1. Originally Posted by simplependulum
Problem :

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

Is it related to INTEGRAL ? If you think it is not , i still have another problem
The interesting thing of the problem is

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

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

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

2. 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
$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ ...
$

converges. Show that the rearranged series
$
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

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

The harmonic series
$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ ...
$

converges. Show that the rearranged series
$
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"! so let's just keep it that way please.

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

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

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:

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

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

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

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

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

Have fune!

4. Originally Posted by NonCommAlg

1) $\int_0^1 (1-x)e^{-x} \ln x \ dx$
Rewrite the integral as $-\int_0^1\int_x^1\frac{(1-x)e^{-x}}t\,dt\,dx,$ after reverse integration order and by straightforward computations we'll get that the integral is $-\int_0^1e^{-t}\,dt=\frac1e-1.$

5. Originally Posted by NonCommAlg

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

----------

See below for full answer. I did this when I was in college, so no time to do anything else!

6. Originally Posted by Krizalid
Start with this simple fact: $\frac{\sin(\ln x)}{\ln x}=\int_0^1\cos(t\ln x)\,dt.$ Thus the integral becomes (after reversing integration order) $\int_0^1\int_0^1x(x+1)\cos(t\ln x)\,dx\,dt,$ so it remains to compute $\int_0^1\left(\frac3{t^2+9}+\frac2{t^2+4}\right)\, dt$ and its value is $\arctan\left(\frac13\right)+\arctan\left(\frac12\r ight)$ which is the answer of the integral.
and: $\tan^{-1}\left(\frac13\right)+\tan^{-1}\left(\frac12\right)=\tan^{-1} \left(\frac{\frac{1}{2}+\frac{1}{3}}{1 - \frac{1}{6}} \right) = \frac{\pi}{4}.$

ok, two down, three to go:

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

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

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

7. I attack $
I=\int_0^{\infty} \sin (x^a) \ dx, \ \ a > 1.
$

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

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

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

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

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

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

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

8. 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 $z$ (possibly complex) the following integral exists, and evaluate it for those $z$:

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

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

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

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

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

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

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

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

which requires Re $(z+2)>1$, i.e. Re $(z)>-1$ to exist.

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

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

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

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

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

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

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

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

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

4) $\int_0^{\pi} (\sin x)^a \sin(ax) \ dx, \ \ a > -1$
no.3 the answer is $\frac{\pi}{2}(\frac{1}{a}+\frac{1}{b})\ln(a+b) - \frac{{\pi}}{2}(\frac{\ln{b}}{a} + \frac{\ln{a}}{b})$

no.4 $\int_0^{\pi} e^{iax} (\sin(x))^a ~dx = \frac{\pi}{2^a}e^{i\frac{\pi}{2}a}$

12. Originally Posted by luobo
which is also
$
\frac{\Gamma\left(\frac{1}{a}\right)\sin\frac{\pi} {2a}}{a}$
When $a = 2$ , the integral becomes the Frensel Integral and the result is $\sqrt{\pi}/\sqrt{8}$

yes you are right , it is $\frac{\sqrt{\pi}}{\sqrt{8}}$

13. Originally Posted by simplependulum
When $a = 2$ , the integral becomes the Frensel Integral and the result is $\sqrt{\pi}/\sqrt{2}$
Fresnel integrals: $\frac{\sqrt{\pi}}{2\sqrt{2}}$

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

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

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

Ok, this problem is nice:

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

Note: Here by $f^r(x)$ we mean $(f(x))^r.$ The solution I know doesn't use any known integral inequalities. Instead, it uses a nice integration technique and it's fairly short!

15. Originally Posted by NonCommAlg
Problem: Suppose $f,g : [a,b] \longrightarrow (0, \infty)$ are continuous, $f$ is decreasing and $g$ is increasing. Prove that for all real numbers $c \geq d > 0: \ \ \frac{\int_a^b f^c(x) \ dx}{\int_a^b f^d(x) \ dx} \geq \frac{\int_a^b g(x)f^c(x) \ dx}{\int_a^b g(x) f^d(x) \ dx}.$
Since $c\geq d$ we know that $\tfrac{f^d(x)}{f^c(x)}$ must be non-decreasing - because f is non-increasing- (we can divide by $f$ since it is non-zero in our interval), thus we must have $
\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: $
\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: $
f^c \left( x \right) \cdot f^c \left( y\right) > 0
$

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

Page 4 of 7 First 1234567 Last