Vrftsjtryk

I observed graphically that $$f(z)=int_0^infty lnleft(1+frac{z}{cosh(x)}right)dx=frac{pi^2}{8}+frac{(cosh^{-1}(z))^2}{2},zge1$$

Can anyone explain why this holds? I tried differentiating with respect to $z$ but I didn't really know how to continue further.

Also, if anyone knows how to obtain a closed form of $f(z)$ for $-1<z<1$ it would be greatly appreciated as I'm curious about specific values such as $f(-frac{1}{2})=-frac{7pi^2}{72}$ and $f(frac{1}{2})=frac{5pi^2}{72}.$

Substitute $z=frac{1-w^2}{1+w^2}$, the original integral turns into
$$I(w)=int_0^inftylnleft(1+frac{1-w^2}{1+w^2}operatorname{sech} xright)dx.$$
Use Feymann's trick, $$I'(w)=int_0^inftyfrac{-4w}{1-w^4+(1+w^2)^2cosh x}dx$$
Recall $cosh x=frac12(e^x+e^{-x})$,
$$I'(w)=int_0^inftyfrac{-4w}{1-w^4+(1+w^2)^2frac12(e^x+e^{-x})}dx\
=int_0^inftyfrac{-4we^x}{(1-w^4)e^x+(1+w^2)^2frac12(e^{2x}+1)}dx\
=int_0^inftyfrac{-4w}{(1-w^4)e^x+(1+w^2)^2frac12(e^{2x}+1)}de^x\
=-4frac{arctan w}{1+w^2}$$
$$I(w)=I(0)+int_0^w -4frac{arctan l}{1+l^2}dl\
=int_0^inftyln(1+operatorname{sech}(x))dx-2arctan^2w\
=frac{pi^2}8-2arctan^2w$$
Now, solve $z=frac{1-w^2}{1+w^2}$ w.r.t. $w$ and use $-2arctan^2sqrt{frac{1-z}{1+z}}=frac12operatorname{arccosh}^2 z$, we get $$f(z)=frac{pi^2}{8}+frac{(cosh^{-1}(z))^2}{2},z>-1.$$
The formula is valid when $zin[-1,1]$ if we consider $cosh^{-1}$ as a complex-valued function. If you don't like this representation, just deduce $f(z)=frac{pi^2}{8}-frac{(arccos(z))^2}{2}$ using some algebra.

Here is a more direct way.

The first thing to note is the function $f(z)$ is continuous for all $z > - 1$, including at the point $z = 1$.

From
$$f(z) = int_0^infty ln (1 + z operatorname{sech} x) , dx,$$
on differentiating with respect to $z$ we have
begin{align}
f'(z) &= int_0^infty frac{operatorname{sech} x}{1 + z operatorname{sech}} , dx\
&= int_0^infty frac{dx}{cosh x + z}\
&= 2 int_0^infty frac{e^x}{e^{2x} + 2z e^x + 1} , dx.
end{align}
In the last line we have used the fact that $cosh x = (e^{x} + e^{-x})/2$.

Making a substitution of $u = e^x, du = e^x , dx$ leads to
$$f'(z) = 2 int_1^infty frac{du}{(u + z)^2 + 1 - z^2}. qquad (*)$$

Case 1: $-1 < z < 1$

For this case, as the term $(1 - z^2)$ will be positive we can write ($*$) as:
begin{align}
f'(z) &= 2 int_1^infty frac{du}{(u + z)^2 + (sqrt{1 - z^2})^2}\
&= frac{2}{sqrt{1 - z^2}} left [tan^{-1} left (frac{u + z}{sqrt{1 - z^2}} right ) right ]_1^infty\
&= frac{pi}{sqrt{1 - z^2}} - frac{2}{sqrt{1 - z^2}} tan^{-1} left (frac{1 + z}{sqrt{1 - z^2}} right )\
&= frac{pi}{sqrt{1 - z^2}} - frac{2}{sqrt{1 - z^2}} left [frac{pi}{2} - tan^{-1} left (frac{sqrt{1 - z^2}}{1 + z} right ) right ]\
&= frac{2}{sqrt{1 - z^2}} tan^{-1} left (frac{sqrt{1 - z^2}}{1 + z} right ), qquad (**)
end{align}
where we have made use of the result $tan^{-1} (x) = pi/2 - tan^{-1} (1/x)$ for $x > 0$.

Noting that
$$tan left (frac{1}{2} cos^{-1} z right ) = frac{sqrt{1 - z^2}}{1 + z},$$
the result in ($**$) can be rewritten as
$$f'(z) = frac{1}{sqrt{1 - z^2}} cos^{-1} z.$$
Integrating up with respect to $z$ it is readily seen that
$$f(z) = int frac{cos^{-1} z}{sqrt{1 - z^2}} , dz = -frac{1}{2} (cos^{-1} z )^2 + C.$$
To find the constant $C$, choosing $z= 0$ we see that $f(0) = 0$. Thus $C = pi^2/8$, leading to
$$f (z) = frac{pi^2}{8} - frac{1}{2} (cos^{-1} z)^2, , -1 < z < 1. qquad (dagger)$$

Case 2: $z > 1$

For this case, as the term $(1 - z^2)$ will be negative we can write ($*$) as:
begin{align}
f'(z) &= 2 int_1^infty frac{du}{(u + z)^2 - (sqrt{z^2 - 1})^2}\
&= -frac{2}{sqrt{z^2 - 1}} left [coth^{-1} left (frac{u + z}{sqrt{z^2 - 1}} right ) right ]_1^infty\
&= frac{2}{sqrt{z^2 - 1}} coth^{-1} left (frac{1 + z}{sqrt{z^2 - 1}} right )\
&= frac{2}{sqrt{z^2 - 1}} tanh^{-1} left (frac{sqrt{z^2 - 1}}{z+ 1} right ), qquad (***)
end{align}
where we have made use of the result $coth^{-1} (x) = tanh^{-1} (1/x), x neq 0$.

Noting that
$$tanh left (frac{1}{2} cosh^{-1} z right ) = frac{sqrt{z^2 - 1}}{z + 1},$$
the result in ($***$) can be written as
$$f'(z) = frac{1}{sqrt{z^2 - 1}} cosh^{-1} z.$$
Integrating up with respect to $z$ it is readily seen that
$$f(z) = int frac{cosh^{-1} z}{sqrt{z^2 - 1}} , dz = frac{1}{2} (cosh^{-1} z)^2 + C.$$
To find the unknown constant $C$, as the function $f(z)$ is continuous for all $z > - 1$, for $f$ to be continuous at $z = 1$, from ($dagger$) we require $f(1) = pi^2/8$. Thus $C = pi^2/8$, leading to
$$f(z) = frac{pi^2}{8} + frac{1}{2} (cosh^{-1} z)^2, qquad z > 1.$$

First,
$$
begin{align}
int_0^{pi/2}frac{mathrm{d}theta}{1+alphasin(theta)}
&=int_0^1frac{2,mathrm{d}z}{1+2alpha z+z^2}tag1\
&=frac1{sqrt{alpha^2-1}}int_0^1left(frac1{z+alpha-sqrt{alpha^2-1}}-frac1{z+alpha+sqrt{alpha^2-1}}right)mathrm{d}ztag2\
&=frac1{sqrt{alpha^2-1}}left[logleft(frac{z+alpha-sqrt{alpha^2-1}}{z+alpha+sqrt{alpha^2-1}}right)right]_0^1tag3\
&=frac1{sqrt{alpha^2-1}}logleft(alpha+sqrt{alpha^2-1}right)tag4\
&=left{begin{array}{}
frac1{sqrt{1-alpha^2}}cos^{-1}(alpha)&text{if }|alpha|le1\
frac1{sqrt{alpha^2-1}}cosh^{-1}(alpha)&text{if }alphage1
end{array}right.tag5
end{align}
$$
Explanation:
$(1)$: substitute $z=tan(theta/2)$
$(2)$: partial fractions
$(3)$: integrate
$(4)$: evaluate
$(5)$: simplify in different cases

Therefore,
$$
begin{align}
int_0^inftylogleft(1+frac{alpha}{cosh(x)}right),mathrm{d}x
&=int_1^inftylogleft(1+frac{2alpha}{u+frac1u}right)frac{mathrm{d}u}utag6\
&=int_0^1logleft(1+frac{2alpha}{u+frac1u}right)frac{mathrm{d}u}utag7\
&=int_0^{pi/4}logleft(1+alphasin(2theta)right)frac{2,mathrm{d}theta}{sin(2theta)}tag8\
&=int_0^{pi/2}logleft(1+alphasin(theta)right)frac{mathrm{d}theta}{sin(theta)}tag9\
&=left{begin{array}{}
frac{pi^2}8-frac12cos^{-1}(alpha)^2&text{if }|alpha|le1\
frac{pi^2}8+frac12cosh^{-1}(alpha)^2&text{if }alphage1
end{array}right.tag{10}
end{align}
$$
Explanation:
$phantom{1}(6)$: substitute $u=e^x$
$phantom{1}(7)$: substitute $umapsto1/u$
$phantom{1}(8)$: substitute $u=tan(theta)$
$phantom{1}(9)$: substitute $thetamapstotheta/2$
$(10)$: apply the integral of $(5)$