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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.

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

In[1] := Integrate[Integrate[Gudermannian[x+y]/Exp[x+y],{x,0,Infinity}],{y,0,1}]//Simplify//InputForm
Out[1]//InputForm= 1 - Pi^2/24 - Gudermannian[1]/E + Log[2/(1 + E^2)] - PolyLog[2, -E^2]/2
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  • $\begingroup$ Many thanks for your answer. $\endgroup$ – user243301 May 8 '18 at 17:11

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