/4)-\psi(3/2))$ $=\frac{2\Gamma^2(3/4)}{\sqrt{\pi}}\left(\frac{\pi}{2}-\ln 2-2\right).$ We now find $K.$ Let $\psi_1$ be the trigamma function. We have
$2K=\int_0^1t^{-1/4}(1-t)^{-1/4}(\text{Li}_2(t)+\text{Li}_2(1-t)) \ dt=\int_0^1t^{-1/4}(1-t)^{-1/4}\left(\frac{\pi^2}{6}-\ln t \ln(1-t)\right) dt$ $=\frac{\pi^2}{6}B(3/4,3/4)-\int_0^1t^{-1/4}(1-t)^{-1/4}\ln t \ln(1-t) \ dt=\frac{\pi \sqrt{\pi}}{3}\Gamma^2(3/4)-\frac{\partial^2}{\partial x \partial y}B(x,y) \Big\vert_{x=y=3/4}$ $=\frac{\pi \sqrt{\pi}}{3}\Gamma^2(3/4)-B(3/4,3/4)((\psi(3/4)-\psi(3/2))^2-\psi_1(3/2))$ $=\frac{\pi \sqrt{\pi}}{3}\Gamma^2(3/4)-\frac{2\Gamma^2(3/4)}{\sqrt{\pi}}\left( \left(\frac{\pi}{2} - \ln 2 - 2\right)^2-\frac{\pi^2}{2}+4\right).$ Now that we have both $J,K,$ we put them in $(*)$ and use the fact that $\Gamma(1/4)\Gamma(3/4)=\sqrt{2}\pi$ to get
$I=\frac{\sqrt{2}\pi(5\pi^2+12\pi\ln 2 - 12\ln^22)}{96}.$ ..........................
Similarly, we can find your other integral, i.e. $\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx,$ which is by the way a little easier than $\int_0^{\pi/2}x^2\sqrt{\tan x} \ dx.$ So, again, we put $\sin x = \sqrt{t}$ to get
$L:=\int_0^{\pi/2} x^2\sqrt{\cot x} \ dx = \frac{1}{2}\int_0^1 (\sin^{-1}\sqrt{t})^2t^{-3/4}(1-t)^{-1/4}dt$ and then the Maclaurin series of $(\sin^{-1}\sqrt{t})^2$ gives
$L=\sum_{n \ge 1}\frac{4^{n-1}}{n^2\binom{2n}{n}}\int_0^1 t^{n-3/4}(1-t)^{-1/4}dt=\sum_{n \ge 1} \frac{4^{n-1}}{n^2\binom{2n}{n}} \cdot \frac{\Gamma(n+1/4)\Gamma(3/4)}{\Gamma(n+1)}$ $=\frac{\sqrt{\pi}\Gamma(3/4)}{4}\sum_{n\ge 1} \frac{\Gamma(n+1/4)}{n^2\Gamma(n+1/2)}=\frac{\sqrt{\pi}\Gamma(3/4)}{4\Gamma(1/4)}\sum_{n\ge 1}\frac{1}{n^2}B(n+1/4, 1/4)$ $=\frac{\sqrt{\pi}\Gamma(3/4)}{4\Gamma(1/4)}\sum_{n \ge 1}\frac{1}{n^2}\int_0^1t^{n-3/4}(1-t)^{-3/4} dt$ $=\frac{\sqrt{\pi}\Gamma(3/4)}{4\Gamma(1/4)}\int_0^1t^{-3/4}(1-t)^{-3/4}\text{Li}_2(t)dt.$ So if we let $M:=\int_0^1t^{-3/4}(1-t)^{-3/4}\text{Li}_2(t)dt,$ then
$L=\frac{\sqrt{\pi}\Gamma(3/4)}{4\Gamma(1/4)}M. \ \ \ \ \ \ \ (**)$ Now we have
$2M=\int_0^1t^{-3/4}(1-t)^{-3/4}(\text{Li}_2(t)+\text{Li}_2(1-t)) \ dt=\int_0^1t^{-3/4}(1-t)^{-3/4}\left(\frac{\pi^2}{6}-\ln t \ln(1-t)\right) dt$ $=\frac{\pi^2}{6}B(1/4,1/4)-\int_0^1t^{-3/4}(1-t)^{-3/4}\ln t \ln(1-t) \ dt=\frac{\pi \sqrt{\pi}}{6}\Gamma^2(1/4)-\frac{\partial^2}{\partial x \partial y}B(x,y) \Big\vert_{x=y=1/4}$ $=\frac{\pi \sqrt{\pi}}{6}\Gamma^2(1/4)-B(1/4,1/4)((\psi(1/4)-\psi(1/2))^2-\psi_1(1/2))$ $=\frac{\pi \sqrt{\pi}}{6}\Gamma^2(1/4)-\frac{\Gamma^2(1/4)}{\sqrt{\pi}}\left( \left(\frac{\pi}{2} +\ln 2\right)^2-\frac{\pi^2}{2}\right).$ Now that we have $M,$ we put it in $(**)$ and use the reflection formula $\Gamma(1/4)\Gamma(3/4)=\sqrt{2}\pi$ to get
$L=\frac{\sqrt{2}\pi(5\pi^2-12\pi \ln 2-12\ln^22)}{96}.$

This post has been edited 1 time. Last edited by ysharifi, Jan 13, 2019, 7:22 PM
Reason: added solution for the integral of x^2*sqrt(cot(x))

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a