I need to calculate $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)^n} e^{-\rho x^2 + a x} dx$$ where $n \in \mathbb{N}$, $\rho > 0$ and $a \in \mathbb{R}$, but I don't know how to follow. I've tried to include the expression in symbolic software trying to get a result with respect other functions, but nothing. I start thinking about approximating the integral using numerical integration, but before, I would like to be sure that the integral can not be expressed with respect other functions. Does anybody knows if I should go directly for numerical integration?

I am really lost, thank you in advance.


Some particular cases can be computed using Wolfram Alpha. It seems that

$$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 + x} dx = \frac{\sqrt{\pi/\rho}}{2}$$

and $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 + 3 x} dx = \int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 - 3 x} dx = (2 e^{1/\rho}-1)\frac{\sqrt{\pi/\rho}}{2}.$$

More generally (thank you JanG) we have $$\int_{-\infty}^{+\infty} \frac{1}{\left(e^x+ e^{-x}\right)} e^{-\rho x^2 + (2m+1) x} dx = \frac{\sqrt{\pi/\rho}}{2} (-1)^m \left( 1 + 2 \sum_{\ell=1}^m (-1)^\ell e^{\ell^2/\rho}\right)$$ for $m$ non-negative integer.

  • 3
    $\begingroup$ have you tried contour integration? $\endgroup$ – user159517 Nov 19 '16 at 13:56
  • $\begingroup$ I've been thinking with contour integration as suggest, but the function has singularities in $z=\frac{2 k +1}{2} \pi i$, for $k \in \mathbb{Z}$. I've seen examples where the function has a finite number of singularities, but I don't know how to follow in this case. $\endgroup$ – marc1s Nov 20 '16 at 19:03
  • 2
    $\begingroup$ Since $a^{2n+1}+b^{2n+1} = (a+b)(a^{2n}-a^{2n-1}b + a^{2n-2}b^2 +\dots + b^{2n})$ the integral can be evaluated for $a = 2n+1, n$ not a negative integer. I got the value \begin{equation*} \dfrac{\sqrt{\pi/\rho}}{2}(-1)^{n}\left(1+2\sum_{k=1}^{n}(-1)^{k}e^{k^{2}/\rho}\right). \end{equation*} $\endgroup$ – JanG Nov 23 '16 at 11:58
  • $\begingroup$ Nice JanG! That generalise the integral resolution a little more. $\endgroup$ – marc1s Nov 23 '16 at 14:06
  • 2
    $\begingroup$ Ramanujan considered very similar looking integrals in his notebooks...math.stackexchange.com/questions/1987764/… $\endgroup$ – tired Nov 28 '16 at 14:00

Integral in general is terrible, so in practice I would suggest numerical methods. However, it can be represented as a series in a complicated way.

First, let's deal with the power in the denominator. We have:

$$\int_{-\infty}^{+\infty} \frac{e^{-\rho x^2 + a x}}{\left(e^x+ e^{-x}\right)^n} dx=\int_{-\infty}^{+\infty} \frac{e^{-\rho x^2 + (a+n) x}}{\left(1+ e^{2x}\right)^n} dx$$

Let's introduce a new parameter $b$:

$$I_n(\rho,a,b)=\int_{-\infty}^{+\infty} \frac{e^{-\rho x^2 + (a+n) x}}{\left(b+ e^{2x}\right)^n} dx$$

Introducing a new integral:

$$J(\rho,a,b,n)=\int_{-\infty}^{+\infty} \frac{e^{-\rho x^2 + (a+n) x}}{b+ e^{2x}} dx$$

We have:

$$I_n=\frac{(-1)^{n-1}}{(n-1)!} \frac{\partial^{n-1} }{\partial~ b^{n-1}} J \tag{1}$$

Thus, we have simplified the problem to finding $J$ in analytic form. It's still too complicated to obtain a closed form (as far as I know), but the series solution is possilbe.

Let $y=2x$, $\rho=4R$ and $a+n=2A$:

$$J=\frac{1}{2} \int_{-\infty}^{+\infty} \frac{e^{-R y^2 + A y}}{b+ e^y} dy=\frac{1}{2} \int_{-\infty}^{\ln b} \frac{e^{-R y^2 + A y}}{b+ e^y} dy+\frac{1}{2} \int_{\ln b}^{+\infty} \frac{e^{-R y^2 + A y}}{b+ e^y} dy$$

Separation of the limits allows us to represent each denominator as a series:

$$\frac{1}{b+ e^y}=\sum_{k=0}^\infty (-1)^k b^{-k-1} e^{ky}, \qquad y< \ln b$$

$$\frac{1}{b+ e^y}=\sum_{l=0}^\infty (-1)^l b^l e^{-(l+1)y}, \qquad y> \ln b$$

Skipping the standard integration we obtain:

$$J=\frac{1}{4} \sqrt{\frac{\pi }{R}} \left[\sum _{k=0}^{\infty } (-1)^k b^{-k-1} e^{\frac{(A+k)^2}{4 R}} \left(\text{erf}\left[\sqrt{R} \ln b-\frac{A+k}{2 \sqrt{R}}\right]+1\right)+ \\ + \sum _{l=0}^{\infty } (-1)^l b^l e^{\frac{(A-l-1)^2}{4 R}} \left(\text{erf}\left[\frac{A-l-1}{2 \sqrt{R}}-\sqrt{R} \ln b \right]+1\right)\right] \tag{2}$$

Here $\text{erf}$ is the error function.

Using $(1)$ and $(2)$ we obtain a (terrible) analytic expression for the general integral $I_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.