# On $\int_0^1\left(\int_0^\infty\frac{\operatorname{gd}(x+y)}{\exp(x+y)}dx\right)dy$, being $\operatorname{gd}(u)$ the Gudermannian function

While I was playing with Wolfram Alpha online calculator, to create double integrals involving negative exponentials and the so-called Gudermannian function, denoted in this post as $\operatorname{gd}(u)$, I wondered that should be possible to get the closed-form of $$\int_0^1\left(\int_0^\infty\frac{\operatorname{gd}(x+y)}{e^{x+y}}dx\right)dy.\tag{1}$$ I believe that $(1)$ hasn't a very nice closed-form (I was trying to define integrals involving these functions with a nice closed-form).

Question. Can you justify/calculate the closed-form of $(1)$? Many thanks.

• Many thanks to the user who upvoted the post. – user243301 May 8 '18 at 14:08
• You could also accept my answer. – Somos May 8 '18 at 17:30
• @Somos I am going to wait if there are more contributions, before accepting an answer. Many thanks. – user243301 May 8 '18 at 17:47
• Sorry, that is ok. I should have waited at least a few days. – Somos May 8 '18 at 18:13

In[1] := Integrate[Integrate[Gudermannian[x+y]/Exp[x+y],{x,0,Infinity}],{y,0,1}]//Simplify//InputForm