# MHF Integral Bee!

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• Aug 16th 2009, 02:14 AM
NonCommAlg
MHF Integral Bee!
This thread is for pros! (Rock) Here is how it works: it starts with me giving you an integral to solve. If you have a complete and correct solution, post it. Otherwise do not post anything please!

So you post your solution and also your own problem (integral of course!) for others to solve. It goes on like this until this thread has 100 posts. If after 48 hours nobody solved your problem, which
is not very likely, you'll post your solution and again it'll be your turn to challenge us with a new problem. Finally whoever earns the right to post in this thread can also post his solution, if he has a

different one of course, to any problem that has already been solved in this thread. Make sure your problem is interesting and definitely not very easy to solve. Do not give any hint please!

Let's get started: Evaluate $I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{4x - \pi}{1 - \cot x} \ dx.$
• Aug 16th 2009, 05:34 AM
luobo
Quote:

Originally Posted by NonCommAlg
This thread is for pros! (Rock) Here is how it works: it starts with me giving you an integral to solve. If you have a complete and correct solution, post it. Otherwise do not post anything please!

So you post your solution and also your own problem (integral of course!) for others to solve. It goes on like this until this thread has 100 posts. If after 48 hours nobody solved your problem, which
is not very likely, you'll post your solution and again it'll be your turn to challenge us with a new problem. Finally whoever earns the right to post in this thread can also post his solution, if he has a

different one of course, to any problem that has already been solved in this thread. Make sure your problem is interesting and definitely not very easy to solve. Do not give any hint please!

Let's get started: Evaluate $I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{4x - \pi}{1 - \cot x} \ dx.$

Hi, good morning.
Let $t=\frac{\pi}{4}-x$, then the integral becomes $I=2 \int_{0}^{\frac{\pi}{2}} (x\cot x-x)\, dx = 2 \int_{0}^{\frac{\pi}{2}} x\cot x\, dx - \frac{\pi^2}{4}=2 \int_{0}^{\frac{\pi}{2}} x\, d\ln(\sin x)- \frac{\pi^2}{4}$

So, $I= - 2 \int_{0}^{\frac{\pi}{2}} \ln(\sin x)\, dx- \frac{\pi^2}{4} = \pi \ln2- \frac{\pi^2}{4}$
• Aug 16th 2009, 01:51 PM
NonCommAlg
luobo, you solved the integral i gave (good work, by the way!) but you forgot to post an integral for us to solve! (Thinking) (see the rules of this thread again!)
• Aug 16th 2009, 02:36 PM
luobo
Quote:

Originally Posted by NonCommAlg
luobo, you solved the integral i gave (good work, by the way!) but you forgot to post an integral for us to solve! (Thinking) (see the rules of this thread again!)

Sorry. See below.

$I=\int_{0}^{1} \frac{x^\alpha-1}{\ln x} \, dx=?$ where $\alpha > -1$
• Aug 16th 2009, 04:00 PM
pankaj
$
\frac{dI}{d\alpha}=\int_{0}^{1}\frac{x^{\alpha}\ln x}{\ln x}dx=\frac{1}{\alpha+1}
$

$
I=\ln (\alpha+1)+c
$

Challenge Problem
If $f$ is a continuous function on $[0,1]$ such that $f(x)+f(x+\frac{1}{2})=1$,then evaluate $
\int_{0}^{1}f(x)dx
$
• Aug 16th 2009, 04:20 PM
luobo
Quote:

Originally Posted by pankaj
$
\frac{dI}{d\alpha}=\int_{0}^{1}\frac{x^{\alpha}\ln x}{\ln x}dx=\frac{1}{\alpha+1}
$

$
I=\ln (\alpha+1)+c
$

Challenge Problem
If $f$ is a continuous function on $[0,1]$ such that $f(x)+f(x+\frac{1}{2})=1$,then evaluate $
\int_{0}^{1}f(x)dx
$

$\int_{0}^{1}f(x)dx=\int_{0}^{\frac{1}{2}}f(x)dx + \int_{\frac{1}{2}}^{1}f(x)dx=$ $\int_{0}^{\frac{1}{2}}f(x)dx + \int_{0}^{\frac{1}{2}}f(x+\frac{1}{2})dx=\frac{1}{ 2}$

Relay Problem:
$\int_{0}^{\infty} \frac{x^3}{e^x-1} \, dx$
• Aug 16th 2009, 05:58 PM
NonCommAlg
Quote:

Originally Posted by luobo

$\int_{0}^{\infty} \frac{x^3}{e^x-1} \, dx$

let $I(a)=\int_0^{\infty} \frac{x^a}{e^x - 1} \ dx, \ a > 0.$ we have:

$I(a)=\int_0^{\infty}\frac{x^ae^{-x}}{1-e^{-x}} \ dx=\int_0^{\infty}x^ae^{-x}\sum_{n \geq 0}e^{-nx} \ dx=\sum_{n \geq 0} \int_0^{\infty}x^ae^{-(n+1)x} \ dx$

$=\Gamma(a+1)\sum_{n \geq 0} \frac{1}{(n+1)^{a+1}}=\Gamma(a+1) \zeta(a+1).$

so, for example $I(3)=\Gamma(4)\zeta(4)=\frac{\pi^4}{15}.$

Problem: $\int_0^{\infty} \frac{\sin(\tan x)}{x} \ dx = ?$
• Aug 17th 2009, 08:48 AM
NonCommAlg
Quote:

Originally Posted by NonCommAlg

Problem: $\int_0^{\infty} \frac{\sin(\tan x)}{x} \ dx = ?$

forget about this one for now and see if you can prove the following nice result, which will also solve the above problem:

Problem: Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be such that $f(x+\pi)=-f(x), \ f(x)=f(-x), \ \forall x \in \mathbb{R}.$ Assuming that $I=\int_0^{\infty} f(x) \frac{\sin x}{x} \ dx$ exists, prove that $I= \int_0^{\frac{\pi}{2}} f(x) \cos x \ dx.$
• Aug 17th 2009, 12:46 PM
luobo
Quote:

Originally Posted by NonCommAlg
forget about this one for now and see if you can prove the following nice result, which will also solve the above problem:

Problem: Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and suppose that $f(x+\pi)=-f(x), \ f(x)=f(-x), \ \forall x \in \mathbb{R}.$ Assuming that $I=\int_0^{\infty} f(x) \frac{\sin x}{x} \ dx$ exists, prove that $I= \int_0^{\frac{\pi}{2}} f(x) \cos x \ dx.$

$
\int_{0}^{\infty} f(x)\frac{\sin x}{x}\,dx=
$

$
\int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x}\,dx+\sum_{n=1}^{\infty}\int_{n\pi-\frac{\pi}{2}}^{n\pi+\frac{\pi}{2}} f(x)\frac{\sin x}{x}\,dx=
$

$
\int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x}\,dx+\sum_{n=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x+n\pi)\frac{\sin (x+n\pi)}{x+n\pi}\,dx=
$
{Note: $t= x-n\pi$}

$
\int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x}\,dx+\sum_{n=1}^{\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x)\frac{\sin (x)}{x+n\pi}\,dx=
$
{Note: $f(x+n\pi)=(-1)^n f(x), \sin(x+n\pi)=(-1)^n \sin x$}

$
\int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x}\,dx+\sum_{n=1}^{\infty}\left(\int_{-\frac{\pi}{2}}^{0} f(x)\frac{\sin (x)}{x+n\pi}\,dx+\int_{0}^{\frac{\pi}{2}} f(x)\frac{\sin (x)}{x+n\pi}\,dx\right)=
$

$
\int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x}\,dx+\sum_{n=1}^{\infty}\left(\int_{0}^{\frac {\pi}{2}} f(x)\frac{\sin (x)}{x-n\pi}\,dx+\int_{0}^{\frac{\pi}{2}} f(x)\frac{\sin (x)}{x+n\pi}\,dx\right)=
$
{Note: Let $t=-x$ for the second integral}

$
\int_{0}^{\frac{\pi}{2}} f(x) \frac{\sin x}{x}\,dx+\sum_{n=1}^{\infty}\int_{0}^{\frac{\pi}{ 2}} f(x)\frac{2x\sin (x)}{x^2-(n\pi)^2} \, dx=
$

$
\int_{0}^{\frac{\pi}{2}} f(x)\sin x \left(\frac{1}{x}+\frac{2x}{\pi^2}\sum_{n=1}^{\inf ty}\frac{1}{{\frac{x^2}{\pi^2}}-n^2}\right) \, dx=
$

$
\int_{0}^{\frac{\pi}{2}} f(x)\sin x \left(\frac{1}{x}+\frac{2x}{\pi^2}\frac{\pi^2}{2} \frac{x-\tan x}{x^2\tan x}\right)\,dx=
$
{Note: $\sum_{n=1}^{\infty} \frac{2z}{z^2-n^2}=\pi \cot\pi z -\frac{1}{z}$}

$
\int_{0}^{\frac{\pi}{2}} f(x)\cos x\,dx
$

I presume this is correct. Sorry for the delay of the challenge problem. It was very busy.

Relay Problem:
Use the equation proven above to finish the rest of the integral posted by NonCommAlg, reposted here:

$
\int_0^{\infty} \frac{\sin(\tan x)}{x} \ dx = ?
$

Note: $f(x)=\frac{\sin(\tan x)}{\sin x}$ is not continuous at $x=k\pi+\frac{\pi}{2}$.
• Aug 17th 2009, 10:16 PM
simplependulum
If $f(x)$ is continous for $x \in [0,\frac{\pi}{2})$ , I think it may also satisify the identity .. because we cut the integral into many pieces $(0,\frac{\pi}{2}) , (\frac{\pi}{2}, \pi) ....$ and thus each improper integrals may exist .

then

$\int_0^{\infty} \frac{ \sin(\tan(x))}{x} ~dx$

$= \sum_{k=0}^{\infty} \int_0^{\infty} \frac{ (-1)^k (\tan(x))^{2k+1}}{(2k+1)! x} ~dx$

$= \sum_{k=0}^{\infty} \int_0^{\infty} \frac{ (-1)^k \sin(x) \sec(x) (\tan(x))^{2k}}{(2k+1)! x} ~dx$

Then $f(x) = \sec(x) (\tan(x))^{2k}$

$= \sum_{k=0}^{\infty} \int_0^{\frac{\pi}{2}} \frac{ (-1)^k (\tan(x))^{2k}}{(2k+1)! } ~dx$

$= \int_0^{\frac{\pi}{2}} \frac{\sin(\tan(x))}{\tan(x)}~dx$

$= \frac{1}{2} \int_{-\infty}^{\infty} \frac{ \sin(t)}{t(t^2+1)}~dt$

Finally,

$= \frac{\pi}{2}(1- e^{-1})$
• Aug 17th 2009, 11:44 PM
NonCommAlg
luobo, the continuity condition for $f$ is unnecessary. i removed it from the problem. your solution is perfect! (Clapping)

here's my solution: assuming that $I=\int_0^{\infty} \frac{\sin(\tan x)}{x} \ dx$ is convergent (i'll leave this open!), we have:

$I=\int_0^{\infty} \frac{\sin(\tan x)}{\sin x} \cdot \frac{\sin x}{x} \ dx = \int_0^{\frac{\pi}{2}} \frac{\sin(\tan x)}{\sin x} \cos x \ dx = \int_0^{\frac{\pi}{2}} \frac{\sin(\tan x)}{\tan x} \ dx.$ now put $\tan x = t$:

$I=\int_0^{\infty} \frac{\sin t}{t(t^2 + 1)} \ dt=\int_0^{\infty} \frac{\sin t}{t} \ dt - \int_0^{\infty} \frac{t \sin t}{t^2 + 1} \ dt=\frac{\pi}{2} + \left[\frac{d}{d \alpha} \int_0^{\infty} \frac{\cos (\alpha t)}{t^2+1} \ dt \right]_{\alpha = 1}$

$=\frac{\pi}{2} + \left[\frac{d}{d \alpha} \left(\frac{\pi}{2} e^{-\alpha} \right) \right]_{\alpha=1}=\frac{\pi}{2}(1 - e^{-1}).$
• Aug 18th 2009, 01:22 AM
simplependulum
Here is the problem :

$\int_a^{a+1} \ln[ \Gamma(x)]~ dx ~~$ $~~ a \geq 0$
• Aug 18th 2009, 11:36 AM
luobo
Quote:

Originally Posted by simplependulum
Here is the problem :

$\int_a^{a+1} \ln[ \Gamma(x)]~ dx ~~$ $~~ a \geq 0$

Let $I(a)=\int_a^{a+1} \ln \Gamma(x) \, dx$, then
$\frac{d I(a)}{da} = \ln \Gamma(a+1)-\ln \Gamma(a)=\ln \frac{\Gamma(a+1)}{\Gamma(a)}=\ln a \Rightarrow I(a)=a\ln a-a +C$
Therefore,
$C=I(0)=\int_0^1 \ln \Gamma(x) \, dx= -\int_0^1 \left[ \ln x + \gamma x +\sum_{k=1}^{\infty} \left( \ln \frac{x+k}{k} - \frac{x}{k} \right)\right] \, dx$

$
=1-\frac{\gamma}{2}-\lim_{n\to\infty} \sum_{k=1}^n \left[(k+1)\ln(k+1)-k\ln k -1 -\ln k -\frac{1}{2k} \right]
$

$
=1-\lim_{n\to\infty} \left[(n+1)\ln(n+1)-n -\ln n! - \frac{1}{2} \ln n\right]
$

$
=1-\lim_{n\to\infty} \ln \frac{(n+1)^{n+1}}{\sqrt{n}\,n!\,e^n}
$

$
=1-\lim_{n\to\infty} \ln \left(1+\frac{1}{n}\right)^n - \lim_{n\to\infty} \ln \frac{n^n(n+1)}{\sqrt{n}\,n!\,e^n}
$

$
=\ln \sqrt{2\pi}
$

Finally,
$
I(a)=a\ln a -a +\ln\sqrt{2\pi}
$

Relay Problem:
$
\int_{-\infty}^\infty x^{2n}e^{-ax^{2}}dx, a>0, n=1,2,3..
$

I have spent too much time and energy on this thread and now become exhausted, so I am going to quit for a while to deal with other things. Good luck!
• Aug 18th 2009, 05:45 PM
Random Variable
I've seen this one before.

Using $\int^{\infty}_{-\infty}e^{-x^{2}} dx = \sqrt{\pi}$

let $x^{2}=au^{2}$

then $2xdx=2audu$

$dx = a\frac{u}{x} du = a\frac{1}{\sqrt{a}} du = \sqrt{a} du$

so $\int^{\infty}_{-\infty}e^{-x^{2}} dx = \sqrt{a}\int^{\infty}_{-\infty}e^{-au^{2}} du = \sqrt{\pi}$

$\int^{\infty}_{-\infty}e^{-au^{2}} du = \sqrt{\frac{\pi}{a}}$

$(-1)^{n} \frac{\partial^{n}}{\partial a^{n}} \int^{\infty}_{-\infty}e^{-au^{2}} du = \int^{\infty}_{-\infty}u^{2n}e^{-au^{2}} du$ $= (-1)^{n} \frac{\partial^{n}}{\partial a^{n}} \sqrt{\frac{\pi}{a}}$

Can I leave my answer in that form?
• Aug 18th 2009, 06:28 PM
simplependulum
(Hi) Luobo , your solution to my problem is very cool

I want to share another method

To find $\int_0^1 \ln[\Gamma(x)]~dx$
we can consider this identity $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$

and take the log. integral $x \in (0,\frac{1}{2})$

for $\Gamma(1-x)$ sub $t = 1 -x$

then from L.H.S.

we can obtain $\int_0^1 \ln[\Gamma(t)]~dt$

and R.H.S $\int_0^{\frac{1}{2}}[ \ln(\pi) - \ln(\sin(\pi x)) ]~dx$

and finally $= \frac{1}{2} \ln(2\pi)$

we can change the integral in terms of gamma function:

$I = 2 \int_0^{\infty} x^{2n} e^{-ax^2}~dx$

Sub $x = \sqrt{ \frac{u}{a} }$
so

$I = \frac{1}{ a^{n+\frac{1}{2} } }\int_0^{\infty} u^{n-\frac{1}{2}} e^{-u} ~du$

$I = \frac{1}{ a^{n+\frac{1}{2}} } \Gamma( n + \frac{1}{2})$

From the Lengendre identity :

$\sqrt{\pi}\Gamma(2n) = 2^{2n-1} \Gamma(n) \Gamma(n+\frac{1}{2})$

Finally

$I = \frac{\sqrt{\pi} (2n-1)! }{ 2^{2n-1} (n-1)! a^{n+\frac{1}{2}}}$

or

$= \frac{\sqrt{\pi} (2n)! }{ 2^{2n} (n)! a^{n+\frac{1}{2}}}$
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