A definite integral of the hyperbolic secant function in terms of the Polygamma function given by: x=PolyGamma1,+where Re(μ)>0.
π/2
∫
0
2
Sech[μTan[x]]
1
π
1
2
μ
π
In[]:=
Plot3D[,{μ,-3,3},{x,-π/2,π/2},AxesLabelAutomatic]
2
Sech[μTan[x]]
Out[]=
In[]:=
fun1[μ_]:=NIntegrate[,{x,0,π/2}],//N
2
Sech[μTan[x]]
PolyGamma1,+
1
2
μ
π
π
In[]:=
fun1closed[μ_]:=Integrate[,{x,0,π/2}],
2
Sech[μTan[x]]
PolyGamma1,+
1
2
μ
π
π
In[]:=
fun1closed[μ]
Out[]=
x,
π
2
∫
0
2
Sech[μTan[x]]
PolyGamma1,+
1
2
μ
π
π
In[]:=
fun1[π+]
Out[]=
{0.27626-0.0783859,0.27626-0.0783859}
In[]:=
fun1[π]
Out[]=
{0.297557,0.297557}
In[]:=
fun1closed[π]
Out[]=
x,
π
2
∫
0
2
Sech[πTan[x]]
-4+
2
π
2
π
In[]:=
fun1[π/4]
Out[]=
{0.809105,0.809105,0.809105}
In[]:=
fun1closed[π/4]//FunctionExpand//FullSimplify
Out[]=
x,-+π
π
2
∫
0
2
SechπTan[x]
1
4
8Catalan
π
We can also take a derivative with respect to μ to get Zeta values
In[]:=
D,,μ/.{μπ/4}//FunctionExpand//FullSimplify
2
Sech[μTan[x]]
PolyGamma1,+
1
2
μ
π
π
In[]:=
NIntegrateTan[x]TanhπTan[x],{x,0,π/2},-π+//N
2
SechπTan[x]
1
4
1
4
56Zeta[3]
2
2
π
Out[]=
{0.268635,0.268635}
In[]:=
IntegrateTan[x]TanhπTan[x],{x,0,π/2},-π+
2
SechπTan[x]
1
4
1
4
56Zeta[3]
2
2
π
Out[]=
Tan[x]TanhπTan[x]x,-π+
π
2
∫
0
2
SechπTan[x]
1
4
1
4
28Zeta[3]
2
π
We can take the infinite sum over μ∈[1,∞) to get a constant value in terms of a definite integral and sum of the polygamma function given by:
In[]:=
∞
∑
μ=1
2
Sech[x]
2
x
2
μ
PolyGamma1,+
1
2
μ
π
μπ
In[]:=
x,//N
∞
∫
0
(-1+πxCoth[πx])
2
Sech[x]
2
2
x
∞
∑
μ=1
PolyGamma1,+
1
2
μ
π
πμ
Out[]=
{1.30636,1.30636}