I want to solve this integral $$\int_{0}^{\frac{\pi}{2}} e^{-(\pi \tan(x) - 1)^2} \, dx$$
I tried Wolfram alpha but it is unable to evaluate
I tried this code
Integrate[Exp[-(Pi*Tan[x] - 1)^2], {x, 0, Pi/2}]
However, Wolfram Alpha is unable to evaluate it.
I have never used any (CAS) or software,If Wolfram Alpha is unable to evaluate it, does that mean Mathematica will also be unable to evaluate it?
We might be able to get a semi-analytic approximation to the integral by noting that the transformation $u=tan(x)$ changes the original integral to the form $$ I = \int_0^\infty du\,\frac{\exp \left(-(\pi\, u-1)^2\right)}{1+u^2}.$$
Alas, this is a form which a closed form is known not to exist (as far as I am aware).
However, we can expand the denominator around $u=0$ in a series expansion $\frac{1}{1+u^2}\simeq 1-u^2+u^4+...$ and then perform the Gaussian integrals analytically with Mathematica:
ser\[Ellipsis]u =
Series[1/(1 + u^2), {u, 0, 30}] // FullSimplify // Normal
(* 1 - u^2 + u^4 - u^6 + u^8 - u^10 + u^12 - u^14 + u^16 - u^18 + u^20 - u^22 + u^24 - u^26 + u^28 - u^30 *)
However, the series is asymptotic and by comparing with the numerical result, it seems the best accuracy is met for 7 terms:
int = NIntegrate[Exp[-(Pi*u - 1)^2]/(1 + u^2), {u, 0, \[Infinity]},WorkingPrecision -> 30]
(* 0.454150685673164197286527404424 *)
and then the comparison:
data = Table[{n,100 (1 - NIntegrate[Exp[-(Pi*u - 1)^2] ser\[Ellipsis]u[[1 ;; n]], {u, 0, \[Infinity]}, WorkingPrecision -> 50]/int)}, {n, 1, Length[ser\[Ellipsis]u]}];
ListLogLogPlot[data // Abs, Frame -> True, FrameStyle ->
Joined -> True, Mesh -> Automatic, FrameLabel -> {"n", "% diff"}]
Keeping then the 7 terms, results in the following approximation (good to about 0.5%):
Integrate[Exp[-(Pi*u - 1)^2] ser\[Ellipsis]u[[1 ;; 7]], {u, 0, \[Infinity]}] // FullSimplify
(* (1/(128 E \[Pi]^13))(363470 - 4 \[Pi]^2 (10895 - 1530 \[Pi]^2 + 260 \[Pi]^4 - 56 \[Pi]^6 + 16 \[Pi]^8) + E Sqrt[\[Pi]] (479851 - 57574 \[Pi]^2 + 8100 \[Pi]^4 - 1384 \[Pi]^6 + 304 \[Pi]^8 - 96 \[Pi]^10 + 64 \[Pi]^12) (1 + Erf[1])) *)
I'm not sure if this accuracy is good for what you want, but it might provide some intuition on the behavior of the integral.
Also, there might be a way to improve the previous mathematical analysis, but that might be a question for another Stack.
Another way of attack: After substitution y = -1+[Pi] Tan[x] the integral is
NIntegrate[Pi/(E^y^2*(Pi^2 + (1 + y)^2)), {y, -1, Infinity}]
(* 0.454151*)
Inserting
Integrate[Exp[(-(Pi^2 + (1 + y)^2))*z], {z, 0, Infinity}]
and exchanging integration variables the form is
(Pi^(3/2)/2)*
NIntegrate[(1 + Erf[1/Sqrt[1 + z]])/(E^(z*(Pi^2 + 1/(1 + z)))*
Sqrt[1 + z]), {z, 0, Infinity}]
which after another substitution x = (1+z)^-1/2 gets to
Pi^(3/2)*
NIntegrate[E^(-1 + Pi^2 - Pi^2/x^2 + x^2)/
x^2 + (E^(-1 + Pi^2 - Pi^2/x^2 + x^2)*Erf[x])/x^2, {x, 0, 1}]
The first term is integrated by Mathematica and we're left with the second integral
Pi^(3/2)*(E^(-1 + Pi^2)/(2*
Sqrt[Pi]) + (I*(DawsonF[1 + I*Pi] - DawsonF[1 - I*Pi]))/(2*
Pi) + E^(-1 + Pi^2)*
NIntegrate[(E^(-(Pi^2/x^2) + x^2)*Erf[x])/x^2, {x, 0, 1}])
which through integration by parts goes to
(-((Erf[Pi - I] + Erf[Pi + I])/(4*Sqrt[Pi])))*Erf[1] +
(1/(2*Pi))*
NIntegrate[(Erf[Pi/x - I*x] + Erf[Pi/x + I*x])/E^x^2, {x, 0, 1}]
Here I' m stuck . Maybe somebody finds the remaining integral . The best I could do was approximating the Erf term with 2:
N[Table[{Erf[Pi/x - I*x] + Erf[Pi/x + I*x]}, {x, 1/100, 1, 1/10}], 10]
(* {{2.000000000 + 0.*10^-10 I}, {2.000000000 +
0.*10^-10 I}, {2.000000000 + 0.*10^-10 I}, {2.000000000 +
0.*10^-10 I}, {2.000000000 + 0.*10^-10 I}, {2.000000000 +
0.*10^-10 I}, {2.000000000 + 0.*10^-10 I}, {1.999999999 +
0.*10^-10 I}, {1.999999846 + 0.*10^-10 I}, {1.999995465 +
0.*10^-10 I}} *)
so an approximative result (better than 1 % relative error) is
(Pi/4)*E^(Pi^2 - 1)*(1 + Erf[1])*(2 - I*Erfi[1 - I*Pi] + I*Erfi[1 + I*Pi])
NIntegrate[Exp[-(Pi*Tan[x] - 1)^2], {x, 0, Pi/2}]produces0.454151. $\endgroup$