# Integrals

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• Jul 21st 2008, 09:49 AM
Mathstud28
I am actually going to use induction on this one, so if I mess it up please forgive me.

$\int_0^{1}\ln^n(x)~dx=(-1)^nn!\quad{n\in\mathbb{N}}$

Base Case

Let us start with the base case $n=0$

$\int_0^1\ln^0(x)~dx$

$=\int_0^1~dx$

$=1$

$=(-1)^00!$

So we have proved the base case.

Inductive Hypothesis

Now lets assume that

$\int_0^1\ln^n(x)~dx=(-1)^nn!\quad{n\in\mathbb{N}}$

This being our inductive hypothesis (IH)

Inductive step

Now lets look at our inductive step

$\int_0^1\ln^{n+1}(x)~dx$

Which by Integration by Parts

$=x\ln^{n+1}(x)\bigg|_0^1-\int_0^{1}\frac{(n+1)\ln(x)^nx}{x}~dx$

$=0-\int_0^1(n+1)\ln^n(x)~dx$

$=-\bigg[n\int_0^1\ln^n(x)~dx+\int_0^1\ln(x)^n~dx\bigg]$

$\underbrace{=}_{\text{IH}}=-\bigg[n(-1)^nn!+(-1)^nn!\bigg]$

$=-(-1)^nn!(n+1)$

$=(-1)^{n+1}(n+1)!$

Thus our inductive step is proven.

$\therefore\quad\boxed{\int_0^1\ln^n(x)~dx=(-1)^nn!\quad{n\in\mathbb{N}}}\quad\blacksquare$

We can make a stronger case of this

$\int_0^1\ln^n(x)~dx=(-1)^n\Gamma(n+1)\quad{n\in\mathbb{R}}$

This is gotten by simple aglaebraic manipulation

$\int_0^{1}\ln^n(x)~dx=(-1)^n\Gamma(n+1)$

$\Rightarrow\int_0^1(-1)^n\ln^n(x)~dx=\Gamma(n+1)$

$\Rightarrow{\int_0^1\left[\ln\left(\frac{1}{x}\right)\right]^n~dx=\Gamma(n+1)}$

Now on the left hand integral if we let $-\ln(x)=t$

We get that

$x=e^{-t}$

$\Rightarrow{dx=-e^{-t}}$

and as $x\to{0}\Rightarrow{t\to\infty}$\

And as $x\to{1}\Rightarrow{t\to{0}}$

So we get

$-\int_{\infty}^{0}t^ne^{-t}~dt=\Gamma(n+1)$

$=\int_0^{\infty}t^{n+1-1}e^{-t}~dt=\Gamma(n+1)$

$\Gamma(n+1)=\Gamma(n+1)\quad\blacksquare$
• Jul 21st 2008, 10:41 AM
Mathstud28
Here is one I found implication but not by itself

$\int_0^1\frac{{\rm{arctanh}}(x)}{x}~dx$

We first do this by seeing that

$\int\frac{dx}{1-x^2}$

$=\int\sum_{n=0}^{\infty}x^{2n}~dx$

$\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$

$={\rm{arctanh}}(x)\quad\forall{x}\backepsilon|x|<1$

so

$\int_0^1\frac{{\rm{arctanh}}(x)}{x}~dx$

$=\int_0^1\frac{dx}{x}\sum_{n=0}^{\infty}\frac{x^{2 n+1}}{2n+1}$

$=\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^1x^{2n}~d x$

$=\sum_{n=0}^{\infty}\frac{1}{2n+1}\cdot\frac{1}{2n +1}$

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

$=\frac{\pi^2}{8}$

$\therefore\quad\boxed{\int_0^1\frac{{\rm{arctanh}} (x)}{x}~dx=\frac{\pi^2}{8}}\quad\blacksquare$

This brings a much simpler calculations of

$\int_0^1\frac{\ln(x)}{1-x^2}~dx$

Because

$\ln(x)=\int_1^x\frac{dy}{y}$

So we have

$\int_0^1\int_1^x\frac{dy}{y(1-x^2)}~dx$

$=-\int_0^1\int_0^y\frac{dx}{y(1-x^2)}~dy$

$=-\int_0^1\frac{dy}{y}\cdot\bigg[{\rm{arctanh}}(x)\bigg]\bigg|_0^1$

$=-\int_0^1\frac{{\rm{arctanh}}(y)}{y}~dy$

$=\frac{-\pi^2}{8}$
• Jul 21st 2008, 12:13 PM
Marine
Quote:

Originally Posted by Mathstud28
Here is one I found implication but not by itself

$\int_0^1\frac{{\rm{arctanh}}(x)}{x}~dx$

We first do this by seeing that

$\int\frac{dx}{1-x^2}$

$=\int\sum_{n=0}^{\infty}x^{2n}~dx$

$\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$

$={\rm{arctanh}}(x)\quad\forall{x}\backepsilon|x|<1$

so

$\int_0^1\frac{{\rm{arctanh}}(x)}{x}~dx$

$=\int_0^1\frac{dx}{x}\sum_{n=0}^{\infty}\frac{x^{2 n+1}}{2n+1}$

$=\sum_{n=0}^{\infty}\frac{1}{2n+1}\int_0^1x^{2n}~d x$

$=\sum_{n=0}^{\infty}\frac{1}{2n+1}\cdot\frac{1}{2n +1}$

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

$=\frac{\pi^2}{8}$

$\therefore\quad\boxed{\int_0^1\frac{{\rm{arctanh}} (x)}{x}~dx=\frac{\pi^2}{8}}\quad\blacksquare$

This brings a much simpler calculations of

$\int_0^1\frac{\ln(x)}{1-x^2}~dx$

Because

$\ln(x)=\int_1^x\frac{dy}{y}$

So we have

$\int_0^1\int_1^x\frac{dy}{y(1-x^2)}~dx$

$=-\int_0^1\int_0^y\frac{dx}{y(1-x^2)}~dy$

$=-\int_0^1\frac{dy}{y}\cdot\bigg[{\rm{arctanh}}(x)\bigg]\bigg|_0^1$

$=-\int_0^1\frac{{\rm{arctanh}}(y)}{y}~dy$

$=\frac{-\pi^2}{8}$

Mathstud28, it a really good work! I find it fantastic how you evaluate integrals (Bow)

(just a small note: the inverse functions of the hyperbolic functions are the area-functions, not arcus functions (as you know of course), which the abbr. name is arsinh, arcosh, artanh, arcoth, arsech, arcsch of - all I mean is, it's not arc... but ar... ;))

Inverse hyperbolic function - Wikipedia, the free encyclopedia

I pretty much like the notaions arc and ar instead of that one with ^-1 cuz the 2nd one is ambiguous - you can think of it as 1/function

best regards, marine
• Jul 21st 2008, 12:24 PM
PaulRS
Show that $\int_0^{\frac{\pi}{2}}\frac{\arctan[\sin(x)]}{\sin(x)}dx=\frac{\pi}{4}\cdot{\ln\left(\frac{\sq rt{2}+1}{\sqrt{2}-1}\right)}$

(Sun) ( I don't have time to post the solution now. It's a nice integral, I like the result)
• Jul 21st 2008, 12:41 PM
Mathstud28
$\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx$

This one is the one that has caused me the most mental anguish. I had nowhere to start and I just layed in bed and thought about it for twenty minutes and then went back to the drawing board. ad infinitum. It was easy once I realized something.

If we make a substitution

Let $x=\frac{\pi}{2}-r\Rightarrow{t=\frac{\pi}{2}-x}$

$\Rightarrow{dx=-dt}$

Then as $x\to{0}\Rightarrow{t\to{\frac{\pi}{2}}}$

and as $x\to\frac{\pi}{2}\Rightarrow{t\to{0}}$

So we get

$-\int_{\frac{\pi}{2}}^0\ln\left(\sin\left(\tiny{\fr ac{\pi}{2}-t}\right)\right)~dt$

$=\int_0^{\frac{\pi}{2}}\ln\left(\cos(t)\right)~dt$

$\Rightarrow\quad\int_0^{\frac{\pi}{2}}\ln\left(\si n(x)\right)~dx=\int_0^{\frac{\pi}{2}}\ln\left(\cos (x)\right)~dx$

So now using this we see the following

$2\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx$

$=\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx+ \int_0^{\frac{\pi}{2}}\ln\left(\cos(x)\right)~dx$

$\int_0^{\frac{\pi}{2}}\ln\left(\cos(x)\sin(x)\righ t)~dx$

Now I wanted $\sin(2x)$ so I did the following

$=\int_0^{\frac{\pi}{2}}\ln\left(\sin(2x)\right)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

Now for the first integral, if you let

$2x=t$ you get

$=\frac{1}{2}\int_0^{\pi}\ln\left(\sin(x)\right)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

Now once again for this first integral let

$u=t-\frac{\pi}{2}$

$\Rightarrow{dt=du}$

And as $t\to{0}\Rightarrow{u\to\frac{-\pi}{2}}$

And as $t\to{\pi}\Rightarrow{u\to\frac{\pi}{2}}$

Giving us

$=\frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\ln\left(\sin\left(u+\frac {\pi}{2}\right)\right)~du-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

$=\frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\ln\left(\cos(u)\right)~du-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

Now seeing that since

$\cos(-x)=\cos(x)$

$\Rightarrow\ln(\cos(-x))=\ln(\cos(x))$

Which means that

$=\int_0^{\frac{\pi}{2}}\ln\left(\cos(u)\right)~du-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

$=\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

So

$2\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx= \int_0^{\frac{\pi}{2}}\sin(x)~dx-\int_0^{\frac{\pi}{2}}\ln(2)~dx$

$\Rightarrow\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\ right)~dx=-\int_0^{\frac{\pi}{2}}\ln(2)$

$=\frac{-\ln(2)\pi}{2}$

So FINALLY we can conclude that

$\boxed{\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\righ t)~dx=\int_0^{\frac{\pi}{2}}\ln\left(\cos(x)\right )~dx=\frac{-\ln(2)\pi}{2}}\quad\blacksquare$

That one was a toughie.
• Jul 21st 2008, 12:46 PM
Mathstud28
Quote:

Originally Posted by Marine
Mathstud28, it a really good work! I find it fantastic how you evaluate integrals (Bow)

(just a small note: the inverse functions of the hyperbolic functions are the area-functions, not arcus functions (as you know of course), which the abbr. name is arsinh, arcosh, artanh, arcoth, arsech, arcsch of - all I mean is, it's not arc... but ar... ;))

Inverse hyperbolic function - Wikipedia, the free encyclopedia

I pretty much like the notaions arc and ar instead of that one with ^-1 cuz the 2nd one is ambiguous - you can think of it as 1/function

best regards, marine

Thank you! I actually always thought it was arc (Thinking). Well..I think I am going to keep using it :D...why break convention for something as petty as looking stupid :p
• Jul 21st 2008, 01:08 PM
Mathstud28
Quote:

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

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

I'll try (Thinking).. $\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\sin(x)\r ight)}{\sin(x)}~dx$

$\Rightarrow\int_0^1\frac{\arctan(u)}{u\sqrt{1-u^2}}~du$

$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1\frac{u^{2n}}{\sqrt{1-u^2}}$

$=\sum_{n=0}^{\infty}\frac{\sqrt{\pi}(-1)^n\Gamma\left(\frac{2n+1}{2}\right)}{2\Gamma\lef t(\frac{2n+1}{2}+\frac{3}{2}\right)}$

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

$=\frac{\pi}{4}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

:D

I will give it some more serious thought later.
• Jul 21st 2008, 01:16 PM
PaulRS
Quote:

Originally Posted by Mathstud28
$\int_0^{\frac{\pi}{2}}\ln\left(\sin(x)\right)~dx$

This one is the one that has caused me the most mental anguish. I had nowhere to start and I just layed in bed and thought about it for twenty minutes and then went back to the drawing board. ad infinitum. It was easy once I realized something.

Here's another way I've found of computing it...(Evilgrin)

Consider the Series: $
\sum\limits_{k = 1}^\infty {\tfrac{{\cos \left( {k \cdot x} \right)}}
{k}} = - \ln \left( 2 \right) - \ln \left[ {\sin \left( {\tfrac{x}
{2}} \right)} \right]
$
for all $x$ such that $
0 < x < 2\pi
$

Integrate from $0$ to $\pi$ now: $
\sum\limits_{k = 1}^\infty {\tfrac{{\int_0^\pi {\cos \left( {k \cdot x} \right)dx} }}
{k}} = - \ln \left( 2 \right) \cdot \pi - \int_0^\pi {\ln \left[ {\sin \left( {\tfrac{x}
{2}} \right)} \right]dx}
$
(assuming we can do that)

But note that the series on the LHS is 0, because each of the terms is 0.

THus $
0 = - \ln \left( 2 \right) \cdot \pi - \int_0^\pi {\ln \left[ {\sin \left( {\tfrac{x}
{2}} \right)} \right]dx}
$

From there it follows easily that: $
\int_0^{\frac{\pi}{2}} {\ln \left[ {\sin \left( x \right)} \right]dx} = - \tfrac{\pi }
{2} \cdot \ln \left( 2 \right)
$
• Jul 21st 2008, 01:19 PM
Mathstud28
Quote:

Originally Posted by PaulRS
Here's another way I've found of computing it...(Evilgrin)

Consider the Series: $
\sum\limits_{k = 1}^\infty {\tfrac{{\cos \left( {k \cdot x} \right)}}
{k}} = - \ln \left( 2 \right) - \ln \left[ {\sin \left( {\tfrac{x}
{2}} \right)} \right]
$
for all $x$ such that $
0 < x < 2\pi
$

Integrate from $0$ to $\pi$ now: $
\sum\limits_{k = 1}^\infty {\tfrac{{\int_0^\pi {\cos \left( {k \cdot x} \right)dx} }}
{k}} = - \ln \left( 2 \right) \cdot \pi - \int_0^\pi {\ln \left[ {\sin \left( {\tfrac{x}
{2}} \right)} \right]dx}
$
(assuming we can do that)

But note that the series on the LHS is 0, because each of the terms is 0.

THus $
0 = - \ln \left( 2 \right) \cdot \pi - \int_0^\pi {\ln \left[ {\sin \left( {\tfrac{x}
{2}} \right)} \right]dx}
$

From there it follows easily that: $
\int_0^\pi {\ln \left[ {\sin \left( x \right)} \right]dx} = - \tfrac{\pi }
{2} \cdot \ln \left( 2 \right)
$

That is assuming the first identity is true (which I am sure it is)...but that is not a common identity...is it?
• Jul 21st 2008, 01:29 PM
Moo
Quote:

but that is not a common identity...is it?
Quote:

$
=\sum_{n=0}^{\infty}\frac{\sqrt{\pi}(-1)^n\Gamma\left(\frac{2n+1}{2}\right)}{2\Gamma\lef t(\frac{2n+1}{2}+\frac{3}{2}\right)}
$

Ah..ah... :/ (Rofl)

And how do you get to this one ? (Surprised)

By the way, $\Gamma\left(\frac{2n+1}{2}+\frac{3}{2}\right)=\Gam ma(n+2)=(n+1)!$

would it simplify ? :D
• Jul 21st 2008, 01:36 PM
Mathstud28
Quote:

Originally Posted by Moo
Ah..ah... :/ (Rofl)

And how do you get to this one ? (Surprised)

By the way, $\Gamma\left(\frac{2n+1}{2}+\frac{3}{2}\right)=\Gam ma(n+2)=(n+1)!$

would it simplify ? :D

Yes sort of! Except the remaining Gamma function in the top (Angry).

and as for your other question I don't know why but I remembered that

$\int_0^1\frac{x^{m}}{(1-x^n)^p}=\frac{\Gamma\left(\frac{m+1}{n}\right)\Gam ma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}$..B....e...t...
• Jul 21st 2008, 01:40 PM
Moo
Quote:

Originally Posted by Mathstud28
Yes sort of! Except the remaining Gamma function in the top (Angry).

and as for your other question I don't know why but I remembered that

$\int_0^1\frac{x^{m}}{(1-x^n)^p}=\frac{\Gamma\left(\frac{m+1}{n}\right)\Gam ma(p+1)}{\Gamma\left(\frac{m+1}{n}+p+1\right)}$..B....e...t...

erm...

I know that the RHS is equal to $\beta\left(\frac{m+1}{n}~,~p+1\right)$ (Rofl) (you cheater...you edited it lol !)
And don't forget dx >.<

You remembered that, but prove it, like for PaulRS !! lool
According to my first thoughts on it, yours looks simpler than Paul's :D :D
• Jul 21st 2008, 01:43 PM
Mathstud28
Quote:

Originally Posted by Moo
erm...

(Rofl) (you cheater...you edited it lol !)

(Wink)
• Jul 21st 2008, 02:09 PM
galactus
$I=\int_{0}^{\frac{\pi}{2}}ln(sin(x))dx$

Now, let's use the rather obscure identity:

$sin(x)=2sin(\frac{x}{2})cos(\frac{x}{2})$

$I=\int_{0}^{\frac{\pi}{2}}ln\left(2sin(\frac{x}{2} )cos(\frac{x}{2})\right)dx$

$=\int_{0}^{\frac{\pi}{2}}ln(2)dx+\int_{0}^{\frac{\ pi}{2}}ln(sin(\frac{x}{2}))dx+\int_{0}^{\frac{\pi} {2}}ln(cos(\frac{x}{2}))dx$

Make the sub $u=\frac{x}{2}$

$I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s in(u))du+2\int_{0}^{\frac{\pi}{4}}ln(cos(u))du$......[1]

Now, use the identity: $cos(u)=sin(\frac{\pi}{2}-u)$

and make the sub $t=\frac{\pi}{2}-u$ in the integral with the $cos(u)$.

We get:

$\int_{0}^{\frac{\pi}{4}}ln(cos(u))du$

$= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(sin(t))dt$

Then, [1] becomes:

$I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s in(u))du+2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(s in(t))dt$

$I=\frac{\pi}{2}ln(2)+2\underbrace{\int_{0}^{\frac{ \pi}{2}}ln(sin(x))dx}_{\text{this is I}}$

$I=\frac{\pi}{2}ln(2)+2I$

Solve for I:

$\boxed{I=\frac{-\pi}{2}ln(2)}$
• Jul 21st 2008, 04:45 PM
Mathstud28
$\int_0^{\infty}\frac{\sin(ax)\cos(bx)}{x}~dx=I\qua d{a\ne{b}}~~(a,b)>0$

We know that

$\sin(ax)\cos(bx)=\frac{1}{2}\bigg[\sin(ax-bx)+\sin(ax+bx)\bigg]$

and

$\frac{1}{x}=\int_0^{\infty}e^{-yx}~dy$

We see that

$I=\frac{1}{2}\int_0^{\infty}\bigg[\sin((a-b)x)e^{-yx}+e^{-yx}\sin((a+b)x)e^{-yx}\bigg]~dy~dx$

$=\frac{1}{2}\int_0^{\infty}\int_0^{\infty}\bigg[\sin((a-b)x)e^{-yx}+\sin((a+b)x)e^{-yx}\bigg]~dx~dy$

Now let $C=a-b$ and $D=a+b$

Then we have

$\frac{1}{2}\int_0^{\infty}\int_0^{\infty}\bigg[\sin\left(Cx\right)e^{-yx}+\sin\left(Dx\right)e^{-yx}\bigg]~dx~dy$

Now when we evaluate this by parts I am sure since I have shown it many times that we get

$\frac{1}{2}\int_0^{\infty}\bigg[\frac{C}{y^2+C^2}+\frac{D^2}{y^2+D^2}\bigg]~dy$

$=\frac{1}{2}\bigg[\arctan\left(\frac{y}{C}\right)+\arctan\left(\frac {y}{D}\right)\bigg]\bigg|_0^{\infty}$

Now we want to back sub to get

$I=\frac{1}{2}\bigg[\arctan\left(\frac{y}{a-b}\right)+\arctan\left(\frac{y}{a+b}\right)\bigg]$

Now let us consider the two cases if

$a>b>0$

Then we have that

$I=\frac{1}{2}\bigg[\frac{\pi}{2}+\frac{\pi}{2}\bigg]$

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

Now if $b>a>0$

This implies that $a-b<0$

So by the oddness of arctangent

$I=\frac{1}{2}\bigg[\frac{-\pi}{2}+\frac{\pi}{2}\bigg]$

$=0$

Now if $a=b$

Then we go back to the beginning and rewite it as this

$\frac{1}{2}\int_0^{\infty}\frac{\sin(2ax)}{x}~dx$

Which we proved in another thread is equal to $\frac{1}{2}\cdot\frac{\pi}{2}$

Giving us

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

Putting htis all together we get

$\boxed{\int_0^{\infty}\frac{\sin(ax)\cos(bx)}{x}~d x=\begin{cases}
0 & \mbox{if}~~0 \frac{\pi}{2} & \mbox{if}~~0 \frac{\pi}{4} &\mbox{if}~~a=b\ne{0}
\end{cases} }$
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