# Symbolic solution to integral needed

Is there a way to get the symbolic result of the following integral?

$$f(x) = \int_{-\infty}^{\infty} \text{Tanh}\left(\frac{u}{2}\right)e^{-\frac{(u-x)^2}{4x}}du$$

Edit:

Where $-4 \leq x \leq 4$ is a real value. It suffice to have $|x|\leq4$, but $x$ can be larger for other applications.

Thanks for helping.

• for: x<=0 integral is divergent. ? – Mariusz Iwaniuk Aug 3 '18 at 15:04
• @Mariusz: Thanks... If I change the range of $x$ to $0 < x \leq 4$? – Arvid Aug 3 '18 at 15:06
• Hard integral they are difficult to solve,or impossible to find closed-form. Only hope is numeric. math.stackexchange.com/questions/1754192/… – Mariusz Iwaniuk Aug 3 '18 at 15:25
• It looks like if you were interested in $x>20$ that $3.545 \sqrt{x}$ (and maybe that is exactly $2 \sqrt{\pi x}$) would be a good approximation. – JimB Aug 3 '18 at 17:51
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Using the identities $\color{red}{\tanh (x)=\frac{\exp (2 x)-1}{\exp (2 x)+1}}$ and $\color{red}{\sum _{k=0}^{\infty } (-1)^k \exp ^k(x)=\frac{1}{1+e^x}}$,

we obtain:

$$\int_{-\infty }^{\infty } \tanh \left(\frac{x}{2}\right) \exp \left(-\frac{(x-a)^2}{4 a}\right) \, dx=\\\int_{-\infty }^{\infty } \frac{(\exp (x)-1) \exp \left(-\frac{(x-a)^2}{4 a}\right)}{\exp (x)+1} \, dx=\\\sum _{k=0}^{\infty } \int_{-\infty }^{\infty } (-1)^k e^{k x-\frac{(-a+x)^2}{4 a}} \left(-1+e^x\right) \, dx=\\\sum _{k=0}^{\infty } -2 (-1)^k \sqrt{a} \left(e^{a k (1+k)}-e^{a (1+k) (2+k)}\right) \sqrt{\pi }$$

Integrate[(-1)^k E^(k x - (-a + x)^2/(4 a)) (-1 + E^x), {x, -Infinity,
Infinity}, Assumptions -> {a \[Element] Reals, k >= 0}, PrincipalValue -> True]
(* -2 (-1)^k Sqrt[a] (E^(a k (1 + k)) - E^(a (2 + 3 k + k^2))) Sqrt[\[Pi]] *)


Closed form of sum probably doesn't exist. The sum is very fast convergent, but for a>0, it is divergent. We can compute only the imaginary part of a for a<0. If a<0, the real part of the sum is 0.

With a->Im[-Infinity], $-2 \sqrt{-\pi a}$ is a good approximation, then is exact formula.

 f[a_] := Sum[-2 (-1)^k Sqrt[a] (E^(a k (1 + k)) - E^(a (2 + 3 k + k^2))) Sqrt[\[Pi]], {k, 0, 10}] // Im
Plot[{f[a], -2 Sqrt[-Pi* a]}, {a, -4, 0}, PlotLegends -> {Sum, -2 Sqrt[-Pi a]}, PlotStyle -> {Red, {Dashed, Black}}]


We can find the first term of the asymptotic for $a\to \infty$:

INT = Integrate[InverseZTransform[(E^(-((a - x)^2/(4 a))) (-1 + E^x))/(b + E^x), b, s], {x, -Infinity, Infinity},
Assumptions -> {s > 0, a > 0}];
ZTransform[INT, s, 1]


$2 \sqrt{a} \sqrt{\pi } \left(-\left(\mathcal{Z}_s\left[e^{a s (1+s)}\right](-1)\right)+\mathcal{Z}_s\left[e^{2 a (1+s)+a s (1+s)}\right](-1)\right)$

then:

$\int_{-\infty }^{\infty } \tanh \left(\frac{x}{2}\right) \exp \left(-\frac{(x-a)^2}{4 a}\right) \, dx\approx2 \sqrt{\pi a}$ if $a\to \infty$

MMA can't find ZTransform, but we have good approximation (see comment by user JimB).

The numerical workaround could be:

i[x_?NumericQ]:= NIntegrate[Tanh[u/2] Exp[-((u - x)^2/(4 x))], {u, -∞, ∞}]
Plot[i[x], {x, 0, 4 }]


which gives you an idea of the unknown shape of the integral .