The integral $\int_{0}^{\infty}~ e^{-x}~ \mbox{sech}^2(x+z)~dx,~~ z \in R$

Question

The integral $$\int_{0}^{\infty}~ e^{-x}~ \mbox{sech}^2(x+z)~dx,~~ z \in R~~~~(1)$$ helps creating an interesting identity for a Gauss Hypergeometric function as $$~_2F_1(2,1;5/2;(1-\tanh z)/2)=3 e^{z} \cosh^2 z ~ (\pi/2-\sin^{-1}(\tanh z)-\mbox{sech}z).~~~(2)$$ The question is: what is the proof for (2) by using (1) or otherwise?

Answer 1

The hypergeometric form is easy enough to obtain:

$$I(z)=4 \int_0^\infty \frac{e^{-x} dx}{(e^{x+z}+e^{-(x+z)})^2}$$

$$x=y-z$$

$$I(z)=4 e^z \int_z^\infty \frac{e^{-y} dy}{(e^y+e^{-y})^2}=4 e^z \int_z^\infty \frac{e^{-3y} dy}{(1+e^{-2y})^2}$$

$$y=- \ln t$$

$$I(z)=4 e^z \int_0^{e^{-z}} \frac{t^2 dt}{(1+t^2)^2}$$

$$t=e^{-z} u$$

$$I(z)=4 e^{-2 z} \int_0^1 \frac{u^2 du}{(1+e^{-2 z}u^2)^2}$$

$$u=\sqrt{v}$$

$$I(z)=2 e^{-2 z} \int_0^1 \frac{v^{1/2} dv}{(1+e^{-2 z} v)^2}=\frac{4}{3} e^{-2 z} {_2 F_1} \left(2, \frac{3}{2}; \frac{5}{2};-e^{-2z} \right)$$

Here we used Euler integral for the Hypergeometric function.

Using one of the Euler transformations, we have:

$${}_{2}F_{1}(a,b;c;p)=(1-p)^{-a}{}_{2}F_{1}\left(a,c-b;c;{\tfrac {p}{p-1}}\right)$$

$${_2 F_1} \left(2, \frac{3}{2}; \frac{5}{2};-e^{-2z} \right)= \frac{e^{2 z}}{(e^z+e^{-z})^2} {_2 F_1} \left(2, 1; \frac{5}{2};\frac{1}{e^{2z}+1} \right)$$

So we can write:

$$I(z)=\frac{1}{3 \cosh^2 z} {_2 F_1} \left(2, 1; \frac{5}{2};\frac{1-\tanh z}{2} \right)$$

Now we need to get the integral in the closed form to prove the identity the OP was interested in, but I leave it for other people.

I suggest using the $t$ integral from above, and integration by parts.

Answer 2

$$I=\int_{0}^{\infty}~ e^{-x} \mbox{sech}^2(x+z)~dx. z \in (0,\infty).~~~~(1)$$ Let us use $e^{-2(x+z)}=t$. to write $$I=\int_{0}^{e^{-2z}} \frac{t^{1/2} dt}{(1+t)^2)}.~~~~(2).$$ Next using An integral representation of $~_2F_1$ [Gradshteyn and Ryzhik [GR], 3.194.1], namely $$\int_{0}^{u} \frac{x^{\mu-1} dx}{(1+\beta x)^{\nu}} =\frac{u^{\mu}}{\mu} ~_2F_1(\nu,\mu;1+\mu,-\beta u).$$ We get $$I=\frac{4}{3} e^{-2z} ~_2F_1[2, 3/2; 5/2; -e^{-2z}]$$ Next using an identity [GR], 9.131.1] that $$_2F_1(a,b;c,t)=(1-t)^{-a} ~_2F_1[a,c-b,c;t/(t-1)],$$ we can write $$I=\frac{1}{3} \mbox{sech}^2z~_2F_1[2, 1; 5/2; \frac{1}{2}(1-\tanh z)].~~~~(3)$$ Alternatively, let us substitute $\tanh (x+z)=y$ in (1), then $$I=e^{z} \int_{\tanh z}^{1} e^{\tanh^{-1}y}~ dy= e^{z} \int_{\tanh z}^{1} \frac{1-y}{\sqrt{1-y^2}} dy=e^{z} [\pi/2-\sin^{-1}(\tanh z)-\mbox{sech}z]~~(4)$$ Finally, by equating $I$ in (3) and (4) we prove the required identity.