1. Originally Posted by PaulRS
Show that $\displaystyle \int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq rt{2}+1}{\sqrt{2}-1}\right)}$

( I don't have time to post the solution now. It's a nice integral, I like the result)
OK, as promised.

Lemma 1: $\displaystyle \int_0^{\tfrac{\pi } {2}} {\sin ^{2n} \left( x \right)dx} = \frac{\pi}{2^{2n+1}}\cdot{\binom{2n}{n}}$

See here

Lemma 2: $\displaystyle \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot x^n } = \tfrac{1} {{\sqrt {1 - 4x} }}$ whenever $\displaystyle \left| x \right| < \tfrac{1} {4}$

See here

Computation: $\displaystyle \int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq rt{2}+1}{\sqrt{2}-1}\right)}$

First we have: $\displaystyle \tfrac{{\arctan \left[ {\sin \left( x \right)} \right]}} {{\sin \left( x \right)}} = 1 - \tfrac{{\sin ^2 \left( x \right)}} {3} + \tfrac{{\sin ^4 \left( x \right)}} {5} \mp ...$ for all $\displaystyle \left| x \right| \leqslant \tfrac{\pi } {2}$ (just use the Taylor Series of $\displaystyle \arctan(x)$ )

Assuming uniform convergence : $\displaystyle \int_0^{\tfrac{\pi } {2}} {\tfrac{{\arctan \left[ {\sin \left( x \right)} \right]}} {{\sin \left( x \right)}}dx} = \int_0^{\tfrac{\pi } {2}} {dx} - \tfrac{{\int_0^{\tfrac{\pi } {2}} {\sin ^2 \left( x \right)dx} }} {3} + \tfrac{{\int_0^{\tfrac{\pi } {2}} {\sin ^4 \left( x \right)dx} }} {5} \mp ...$

By Lemma 1: $\displaystyle \int_0^{\tfrac{\pi } {2}} {\tfrac{{\arctan \left[ {\sin \left( x \right)} \right]}} {{\sin \left( x \right)}}dx} = \pi \cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot \tfrac{{\left( { - 1} \right)^n }} {{2^{2n+1} \cdot \left( {2n + 1} \right)}}}$

By Lemma 2 : $\displaystyle \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot \frac{(-1)^n\cdot x^{2n}}{2^{2n}} } = \tfrac{1} {{\sqrt {1 + x^2} }}$

Integrate from $\displaystyle 0$ to $\displaystyle 1$: $\displaystyle 2\cdot \sum\limits_{n = 0}^\infty {\binom{2n}{n} \cdot \tfrac{{\left( { - 1} \right)^n }} {{2^{2n+1} \cdot \left( {2n + 1} \right)}}} = \int_0^{1}\tfrac{dx} {{\sqrt {1 + x^2} }}= \left. {\text{arcsinh} \left( x \right)} \right|_0^1 = \tfrac{1} {2} \cdot \ln \left( {\tfrac{{\sqrt 2 + 1}} {{\sqrt 2 - 1}}} \right)$

Which yields: $\displaystyle \int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq rt{2}+1}{\sqrt{2}-1}\right)}$

2. $\displaystyle \int\frac{e^{-ax}(1-\cos(px))}{x^2}~dx$

$\displaystyle =\int_0^{\infty}\int_0^{\infty}ye^{-x(a+y)}(1-\cos(px))~dy~dx$

$\displaystyle =\int_0^{\infty}\int_0^{\infty}ye^{-x(a+y)}(1-\cos(px))~dx~dy$

Now

$\displaystyle \int_0^{\infty}ye^{-x(y+a)}(1-\cos(px)~dx$

$\displaystyle =y\bigg[\left(\frac{\cos(px)(y+a)}{(y+a)^2+p^2}-\frac{p\sin(px)}{(y+a)^2+p^2}+\frac{1}{y+a}\right) e^{-x(y+a)}\bigg]\bigg|_0^{\infty}$

$\displaystyle =\frac{-a}{y+a}+\frac{ay+a^2+p^2}{(y+a)^2+p^2}$

So we now need to evaluate

$\displaystyle \int_0^{\infty}\bigg[\frac{-a}{y+a}+\frac{ay+a^2+p^2}{(y+a)^2+p^2}\bigg]~dy$

$\displaystyle =\int_0^{\infty}\bigg[\frac{-a}{y+a}+\frac{ay}{(y+a)^2+p^2}+\frac{a^2+p^2}{(y+a )^2+p^2}\bigg]~dy$

The first term is obviously $\displaystyle a\ln|a+y|$

Now let us examine the other terms

$\displaystyle \int\frac{ay}{(y+a)^2+p^2}~dy$

This can easily be gotten by letting

$\displaystyle z=y+a$

$\displaystyle \Rightarrow{dz=dy}$

So we have

$\displaystyle \int\frac{a(z-a)}{z^2+p^2}$

$\displaystyle =\int\bigg[a\frac{z}{z^2+p^2}-a^2\int\frac{1}{z^2+p^2}\bigg]~dz$

$\displaystyle =\frac{a}{2}\ln(z^2+p^2)-\frac{a^2}{p}\arctan\left(\frac{z}{p}\right)+C$

$\displaystyle \underbrace{=}_{\text{backsub}}\frac{a}{2}\ln\left ((y+a)^2+p^2\right)-\frac{a^2}{p}\arctan\left(\frac{y+a}{p}\right)+C$

And

$\displaystyle \int\frac{a^2+p^2}{(y+a)^2+p^2}~dy$

Similarly with a substitution of $\displaystyle z=y+a$ we get

$\displaystyle (a^2+p^2)\int\frac{dz}{z^2+p^2}$

$\displaystyle =\frac{a^2+p^2}{p}\arctan\left(\frac{z}{p}\right)+ C$

$\displaystyle \underbrace{=}_{\text{backsub}}=\frac{a^2+p^2}{p}\ arctan\left(\frac{y+a}{p}\right)+C$

So now putting this all back together we get

$\displaystyle \int\bigg[\frac{-a}{y+a}+\frac{ay}{(y+a)^2+p^2}-\frac{a^2+p^2}{(y+a)^2+p^2}\bigg]~dy$

$\displaystyle =-a\ln|a+y|+\frac{a}{2}\ln\left((y+a)^2+p^2\right)-\frac{a^2}{p}\arctan\left(\frac{y+a}{p}\right)$$\displaystyle +\frac{a^2+p^2}{p}\arctan\left(\frac{y+a}{p}\right )+C$

$\displaystyle =-a\ln|a+y|+\frac{a}{2}\ln\left((y+a)^2+p^2\right)+p \arctan\left(\frac{y+a}{p}\right)+C$

$\displaystyle =F(y)$

So seeing that

$\displaystyle \int_0^{\infty}\bigg[\frac{a}{y+a}-\frac{ay+a^2+p^2}{(y+a)^2+p^2}\bigg]~dy=F\left(\infty\right)-F(0)$

We need to find

$\displaystyle F(\infty)$ and $\displaystyle F(0)$

So we see that

$\displaystyle F(0)=\frac{a}{2}\ln\left(a^2+p^2\right)+p\arctan\l eft(\frac{a}{p}\right)-a\ln|a|$

and

$\displaystyle F(\infty)=\frac{p{\pi}}{2}$

We put this together to get

$\displaystyle F\left(\infty\right)-F(0)=\frac{-a}{2}\ln(a^2+p^2)-p\arctan\left(\frac{a}{p}\right)+a\ln|a|+\frac{p\p i}{2}$

$\displaystyle =p\,{\rm{arccot}}\left(\frac{a}{p}\right)+a\ln\lef t(\frac{|a|}{\sqrt{a^2+p^2}}\right)$

We can conclude that

$\displaystyle \boxed{\int_0^{\infty}\frac{e^{-ax}(1-\cos(px))}{x^2}~dx=p\,{\rm{arccot}}\left(\frac{a}{ p}\right)+a\ln\left(\frac{|a|}{\sqrt{a^2+p^2}}\rig ht)}\quad\blacksquare$

3. Originally Posted by PaulRS
OK, as promised.
1)That is a really nice integral.

2)Did you start using "\tfrac" because of me, it appears you are more obsessed with making you fractions smaller.

3)How do you prove uniform convergence? Weierstrass test would not work here.

4. Hi! Here are 2 improper integrals for you!

1:

Int[ Abs[ln[Abs[x]]]*a/(Abs[x])^3 * Exp(-a/x^2) ] dx from -oo to +oo

2:

Int[ a(Abs[x])^(i-3) * Exp[-a/x^2) ] dx from -oo to +oo

i<2 ; a>0 for both integrals, and 'i' is NOT the imaginary unit

Abs[x] - is the absolute value of x

I would be thankful, if s.o. could type both integrals with TeX, so that I can check if everything's like it should be

5. Hi! Here are 2 improper integrals for you!
I will attempt to LaTex your integrals. I hope I am interpreting correctly.

Int[ Abs[ln[Abs[x]]]*a/(Abs[x])^3 * Exp(-a/x^2) ] dx from -oo to +oo
$\displaystyle \int_{-\infty}^{\infty}\frac{a|ln|x||e^{\frac{-a}{x^{2}}}}{(|x|)^{3}}$

Int[ a(Abs[x])^(i-3) * Exp[-a/x^2) ] dx from -oo to +oo, i<2 ; a>0 for both integrals, and 'i' is NOT the imaginary unit
I will use k instead of i.

$\displaystyle \int_{-\infty}^{\infty}a(|x|)^{k-3}\cdot e^{\frac{-a}{x^{2}}}, \;\ k<2, \;\ a>0$

6. The function under the integral sign is even. Then we can write :

Since x is positive, we can write :
$\displaystyle I=2 \int_0^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx$

$\displaystyle I=2 \int_0^1 \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx$

Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :

$\displaystyle I=-2 \int_0^1 \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx$

Substitute $\displaystyle u=\frac{1}{x^2} \implies dx=-\frac 12 x^3 \ du$

x=sqrt(u) --> ln(x)=ln(u)/2

The integral is now (after some simplifcations) :
$\displaystyle I=- \int_1^{\infty} a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du+ \int_0^1 a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du$

And now, I think an integration by parts can be done, but there's something worrying me

7. Originally Posted by Moo
The function under the integral sign is even. Then we can write :

Since x is positive, we can write :
$\displaystyle I=2 \int_0^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx$

$\displaystyle I=2 \int_0^1 \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx$

Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :

$\displaystyle I=-2 \int_0^1 \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx$

Substitute $\displaystyle u=\frac{1}{x^2} \implies dx=-\frac 12 x^3 \ du$

x=sqrt(u) --> ln(x)=ln(u)/2

The integral is now (after some simplifcations) :
$\displaystyle I=- \int_1^{\infty} a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du+ \int_0^1 a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du$

And now, I think an integration by parts can be done, but there's something worrying me
I haven't checked it, but assuming your work up to the last step is corect you cannot do integration by parts because natural log is not differentiably redundant. But what you can do is convert e^{-ax} to power series. Assume uniform convergence and switch integral and summation, do integratioy by parts on [math ]\int_0^1x^n\ln(x)~dx[/tex] and then go from there.

You got it?

8. I see you're having lot's of fun with them

To be honest, I have an antiderivative for the first integral, nut this for the second one has series in it

have fun, Marine

9. Originally Posted by Marine
I see you're having lot's of fun with them

To be honest, I have an antiderivative for the first integral, nut this for the second one has series in it

have fun, Marine
THe one Moo did (the first one?) does not have an antiderivative composed entirely of elementary functions. Also the answer is -pi/4

10. Originally Posted by Moo
The function under the integral sign is even. Then we can write :

Since x is positive, we can write :
$\displaystyle I=2 \int_0^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx$

$\displaystyle I=2 \int_0^1 \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a |\ln(x)| e^{\frac{-a}{x^2}}}{x^3} \ dx$

Since ln(x)<0 for x between 0 & 1 and >0 for x>1 :

$\displaystyle I=-2 \int_0^1 \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx+2 \int_1^{\infty} \frac{a \ln(x) e^{\frac{-a}{x^2}}}{x^3} \ dx$

Substitute $\displaystyle u=\frac{1}{x^2} \implies dx=-\frac 12 x^3 \ du$

x=sqrt(u) --> ln(x)=ln(u)/2

The integral is now (after some simplifcations) :
$\displaystyle I=- \int_1^{\infty} a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du+ \int_0^1 a \cdot \frac{\ln(u)}{2} \cdot e^{-au} \ du$

And now, I think an integration by parts can be done, but there's something worrying me
Lets look at the first integral. If we let

then we see that So we get

Now assuming uniform convergence

Now for the second integral once again letting

We get

So putting it all together we get

I don't care enough to simplify, its too late. I am not sure that is right, its hard to switch back and forth from that Moo website to MHF and not make clarical errors. And this is all assuming Moo's integral was teh correct one.

By the way in the step before the solution I didn't type a negative in front of the serise and there was no way i was going back to change it, its fixed in the final solution. And gamma should be negative

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

Not seen in this thread, maybe in another one...?

*code*

Substitute ~ x=\tan \varphi \\
I=\int_0^{\frac \pi 4} \ln \left(1+\tan \varphi\right) ~d \varphi \\
I=\int_0^{\frac \pi 4} \ln \left(\cos \varphi+\sin \varphi\right)- \ln(\cos \varphi) ~d \varphi \\ \\

Let~J=\int_0^{\frac \pi 4} \ln(\cos \varphi+\sin \varphi)~d\varphi \\
We~can~observe~that~ \sin \left(x+\frac \pi 4 \right)=\frac{1}{\sqrt{2}}(\cos(x)+\sin(x)) \\
Thus~\cos \varphi+\sin \varphi=\sqrt{2} \sin \left(\varphi+\frac \pi 4\right) \\ \\

J=\int_0^{\frac \pi 4} \ln(\sqrt{2})+\ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
J=\ln(2) \frac{ \pi }{ 8 } + \int_0^{\frac \pi 4} \ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
Let~\phi=\frac \pi 4 - \varphi \Rightarrow \varphi+\frac \pi 4=\frac \pi 2-\phi \\
J=\ln(2) \frac \pi 8 -\int_{\frac \pi 4}^0 \ln \left(\sin \left(\frac \pi 2-\phi\right)\right)~d \phi \\ \\
J=\ln(2) \frac \pi 8+\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi \\ \\ \\

Therefore \ I=J-\int_0^{\frac \pi 4} \ln(\cos \varphi) \ d \varphi \\
I=\ln(2) \frac \pi 8+\underbrace{\int_0^{\frac \pi 4} \ln(\cos \phi) \ d\phi-\int_0^{\frac \pi 4} \ln(\cos \varphi) \ d\varphi}_{=0} \\ \\
I=\ln(2){\frac \pi 8}

12. Originally Posted by Moo

Not seen in this thread, maybe in another one...?

*picture of the answer* (if you want to do it without seeing it)

*code*

Substitute ~ x=\tan \varphi \\
I=\int_0^{\frac \pi 4} \ln \left(1+\tan \varphi\right) ~d \varphi \\
I=\int_0^{\frac \pi 4} \ln \left(\cos \varphi+\sin \varphi\right)- \ln(\cos \varphi) ~d \varphi \\ \\
Let~J=\int_0^{\frac \pi 4} \ln(\cos \varphi+\sin \varphi)~d\varphi \\
We~can~observe~that~ \sin \left(x+\frac \pi 4 \right)=\frac{1}{\sqrt{2}}(\cos(x)+\sin(x)) \\
Thus~\cos \varphi+\sin \varphi=\sqrt{2} \sin \left(\varphi+\frac \pi 4\right) \\ \\
J=\int_0^{\frac \pi 4} \ln(\sqrt{2})+\ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
J=\ln(2) \frac{ \pi }{ 8 } + \int_0^{\frac \pi 4} \ln \left(\sin \left(\varphi+\frac \pi 4\right)\right)~d\varphi \\
Let~\phi=\frac \pi 4 - \varphi \Rightarrow \varphi+\frac \pi 4=\frac \pi 2-\phi \\
J=\ln(2) \frac \pi 8 -\int_{\frac \pi 4}^0 \ln \left(\sin \left(\frac \pi 2-\phi\right)\right)~d \phi \\ \\
J=\ln(2) \frac \pi 8+\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi \\ \\ \\
Therefore I=J-\int_0^{\frac \pi 4} \ln(\cos \varphi) ~d \varphi \\
I=\ln(2) \frac \pi 8+\underbrace{\int_0^{\frac \pi 4} \ln(\cos \phi) ~d\phi-\int_0^{\frac \pi 4} \ln(\cos \varphi) ~d\varphi}_{=0} \\ \\
I=\ln(2){\frac \pi 8}
This is the integral I solved with the "coo" identity.

My soultion is very simiar to yours except my trig identity was slightly different

13. Ok I have a lot to do...lets start off nice and easy

Number one

$\displaystyle \int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\alpha}(x) }$

This one stumped me for a while unti you see something obvious

Let

$\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\alpha}( x)}$

$\displaystyle \underbrace{=}_{x\mapsto{\frac{\pi}{2}-z}}-\int_{\frac{\pi}{2}}^0\frac{dz}{1+\left(\frac{1}{\ tan(z)}\right)^{\alpha}}$

$\displaystyle =\int_0^{\frac{\pi}{2}}\frac{\tan^{\alpha}(z)}{\ta n^{\alpha}(z)+1}$

$\displaystyle =I$

$\displaystyle \Rightarrow{2}I=\int_0^{\frac{\pi}{2}}\frac{\tan^{ \alpha}(x)}{1+\tan^{\alpha}(x)}dx+\int_0^{\frac{\p i}{2}}\frac{dx}{1+\tan^{\alpha}(x)}$

$\displaystyle =\int_0^{\frac{\pi}{2}}dx$

$\displaystyle =\frac{\pi}{2}$

$\displaystyle 2I=\frac{\pi}{2}$

$\displaystyle \Rightarrow{I=\frac{\pi}{4}}$

$\displaystyle \therefore \boxed{\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{\al pha}(x)}=\frac{\pi}{4}}\quad\blacksquare$

14. This one is just plain easy but requires you to take something as true

$\displaystyle \int_0^{1}\frac{\arctan(x)}{x}~dx$

Now we know that

$\displaystyle \forall{x}\in[0,1]\arctan(x)\to\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}~~\text{uniformly}$

$\displaystyle \therefore\int_0^1\frac{\arctan(x)}{x}~dx$

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

Which by virtue of uniform convergence

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

$\displaystyle =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$

$\displaystyle =K$

Where $\displaystyle K$ is Chatalan's constant $\displaystyle \approx{.9159}$

$\displaystyle \therefore\quad\boxed{\int_0^1\frac{\arctan(x)}{x} ~dx=K}\quad\blacksquare$

15. I must prove this so I can use it later

$\displaystyle \int_0^1{x^m\ln^p(x)}~dx=\frac{(-1)^p\Gamma(p+1)}{(m+1)^{p+1}}$

This is very simple first let us consider a different integral

$\displaystyle \int_0^1{x^m\ln\left(\frac{1}{x}\right)^p~dx}$

Now let

$\displaystyle \ln\left(\frac{1}{x}\right)=z$

then

$\displaystyle x=e^{-z}$

$\displaystyle \Rightarrow{dx=-e^{-z}}$

Now we see that the limits are changed

$\displaystyle 0\mapsto\infty$

$\displaystyle 1\mapsto{0}$

So we get

$\displaystyle -\int_{\infty}^0e^{-mz}z^{p}\frac{dz}{e^{z}}$

$\displaystyle =\int_0^{\infty}{e^{-z(m+1)}z^p}$

Now let $\displaystyle t=z(m+1)$

$\displaystyle \Rightarrow{\frac{t}{m+1}=z}$

$\displaystyle \Rightarrow{\frac{dt}{m+1}=dz}$

Limits stay the same. So we get

$\displaystyle =\int_0^{\infty}e^{-t}\left(\frac{t}{m+1}\right)^p\frac{dt}{m+1}$

$\displaystyle =\frac{1}{(m+1)^{p+1}}\int_0^{\infty}e^{-t}t^{(p+1)-1}~dt$

$\displaystyle =\frac{\Gamma(p+1)}{(m+1)^{p+1}}$

$\displaystyle \Rightarrow\int_0^1{x^m\ln\left(\frac{1}{x}\right) ^p~dx}=\frac{\Gamma(p+1)}{(m+1)^{p+1}}$

Now notice that

$\displaystyle \int_0^{1}x^m\ln\left(\frac{1}{x}\right)^p~dx$

$\displaystyle =\int_0^{1}x^m(-1)^p\ln^p(x)~dx$

So now dividing both sides appropriately gives us the final result

$\displaystyle \boxed{\int_0^1{x^m\ln^p(x)}~dx=\frac{(-1)^p\Gamma(p+1)}{(m+1)^{p+1}}}\quad\blacksquare$

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