Vtyjkyuk

So I have this integral $$
I=int_0^cexpleft(-a^2x^2-fracb^2x^2right),dx, quad(a,b>0)
$$
This is what I tried to write it in terms of error function.
$$
I=e^-2abint_0^cexpleft(-a^2left(x-fracbaxright)^2right),dx
$$
By Glasser's master theorem
$$
I=e^2abint_0^ce^-a^2x^2,dx=fracsqrtpi e^2ab2atexterf(ac)
$$
But according to eq 5 here the correct answer is
$$
I=fracsqrtpi4a(e^2abtexterf(ax+b/x)+e^-2abtexterf(ax-b/x))
$$
What went wrong here and how to get the correct answer?

Without loss of generality we can assume $c>0$ , since $c<0$ only changes the sign of the integral.

As already discussed in the comments, Glasser's master theorem does not apply here. We can instead write the integral as
$$ I = mathrme^2ab int limits_0^c mathrme^-(a x + b/x)^2 , mathrmd x , .$$

Now we would like to let $a x + b/x = t$ . Since this map is not bijective, we need to be careful here! For $0<x<sqrtb/a$ the correct inverse is
$$x = frac12a left(t-sqrtt^2-4abright) , ,$$
while for $x > sqrtb/a$ we have to use
$$x = frac12a left(t+sqrtt^2-4abright) , .$$

For $c leq sqrtb/a , Leftrightarrow , a c - b/c leq 0$ we only need the first version. The substitution yields
$$ I = fracmathrme^2ab2a int limits_ac+b/c^infty left[fractsqrtt^2-4ab - 1right]mathrme^-t^2 , mathrmd t , . $$
Changing variables once more for the first part ($u = sqrtt^2 - 4ab$) and using the complementary error function $operatornameerfc = 1-operatornameerf$ we obtain
$$ I = fracsqrtpi4a left[mathrme^-2ab operatornameerfc(|ac - b/c|) - mathrme^2ab operatornameerfc(ac + b/c)right] , .$$

For $c > sqrtb/a , Leftrightarrow , a c - b/c > 0$ we have to split the integral:
beginalign
I &= mathrme^2ab leftint limits_0^sqrtb/a mathrme^-(a x + b/x)^2 , mathrmd x + int limits_sqrtb/a^c mathrme^-(a x + b/x)^2 , mathrmd x right \
&= fracmathrme^2ab2a leftint limits_2sqrtab^infty left[fractsqrtt^2-4ab - 1right]mathrme^-t^2 , mathrmd t + int limits_2sqrtab^ac+b/c left[1 + fractsqrtt^2-4abright]mathrme^-t^2 , mathrmd t right , .
endalign
Evaluating the remaining integrals as in the other case we find
$$ I = fracsqrtpi4a left[mathrme^-2ab [1+operatornameerf(ac - b/c)] - mathrme^2ab operatornameerfc(ac + b/c)right] , .$$

Both results can be simplified and the final result for arbitrary $c>0$ is
beginalign
I &= fracsqrtpi4a left[mathrme^-2ab operatornameerfc(b/c - ac) - mathrme^2ab operatornameerfc(b/c + ac)right] \
&= fracsqrtpi4a left[mathrme^2ab operatornameerf(ac+b/c) + mathrme^-2ab operatornameerf(ac-b/c) - 2 sinh(2ab)right] , ,
endalign
which agrees with the result from the table (except for a constant, of course). Letting $c to infty$ we obtain the integral
$$ int limits_-infty^infty mathrme^-a^2 x^2 - b^2 /x^2 , mathrmd x = fracsqrtpia mathrme^-2ab , , $$
which can also be computed using the master theorem.