# MHF Integral Bee!

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• Aug 21st 2009, 02:00 AM
simplependulum
Quote:

Originally Posted by simplependulum
Problem :

$\displaystyle L^{-1} [ \sqrt{s + \sqrt{s^2 + a^2}} ]$

Is it related to INTEGRAL ? If you think it is not , i still have another problem (Happy)

The interesting thing of the problem is

$\displaystyle \sqrt{s + \sqrt{s^2 + a^2}} = \frac{1}{2} \sqrt{ 2s + 2\sqrt{(s-ai)(s+ai)}}$

$\displaystyle = \frac{1}{2} \sqrt{ s + ai + 2\sqrt{s+ai}\sqrt{s-ai} + s - ai}$

$\displaystyle = \frac{1}{2} ( \sqrt{ s+ai} + \sqrt{s-ai})$
(Happy)
• Aug 21st 2009, 03:57 AM
luobo
Quote:

Originally Posted by Amer
What is the new integral ?? luobo you should post an integral .

The problem is copied from the internet.

The harmonic series
$\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ ...$
converges. Show that the rearranged series
$\displaystyle 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$
also converges and find its limit
• Aug 21st 2009, 04:39 AM
NonCommAlg
Quote:

Originally Posted by luobo
The problem is copied from the internet.

The harmonic series
$\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ ...$
converges. Show that the rearranged series
$\displaystyle 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$
also converges and find its limit

well, i don't know, this might be related to "integration" somehow but this thread is supposed to be about "techniques of integration"! (Nod) so let's just keep it that way please.

i don't want this thread to go in a different direction. cheers! (Smile)

.................................................. .......................................

I'm going to bend the rules a little bit and give you a few integrals instead of one. Pick whichever you like and share your solution with us if you feel like it: (Clapping)

1) $\displaystyle \int_0^1 (1-x)e^{-x} \ln x \ dx$

2) $\displaystyle \int_0^1 \frac{x(x+1)\sin (\ln x)}{\ln x} \ dx$

3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$

4) $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$

5) $\displaystyle \int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.$ (the answer to this one is in terms of Gamma function!)

Have fune!
• Aug 21st 2009, 10:22 AM
Krizalid
Quote:

Originally Posted by NonCommAlg

1) $\displaystyle \int_0^1 (1-x)e^{-x} \ln x \ dx$

Rewrite the integral as $\displaystyle -\int_0^1\int_x^1\frac{(1-x)e^{-x}}t\,dt\,dx,$ after reverse integration order and by straightforward computations we'll get that the integral is $\displaystyle -\int_0^1e^{-t}\,dt=\frac1e-1.$
• Aug 21st 2009, 10:36 AM
Krizalid
Quote:

Originally Posted by NonCommAlg

2) $\displaystyle \int_0^1 \frac{x(x+1)\sin (\ln x)}{\ln x} \ dx$

Start with this simple fact: $\displaystyle \frac{\sin(\ln x)}{\ln x}=\int_0^1\cos(t\ln x)\,dt.$ Thus the integral becomes (after reversing integration order) $\displaystyle \int_0^1\int_0^1x(x+1)\cos(t\ln x)\,dx\,dt,$ so it remains to compute $\displaystyle \int_0^1\left(\frac3{t^2+9}+\frac2{t^2+4}\right)\, dt$ and its value is $\displaystyle \arctan\left(\frac13\right)+\arctan\left(\frac12\r ight)$ which is the answer of the integral. :D

----------

See below for full answer. I did this when I was in college, so no time to do anything else!
• Aug 21st 2009, 01:17 PM
NonCommAlg
Quote:

Originally Posted by Krizalid
Start with this simple fact: $\displaystyle \frac{\sin(\ln x)}{\ln x}=\int_0^1\cos(t\ln x)\,dt.$ Thus the integral becomes (after reversing integration order) $\displaystyle \int_0^1\int_0^1x(x+1)\cos(t\ln x)\,dx\,dt,$ so it remains to compute $\displaystyle \int_0^1\left(\frac3{t^2+9}+\frac2{t^2+4}\right)\, dt$ and its value is $\displaystyle \arctan\left(\frac13\right)+\arctan\left(\frac12\r ight)$ which is the answer of the integral. :D

and: $\displaystyle \tan^{-1}\left(\frac13\right)+\tan^{-1}\left(\frac12\right)=\tan^{-1} \left(\frac{\frac{1}{2}+\frac{1}{3}}{1 - \frac{1}{6}} \right) = \frac{\pi}{4}.$ (Wink)

ok, two down, three to go:

Quote:

3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$

4) $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$

5) $\displaystyle \int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.$

• Aug 21st 2009, 04:14 PM
Bruno J.
I attack $\displaystyle I=\int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.$

First we substitute $\displaystyle x^a=t$. Then $\displaystyle ax^{a-1}dx=dt$, i.e. $\displaystyle dx = \frac{dt}{ax^{a-1}}=\frac{1}{a}{t^\frac{1-a}{a}}\: dt$.

Now $\displaystyle \int_0^{\infty} \sin (x^a) \ dx = \frac{1}{a}\int_0^{\infty} \sin (t){t^\frac{1-a}{a}}\: dt = \frac{1}{2ia}\int_0^{\infty} (e^{it}-e^{-it}){t^\frac{1-a}{a}}\: dt$.

Now consider $\displaystyle I_1=\frac{1}{2ia}\int_0^{\infty} e^{it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=-u$, so $\displaystyle dt = i\: du$. Then the integral becomes

$\displaystyle \frac{1}{2ia}\int_0^{\infty} e^{-u}(iu)^\frac{1-a}{a}\: i\: du = \frac{i^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{i^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right)$.

Similarily, consider $\displaystyle I_2=\frac{-1}{2ia}\int_0^{\infty} e^{-it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=u$, so $\displaystyle dt = -i\: du$. Then the integral becomes

$\displaystyle \frac{-1}{2ia}\int_0^{\infty} e^{-u}(-iu)^\frac{1-a}{a}\: (-i\: du) = \frac{(-i)^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{(-i)^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right) = \overline{I_1}$.

So finally $\displaystyle I=I_1+I_2=I_1+\overline{I_1} = 2\Re(I_1)=\frac{\Gamma\left(\frac{1}{a}\right)\Re( i^{\frac{1-a}{a}})}{a}=\frac{\Gamma\left(\frac{1}{a}\right)\c os(\frac{\pi}{2}\frac{1-a}{a})}{a}=\frac{\Gamma\left(\frac{1}{a}\right)\si n(\frac{\pi}{2a})}{a}$
• Aug 21st 2009, 04:22 PM
Bruno J.
Here is my integral! I posted a particular case of this one not very long ago but here it is anyways because I think it's very nice.

Determine for which values of $\displaystyle z$ (possibly complex) the following integral exists, and evaluate it for those $\displaystyle z$:

$\displaystyle I(z)=\int_0^1\frac{\log(x)^z\log(1-x)}{x}\ dx$
• Aug 21st 2009, 05:50 PM
luobo
Quote:

Originally Posted by Bruno J.
Here is my integral! I posted a particular case of this one not very long ago but here it is anyways because I think it's very nice.

Determine for which values of $\displaystyle z$ (possibly complex) the following integral exists, and evaluate it for those $\displaystyle z$:
$\displaystyle I(z)=\int_0^1\frac{\log(x)^z\log(1-x)}{x}\ dx$

$\displaystyle I(z)=\int_0^1\frac{\log(x)^z\log(1-x)}{x}\ dx=$

$\displaystyle \int_0^\infty (-u)^z\ln(1-e^{-u})\;du=$ (Note: Let $\displaystyle x=e^{-u}$)

$\displaystyle (-1)^{z+1}\int_0^\infty u^z\sum_{n=1}^{\infty} \frac{e^{-nu}}{n} \; du=$

$\displaystyle (-1)^{z+1}\sum_{n=1}^{\infty} \frac{1}{n} \int_0^\infty u^z e^{-nu} \; du=$

$\displaystyle (-1)^{z+1}\sum_{n=1}^{\infty} \frac{1}{n^{z+2}} \int_0^\infty t^z e^{-t} \; dt=$ (Note: Let $\displaystyle t=nu$)

$\displaystyle (-1)^{z+1}\Gamma(z+1)\sum_{n=1}^{\infty} \frac{1}{n^{z+2}}$

$\displaystyle (-1)^{z+1}\Gamma(z+1)\;\zeta(z+2)$

which requires Re$\displaystyle (z+2)>1$, i.e. Re$\displaystyle (z)>-1$ to exist.
• Aug 21st 2009, 06:33 PM
luobo
Quote:

Originally Posted by Bruno J.
I attack $\displaystyle I=\int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.$

First we substitute $\displaystyle x^a=t$. Then $\displaystyle ax^{a-1}dx=dt$, i.e. $\displaystyle dx = \frac{dt}{ax^{a-1}}=\frac{1}{a}{t^\frac{1-a}{a}}\: dt$.

Now $\displaystyle \int_0^{\infty} \sin (x^a) \ dx = \frac{1}{a}\int_0^{\infty} \sin (t){t^\frac{1-a}{a}}\: dt = \frac{1}{2ia}\int_0^{\infty} (e^{it}-e^{-it}){t^\frac{1-a}{a}}\: dt$.

Now consider $\displaystyle I_1=\frac{1}{2ia}\int_0^{\infty} e^{it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=-u$, so $\displaystyle dt = i\: du$. Then the integral becomes

$\displaystyle \frac{1}{2ia}\int_0^{\infty} e^{-u}(iu)^\frac{1-a}{a}\: i\: du = \frac{i^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{i^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right)$.

Similarily, consider $\displaystyle I_2=\frac{-1}{2ia}\int_0^{\infty} e^{-it}t^\frac{1-a}{a}\: dt$. Set $\displaystyle it=u$, so $\displaystyle dt = -i\: du$. Then the integral becomes

$\displaystyle \frac{-1}{2ia}\int_0^{\infty} e^{-u}(-iu)^\frac{1-a}{a}\: (-i\: du) = \frac{(-i)^{\frac{1-a}{a}}}{2a}\int_0^{\infty} e^{-u}u^{\frac{1}{a}-1}\: du = \frac{(-i)^{\frac{1-a}{a}}}{2a}\Gamma\left(\frac{1}{a}\right) = \overline{I_1}$.

So finally $\displaystyle I=I_1+I_2=I_1+\overline{I_1} = 2\Re(I_1)=\frac{\Gamma\left(\frac{1}{a}\right)\Re( i^{\frac{1-a}{a}})}{a}=\frac{\Gamma\left(\frac{1}{a}\right)\c os(\frac{\pi}{2}\frac{1-a}{a})}{a}$

which is also
$\displaystyle \frac{\Gamma\left(\frac{1}{a}\right)\sin\frac{\pi} {2a}}{a}$ (Happy)
• Aug 21st 2009, 07:01 PM
simplependulum
Quote:

Originally Posted by NonCommAlg
3) $\displaystyle \int_0^{\infty}\cot^{-1}(ax) \cot^{-1}(bx) \ dx, \ \ a > 0, \ b > 0$

4) $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$

no.3 the answer is $\displaystyle \frac{\pi}{2}(\frac{1}{a}+\frac{1}{b})\ln(a+b) - \frac{{\pi}}{2}(\frac{\ln{b}}{a} + \frac{\ln{a}}{b})$

no.4 $\displaystyle \int_0^{\pi} e^{iax} (\sin(x))^a ~dx = \frac{\pi}{2^a}e^{i\frac{\pi}{2}a}$
• Aug 21st 2009, 07:21 PM
simplependulum
Quote:

Originally Posted by luobo
which is also
$\displaystyle \frac{\Gamma\left(\frac{1}{a}\right)\sin\frac{\pi} {2a}}{a}$ (Happy)

When $\displaystyle a = 2$ , the integral becomes the Frensel Integral and the result is $\displaystyle \sqrt{\pi}/\sqrt{8}$ (Happy)

yes you are right , it is $\displaystyle \frac{\sqrt{\pi}}{\sqrt{8}}$
• Aug 21st 2009, 07:26 PM
luobo
Quote:

Originally Posted by simplependulum
When $\displaystyle a = 2$ , the integral becomes the Frensel Integral and the result is $\displaystyle \sqrt{\pi}/\sqrt{2}$ (Happy)

Fresnel integrals: $\displaystyle \frac{\sqrt{\pi}}{2\sqrt{2}}$
• Aug 22nd 2009, 12:56 AM
NonCommAlg
Quote:

Originally Posted by simplependulum
no.3 the answer is $\displaystyle \frac{\pi}{2}(\frac{1}{a}+\frac{1}{b})\ln(a+b) - \frac{{\pi}}{2}(\frac{\ln{b}}{a} + \frac{\ln{a}}{b})$

correct!

Quote:

no.4 $\displaystyle \int_0^{\pi} e^{iax} (\sin(x))^a ~dx = \frac{\pi}{2^a}e^{i\frac{\pi}{2}a}$
and that gives us: $\displaystyle \int_0^{\pi} (\sin x)^a \sin(ax) \ dx = \frac{\pi}{2^a} \sin \left(\frac{\pi a}{2} \right),$ which is the correct answer. i hope you'll also post your solutions! (Clapping)

---------------------------------------------

Ok, this problem is nice:

Problem: Suppose $\displaystyle f,g : [a,b] \longrightarrow (0, \infty)$ are continuous, $\displaystyle f$ is decreasing and $\displaystyle g$ is increasing. Prove that for all real numbers $\displaystyle c \geq d > 0: \ \ \frac{\int_a^b f^c(x) \ dx}{\int_a^b f^d(x) \ dx} \geq \frac{\int_a^b g(x)f^c(x) \ dx}{\int_a^b g(x) f^d(x) \ dx}.$

Note: Here by $\displaystyle f^r(x)$ we mean $\displaystyle (f(x))^r.$ The solution I know doesn't use any known integral inequalities. Instead, it uses a nice integration technique and it's fairly short! (Evilgrin)
• Aug 22nd 2009, 08:12 AM
PaulRS
Quote:

Originally Posted by NonCommAlg
Problem: Suppose $\displaystyle f,g : [a,b] \longrightarrow (0, \infty)$ are continuous, $\displaystyle f$ is decreasing and $\displaystyle g$ is increasing. Prove that for all real numbers $\displaystyle c \geq d > 0: \ \ \frac{\int_a^b f^c(x) \ dx}{\int_a^b f^d(x) \ dx} \geq \frac{\int_a^b g(x)f^c(x) \ dx}{\int_a^b g(x) f^d(x) \ dx}.$

Since $\displaystyle c\geq d$ we know that $\displaystyle \tfrac{f^d(x)}{f^c(x)}$ must be non-decreasing - because f is non-increasing- (we can divide by $\displaystyle f$ since it is non-zero in our interval), thus we must have $\displaystyle \left( {\tfrac{{f^d \left( x \right)}} {{f^c \left( x \right)}} - \tfrac{{f^d \left( y \right)}} {{f^c \left( y \right)}}} \right) \cdot \left( {g\left( x \right) - g\left( y \right)} \right) \geqslant 0$

Expand: $\displaystyle \tfrac{{f^d \left( x \right)}} {{f^c \left( x \right)}} \cdot g\left( x \right) + \tfrac{{f^d \left( y \right)}} {{f^c \left( y \right)}} \cdot g\left( y \right) \geqslant \tfrac{{f^d \left( x \right)}} {{f^c \left( x \right)}} \cdot g\left( y \right) + \tfrac{{f^d \left( y \right)}} {{f^c \left( y \right)}} \cdot g\left( x \right)$ the inequality will be preserved if we multiply by: $\displaystyle f^c \left( x \right) \cdot f^c \left( y\right) > 0$

Thus: $\displaystyle f^d \left( x \right) \cdot g\left( x \right) \cdot f^c \left( y \right) + f^d \left( y \right) \cdot g\left( y \right) \cdot f^c \left( x \right) \geqslant f^c \left( y \right) \cdot g\left( y \right) \cdot f^d \left( x \right) + f^c \left( x \right) \cdot g\left( x \right) \cdot f^d \left( y \right)$

Now integrate in $\displaystyle \left[ {a,b} \right] \times \left[ {a,b} \right]$ and the inequality follows since $\displaystyle g(x),f(x)>0$ in $\displaystyle \left[ {a,b} \right]$ -and so the integrals are positive, and we can divide-
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