1. ## MHF Integral Bee!

This thread is for pros! 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 $\displaystyle I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{4x - \pi}{1 - \cot x} \ dx.$

2. Originally Posted by NonCommAlg
This thread is for pros! 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 $\displaystyle I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{4x - \pi}{1 - \cot x} \ dx.$
Hi, good morning.
Let $\displaystyle t=\frac{\pi}{4}-x$, then the integral becomes $\displaystyle I=2 \int_{0}^{\frac{\pi}{2}} (x\cot x-x)\, dx = 2 \int_{0}^{\frac{\pi}{2}} x\cot x\, dx - \frac{\pi^2}{4}=2 \int_{0}^{\frac{\pi}{2}} x\, d\ln(\sin x)- \frac{\pi^2}{4}$

So, $\displaystyle I= - 2 \int_{0}^{\frac{\pi}{2}} \ln(\sin x)\, dx- \frac{\pi^2}{4} = \pi \ln2- \frac{\pi^2}{4}$

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

4. 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! (see the rules of this thread again!)
Sorry. See below.

$\displaystyle I=\int_{0}^{1} \frac{x^\alpha-1}{\ln x} \, dx=?$ where $\displaystyle \alpha > -1$

5. $\displaystyle \frac{dI}{d\alpha}=\int_{0}^{1}\frac{x^{\alpha}\ln x}{\ln x}dx=\frac{1}{\alpha+1}$

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

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

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

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

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

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

7. Originally Posted by luobo

$\displaystyle \int_{0}^{\infty} \frac{x^3}{e^x-1} \, dx$
let $\displaystyle I(a)=\int_0^{\infty} \frac{x^a}{e^x - 1} \ dx, \ a > 0.$ we have:

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

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

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

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

8. Originally Posted by NonCommAlg

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

9. 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 $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and suppose that $\displaystyle f(x+\pi)=-f(x), \ f(x)=f(-x), \ \forall x \in \mathbb{R}.$ Assuming that $\displaystyle I=\int_0^{\infty} f(x) \frac{\sin x}{x} \ dx$ exists, prove that $\displaystyle I= \int_0^{\frac{\pi}{2}} f(x) \cos x \ dx.$
$\displaystyle \int_{0}^{\infty} f(x)\frac{\sin x}{x}\,dx=$

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

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

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

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

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

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

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

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

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

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:

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

Note: $\displaystyle f(x)=\frac{\sin(\tan x)}{\sin x}$ is not continuous at $\displaystyle x=k\pi+\frac{\pi}{2}$.

10. If $\displaystyle f(x)$ is continous for $\displaystyle x \in [0,\frac{\pi}{2})$ , I think it may also satisify the identity .. because we cut the integral into many pieces $\displaystyle (0,\frac{\pi}{2}) , (\frac{\pi}{2}, \pi) ....$ and thus each improper integrals may exist .

then

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

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

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

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

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

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

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

Finally,

$\displaystyle = \frac{\pi}{2}(1- e^{-1})$

11. luobo, the continuity condition for $\displaystyle f$ is unnecessary. i removed it from the problem. your solution is perfect!

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

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

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

$\displaystyle =\frac{\pi}{2} + \left[\frac{d}{d \alpha} \left(\frac{\pi}{2} e^{-\alpha} \right) \right]_{\alpha=1}=\frac{\pi}{2}(1 - e^{-1}).$

12. Here is the problem :

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

13. Originally Posted by simplependulum
Here is the problem :

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

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

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

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

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

$\displaystyle =1-\lim_{n\to\infty} \ln \left(1+\frac{1}{n}\right)^n - \lim_{n\to\infty} \ln \frac{n^n(n+1)}{\sqrt{n}\,n!\,e^n}$
$\displaystyle =\ln \sqrt{2\pi}$

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

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

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!

14. I've seen this one before.

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

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

then $\displaystyle 2xdx=2audu$

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

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

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

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

Can I leave my answer in that form?

15. Luobo , your solution to my problem is very cool

I want to share another method

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

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

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

then from L.H.S.

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

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

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

we can change the integral in terms of gamma function:

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

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

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

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

From the Lengendre identity :

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

Finally

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

or

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

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