1. 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

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

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

$\Gamma(n+1)=\Gamma(n+1)\quad\blacksquare$

2. 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

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

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

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

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

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

3. 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

(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

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

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

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

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.

6. Originally Posted by Marine
Mathstud28, it a really good work! I find it fantastic how you evaluate integrals

(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 . Well..I think I am going to keep using it ...why break convention for something as petty as looking stupid

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

( I don't have time to post the solution now. It's a nice integral, I like the result)
I'll try .. $\int_0^{\frac{\pi}{2}}\frac{\arctan\left(\sin(x)\r ight)}{\sin(x)}~dx$

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

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

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

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

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

I will give it some more serious thought later.

8. 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...

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)
$

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

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?

10. but that is not a common identity...is it?
$
=\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... :/

And how do you get to this one ?

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

would it simplify ?

11. Originally Posted by Moo
Ah..ah... :/

And how do you get to this one ?

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

would it simplify ?
Yes sort of! Except the remaining Gamma function in the top .

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...

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

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 $\beta\left(\frac{m+1}{n}~,~p+1\right)$ (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

13. Originally Posted by Moo
erm...

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

14. $I=\int_{0}^{\frac{\pi}{2}}ln(sin(x))dx$

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:

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

15. $\int_0^{\infty}\frac{\sin(ax)\cos(bx)}{x}~dx=I\qua d{a\ne{b}}~~(a,b)>0$

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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