1. Now this is one of the ones that I am most proud of

$\int_0^{\infty}\ln\left(\frac{e^x+1}{e^x-1}\right)~dx$

First let

$\ln(u)=x$

$\Rightarrow{\frac{du}{u}=dx}$

And we can see that
$0\mapsto{1}$

$\infty\mapsto\infty$

So we get

$=\int_1^{\infty}\frac{\ln\left(\frac{u+1}{u-1}\right)}{u}~dx$

Now here is the next step

Let

$u=\frac{1}{z}$

$\Rightarrow{du=\frac{-dz}{z^2}}$

$\infty\mapsto{0}$

$1\mapsto{1}$

$=-\int_{1}^{0}\frac{\ln\left(\frac{\frac{1}{z}+1}{\f rac{1}{z}-1}\right)}{\frac{1}{z}}\cdot\frac{dz}{z^2}$

Now with a little mutliplicative identity

$=\int_0^1\frac{\ln\left(\frac{1+z}{1-z}\right)}{z}~dz$

$=\int_0^1\frac{\ln(1+z)}{z}~dz-\int_0^{1}\frac{\ln(1-z)}{z}~dz$

Now imputting the uniformly convergent power seires

$\ln(1+z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{n+1}}{n+1}~\wedge~\ln(1-z)=-\sum_{n=0}^{\infty}\frac{z^{n+1}}{n+1}$

We ultimately arrive at (I've doen these earlier in the thread if you are curious)

$=\frac{\pi^2}{12}+\frac{\pi^2}{6}$

$=\frac{\pi^2}{4}$

So now industrious as ever we have found that

$\boxed{\int_0^{\infty}\ln\left(\frac{e^x+1}{e^x-1}\right)~dx=\int_1^{\infty}\frac{\ln\left(\frac{x +1}{x-1}\right)}{x}~dx=\int_0^1\frac{\ln\left(\frac{1+x} {1-x}\right)}{dx}~dx=\frac{\pi^2}{4}}\quad\blacksquar e$

2. $\int_0^{\frac{\pi}{4}}\ln\left(1+\tan(x)\right)~dx$

Here is another classic trick, first note that

$\int_0^{\frac{\pi}{4}}\ln\left(1+\tan(x)\right)~dx$

$=\int_0^{\frac{\pi}{4}}\ln\left(\frac{\cos(x)+\sin (x)}{\cos(x)}\right)~dx$

$=\int_0^{\frac{\pi}{4}}\ln\left(\cos(x)+\sin(x)\ri ght)~dx-\int_0^{\frac{\pi}{4}}\ln(\cos(x))~dx$

Now as I stated earlier once

$a\cos(x)+b\sin(x)=\sqrt{a^2+b^2}\cos\left(\arctan\ left(\frac{b}{a}\right)-x\right)$

So

$\cos(x)+\sin(x)=\sqrt{1^2+1^2}\cos\left(\arctan\le ft(\frac{1}{1}\right)-x\right)~dx$

$=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)$

So we have that

$=\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\cos\left( \frac{\pi}{4}-x\right)\right)~dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos(x)\right)~dx$

Now in the first integral we see thta if we let

$\frac{\pi}{4}-x=z$

$\Rightarrow{dx=-dz}$

Then

$0\mapsto\frac{\pi}{4}$

$\frac{\pi}{4}\mapsto{0}$

So we get

$=-\int_{\frac{\pi}{4}}^0\ln\left(\sqrt{2}\cos(z)\rig ht)~dz$

$=\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\cos(x)\ri ght)~dx$

So now we see that

$=\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\cos\left( \frac{\pi}{4}-x\right)\right)~dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos(x)\right)~dx$

$=\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\cos(x)\ri ght)~dx-\int_0^{\frac{\pi}{4}}\ln\left(\cos(x)\right)~dx$

$=\int_0^{\frac{\pi}{4}}\ln\left(\frac{\sqrt{2}\cos (x)}{\cos(x)}\right)~dx$

$=\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\right)~dx$

$=\frac{\ln(2)\pi}{8}$

$\therefore\quad\boxed{\int_0^{\frac{\pi}{4}}\ln\le ft(1+\tan(x)\right)~dx=\frac{\ln(2)\pi}{8}}\quad\b lacksquare$

3. As Moo showed earlier the best way I have found to compute

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

Is to let

$x=\tan(z)$

$\Rightarrow{dx=\sec^2(z)dz}$

$1\mapsto\frac{\pi}{4}$

$0\mapsto{0}$

So we get

$=\int_0^{\frac{\pi}{4}}\frac{\ln\left(1+\tan(z)\ri ght)}{1+\tan^2(z)}\cdot\sec^2(z)dz$

$=\int_0^{\frac{\pi}{4}}\frac{\ln\left(1+\tan(z)\ri ght)}{\sec^2(z)}\cdot\sec^2(z)dz$

$=\int_0^{\frac{\pi}{4}}\ln\left(1+\tan(z)\right)~d z$

Now that looks familar

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

4. Here is another one I have done recently

$\int_0^{\infty}\frac{x}{\sinh(x)}~dx$

Now rewrite it as
$=2\int_0^{\infty}\frac{x}{e^{x}-e^{-x}}~dx$

$=2\int_0^{\infty}\frac{xe^{-x}}{1-\left(e^{-x}\right)^2}~dx$

Now if we let

$z=e^{-x}$

$\Rightarrow{-\ln(z)=x}$

$\Rightarrow{\frac{-dz}{x}=dx}$

And

$\infty\mapsto{0}$

$0\mapsto{1}$

So we have

$=-2\int_0^1\frac{\ln(z)}{1-z^2}~dz$

Now we have done this one before, but all you would do is either use the uniformly convergent Maclaurin series on the denominator and go from there, or use PFD on denominator and do two seperately...either way

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

$=\frac{\pi^2}{4}$

$\therefore\quad\boxed{\int_0^{\infty}\frac{x}{\sin h(x)}~dx=\frac{\pi^2}{4}}\quad\blacksquare$

5. $\int_0^{\infty}\frac{dx}{\cosh(x)}$

Same trick here

$=2\int_0^{\infty}\frac{dx}{e^{x}+e^{-x}}~dx$

$=2\int_0^{\infty}\frac{e^{-x}}{1+\left(e^{-x}\right)^2}~dx$

Now letting

$z=e^{-x}$

$\Rightarrow{dz=-e^{-x}dx}$

and

$\infty\mapsto{0}$

$0\mapsto{1}$

So we get

$=2\int_0^1\frac{dz}{1+z^2}$

$=2\left\{\arctan(1)-\arctan(0)\right\}$

$=2\cdot\frac{\pi}{4}$

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

$\therefore\quad\boxed{\int_0^{\infty}\frac{dx}{\co sh(x)}=\frac{\pi}{2}}\quad\blacksquare$

6. Give this one a go:

$\int_{0}^{1}\frac{(sin^{-1}(x))^{2}}{x}$

7. Originally Posted by galactus
Give this one a go:

$\int_{0}^{1}\frac{(sin^{-1}(x))^{2}}{x}$
I can, but I have to give a lot of stuff that I am not willing to prove

Firstly

$\left(\arcsin(x)\right)^2=\sum_{n=1}^{\infty}\frac {2^{2n-1}x^{2n}}{n^2\binom{2n}{n}}$

Weierstrass M-test shows its uniform on $(0,1)$

So

$\int_0^1\frac{\arcsin^2(x)}{x}=\sum_{n=1}^{\infty} \frac{2^{2n-1}}{(2n+1)n^2\binom{2n}{n}}$

Which is equal to

$\frac{\pi^2\ln(2)}{4}-\frac{7}{8}\zeta(3)$

I mean that stuff is tedious to prove...were you looking for an answer or an explanation?

8. No, not as far as proving the series for arcsine. Just an explanation going from there to the zeta solution. I got it to the series you have for arcsin. But I think there are some identities I am not aware of to proceed further. Deriving that zeta(3) out of it is where I am having trouble.

I got this from trying to evaluate $\int_{0}^{\frac{\pi}{2}}xln(sin(x))dx$....[1]

Done some fanaggling and found it is equivalent to $\frac{-1}{2}\int_{0}^{1}\frac{(arcsin(x))^{2}}{x}dx$

I used the series for arcsin and proceeded as follows.

I used the series for arcsin as $\sum_{n=0}^{\infty}\frac{(2n)!\cdot x^{2n+1}}{(2^{n}n!)^{2}(2n+1)}$

I squared that, divided by x and got:

$\int_{0}^{1}\sum_{n=0}^{\infty}\frac{[(2n)!]^{2}x^{4n+1}}{(2^{n}n!)^{4}(2n+1)^{2}}dx$

Integrate over 0 to 1 and it becomes:

$\sum_{n=0}^{\infty}\frac{[(2n)!]^{2}(\frac{1}{16})^{n}}{2(2n+1)^{3}(n!)^{4}}$

Now, I was going to proceed with gamma/beta or something, but haven't finished. I see what you mean by tedious. This is where I wasn't totally sure what the best way to go was. I have some ideas, but have not had time to go any further. I was going to look at it after bit. I was asking for your input.

I also found that $\frac{-1}{2}\int_{0}^{\frac{\pi}{2}}x^{2}cot(x)dx$

is also an equivalent integral.

Playing around with it, I found many approaches.

I was trying to evaluate [1] and started with a u sub of $u=sin(x)$. This lead to

$\int_{0}^{1}\frac{arcsin(u)ln(u)}{\sqrt{1-u^{2}}}du$

Then, I used the series for $\frac{arcsin(x)}{\sqrt{1-x^{2}}}$ and went from there.

gives $\int_{0}^{1}\sum_{n=0}^{\infty}\frac{2^{2n}(n!)^{2 }ln(u)}{(2n+1)!}u^{2n+1}$

Using the aforementioned series and noting that $\int_{0}^{1}u^{2n+1}ln(u)du=\frac{-1}{4(n+1)^{2}}$

I got $\frac{-1}{4}\sum_{n=0}^{\infty}\frac{2^{2n}(n!)^{2}}{(n+1 )^{2}(2n+1)!}$

I then used the gamma/beta thing and whittled it down to

$\frac{-1}{4}\sum_{n=0}^{\infty}\frac{2^{2n}{\Gamma}(n+1){ \Gamma}(n+1)}{(n+1)^{2}{\Gamma}((n+1)+(n+1))}$

Now using the beta integral $B(n+1,n+1)=\int_{0}^{1}x^{n}(1-x)^{n}dx$, I eventually get:

$\frac{-1}{4}\int_{0}^{1}\frac{2^{2n}}{(n+1)^{2}}\sum_{n=0 }^{\infty}(x-x^{2})^{n}dx$

Then I got hung up. I can find a closed form for the geometric series. That $\frac{2^{2n}}{(n+1)^{2}}$ is what I didn't know what to do with. It diverges, so I must've erred somewhere.

Any thoughts there, Big M

9. Originally Posted by galactus
I squared that, divided by x and got:

$\int_{0}^{1}\sum_{n=0}^{\infty}\frac{[(2n)!]^{2}x^{4n+1}}{(2^{n}n!)^{4}(2n+1)^{2}}dx$
Uh-oh ...Cauchy-Schwarz inequality...your on the right track...I have to go soon but later I will come on and show some of the steps

10. Uh-oh?. Did I make a booboo?. I am sure it is something easy.

I amended my last post so I added more to peruse. Maybe I gave you some ideas. You can surely give me some.

11. Originally Posted by galactus
Uh-oh?. Did I make a booboo?. I am sure it is something easy.

I amended my last post so I added more to peruse. Maybe I gave you some ideas. You can surely give me some.
Yes use the series that i have in my post (I corrected it)

$\left(\sum_{n=a}^{\infty}x^n\right)^2\ne\sum_{n=0} ^{\infty}x^{2n}$

I will help more later

12. , I knew that. I knew there was some silly oversight.

Another option we may have is by using the series for ln(1-x) and replacing x with $cos^{2}(x)$ and noting that

$sin(x)=\sqrt{1-cos^{2}(x)}$

Then, we have $\frac{x}{2}ln(1-cos^{2}(x))$

So, the series is $\frac{-1}{2}\sum_{n=1}^{\infty}\frac{xcos^{2n}(x)}{n}$

13. Hey MS, you're back. I am sitting here bored. May as well play with this again.

Starting with the series you gave for $\int_{0}^{1}\frac{(sin^{-1}(x))^{2}}{x}dx$

We have $\sum_{n=1}^{\infty}\frac{2^{2n-1}(n!)^{2}}{(2n+1)n^{2}(2n)!}$

Now, $\frac{(n!)^{2}}{(2n)!}=\frac{n{\Gamma}(n){\Gamma}( n+1)}{{\Gamma}(2n+1)}$

$=n\cdot B(n,n+1)=\int_{0}^{1}nx^{n-1}(1-x)^{n}dx$

So, we get:

$\sum_{n=1}^{\infty}\frac{n2^{2n}}{2n^{2}(n+1)}\int _{0}^{1}x^{n-1}(1-x)^{n}dx$

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

Now, MS, what are your thoughts?. Where is that pesky zeta(3).

I know there is some silly thing I am overlooking.

14. Here is one that I found, I am sure it has been done before, but I came up with independently and I am crossing my fingers I am lucky enough to notice it first!

I am going to show that

$\forall{p>-1}~~\int_0^1\frac{\ln^p\ln(1+x)}{x}dx=(-1)^p\Gamma(p+1)\eta(p+2)$

Where $\eta(x)$ is Diriclet's Eta funcion given by $\eta(x)=\left(1-2x^{1-x}\right)\zeta(x)$

Proof: Consider that

$\forall{x}\in(0,1]~\ln(1+x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}$

Now note that it is uniformly for every interior point of its interval of convergence. Or seeing that if we define $u_n(x)=\frac{(-1)^n}{n+1}\cdot{x^n}=a_n\cdot{f_n(x)}$ that $\sum{a_n}$ converges and that $\forall{x}\in(0,1]~~0\leqslant{f_n(x)}\leqslant{1}$. And finally noting that $\forall{x}\in(0,1]~~f_n(x)\in\downarrow$. So we can conclude that this series is uniformly convergent by Abel's Test for Uniform Convergence.

So now we rewrite our integral as follows

\begin{aligned}\int_0^1\frac{\ln^p(x)\ln(1+x)}{x}d x&=\int_0^1\left\{\frac{\ln^p(x)}{x}\sum_{n=0}^{\i nfty}\frac{(-1)^nx^{n+1}}{n+1}\right\}dx\\
&=\int_0^1\sum_{n=0}^{\infty}\frac{(-1)^n\ln^p(x)x^n}{n+1}dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^1\ln^p(x)x^ndx
\end{aligned}

Now let us compute

$\int_0^1x^n(-\ln(x))^pdx$

Now letting $-\ln(x)\mapsto{\varphi}$ we get

$\int_0^1x^n(-\ln(x))^pdx\overbrace{\mapsto}^{-\ln(x)=\varphi}\int_0^{\infty}e^{-(n+1)\varphi}\varphi^pd\varphi$

Now letting $(n+1)\varphi=\psi$ gives

\begin{aligned}\int_0^{\infty}e^{-(n+1)\varphi}\varphi^pd\varphi&\overbrace{\mapsto} ^{(n+1)\varphi=\psi}\int_0^{\infty}e^{-\psi}\left(\frac{\psi}{n+1}\right)^p\cdot\frac{d\p si}{n+1}\\
&=\frac{1}{(n+1)^{p+1}}\int_0^{\infty}e^{-\psi}\psi^pd\psi\\
&=\frac{\Gamma(p+1)}{(n+1)^{p+1}}
\end{aligned}

So if

$\int_0^1x^n(-\ln(x))^pdx=\frac{\Gamma(p+1)}{(n+1)^{p+1}}$

We can see by dividing both sides by $(-1)^p$ that

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

So back to our integral we have

\begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^1x^n\ln^p(x)dx&=\sum_{n=0}^{\inft y}\frac{(-1)^n}{n+1}\cdot\frac{(-1)^p\Gamma(p+1)}{(n+1)^{p+1}}\\
&=(-1)^p\Gamma(p+1)\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^{p+2}}\\
&=(-1)^p\Gamma(p+1)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{p+2}}\\
&=(-1)^p\Gamma(p+1)\eta\left(p+2\right)\end{aligned}

So we can conclude that

$\boxed{\int_0^1\frac{\ln^p(x)\ln(1+x)}{x}dx=(-1)^p\Gamma(p+1)\eta(p+2)\quad{p>-1}}\quad\blacksquare$

Note the restriction on $p$ may be extended to fit any points where both $\Gamma(p+1)$ and $\eta(p+2)$ are both well-defined.

Also note that a slight change gives

$\int_0^1\frac{\ln^p(x)\ln(1-x)}{x}dx=(-1)^{p+1}\Gamma(p+1)\zeta(p+2)$

15. so many integrals... i think i'm gonna cry.

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)
the first time i saw this integral, i couldn't solve it, but now i came up with a solution and i have to say that this one is a really nice integral.

------

straightforward computations show that $\int_{0}^{\frac{\pi }{2}}{\frac{dx}{1+t^{2}\sin ^{2}x}}=\frac{\pi }{2\sqrt{1+t^{2}}},$ and $\int_{0}^{1}{\frac{dt}{\sqrt{1+t^{2}}}}=\ln \left( 1+\sqrt{2} \right).$ (1)

now $\int_{0}^{\frac{\pi }{2}}{\frac{\arctan (\sin x)}{\sin x}\,dx}=\int_{0}^{\frac{\pi }{2}}{\int_{0}^{1}{\frac{dt\,dx}{1+t^{2}\sin ^{2}x}}},$ so swap the integration order and by applying (1) the integral equals $\frac{\pi }{2}\ln \left( \sqrt{2}+1 \right)=\frac{\pi }{4}\ln \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right).$

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