Integrating $\int_{0}^{\infty} \mathrm{d}x \frac{x^{2 k +1}}{\cosh \left( x- y\right)} $

Question

For $y>0$, I'm trying to compute in a closed form the following integral: $$ I =\int_{0}^{\infty} \frac{x^{2 k +1}}{\cosh \left( x- y\right)} \mathrm{d} x$$

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Here is my attempt to a solution:

$$I = I_1 + I_2 = \int_{0}^{y}\mathrm{d}x \frac{x^{2 k +1}}{\cosh \left( x- y\right)} + \int_{y}^{\infty} \mathrm{d}x \frac{x^{2 k +1}}{\cosh \left( x- y\right)} $$

I believe that you can compute $I_2$ by expanding the denominator as follows:

\begin{align}I_2&= 2\int_{y}^{\infty} \mathrm{d}x \frac{x^{2 k +1}}{e^{\left( x- y\right)} + e^{-\left( x- y\right)}} \\&= 2\int_{y}^{\infty} \mathrm{d}x e^{-\left( x- y\right)} \frac{x^{2 k +1}}{1 + e^{- 2\left( x- y\right)}} \\&= 2 \int_{0}^{\infty} \mathrm{d}z e^{-z} \frac{(y + z)^{2 k +1}}{1 + e^{-2 z}} \\&= 2 \sum_{n=0}^{2 k + 1} \binom{2k +1}{ n} y^{2k + 1- n} \int_{0}^{\infty} \mathrm{d}z e^{-z} \frac{z^n}{1 + e^{-2 z}}\end{align}

Now we recall the definition of the Lerch Transcendent: $$\Phi(z, s, a) = \sum_{m=0}^{\infty} \frac{z^m}{(z + a)^s}$$

And we obtain:

\begin{align}I_2 &= 2 \sum_{n=0}^{2 k + 1} \binom{2k +1}{ n} y^{2k + 1- n} \sum_{m=0}^{\infty} (-1)^m \int_{0}^{\infty} \mathrm{d}z e^{-(1 + 2m)z} z^n \\&= 2 \sum_{n=0}^{2 k + 1} \binom{2k +1}{ n} \Gamma (n + 1) 2^{1+ n} y^{2k + 1- n} \sum_{m=0}^{\infty} \frac{(-1)^m}{(n + 1/2)^{n +1}} \\&= 2 \sum_{n=0}^{2 k + 1} \binom{2k+1}{ n} \Gamma (n + 1) 2^{1+ n} y^{2k + 1- n} \Phi(-1, n+ 1, 1/2)\end{align}

Therefore the second piece evaluates to a polynomial in $y$. My main issue is about $I_1$: I can not understand what is the best strategy to tackle the integration, since a similar treatment to $I_2$ does not seem to be available. I will be very grateful to anyone helping on this problem.

Answer 1

Alright, let's focus on $I_1$:

$$ I_1 = \int_{0}^{y}\mathrm{d}x \frac{x^{2 k +1}}{\cosh \left( x- y\right)} $$

Given the nature of $\cosh x$, we can rewrite the integral as:

$$I_1 = \int_{0}^{y}\mathrm{d}x \frac{x^{2 k +1}}{\frac{e^{x-y} + e^{-x+y}}{2}}$$

Simplifying further:

$$I_1 = 2 \int_{0}^{y}\mathrm{d}x x^{2 k +1} \frac{1}{e^{x-y} + e^{-x+y}}$$

Now, make a substitution to transform the integral in a way that's easier to deal with:

Let's use $z = x - y$, which implies $x = z + y$.

When $x = 0 $, $z = -y $ and when $x = y$, $z = 0 $. So the new limits of integration are $[-y, 0]$.

This gives:

$$ dx = dz $$ $$x^{2 k +1} = (z + y)^{2 k +1}$$

Plug these in to get:

$$I_1 = 2 \int_{-y}^{0} \mathrm{d}z (z + y)^{2 k +1} \frac{1}{e^z + e^{-z}} $$

This form should be easier to work with, though it's still a challenge due to the nature of the exponential functions in the denominator.

To proceed, consider the series expansion:

$$\frac{1}{e^z + e^{-z}} = \frac{e^{-z}}{1 + e^{-2z}}$$

You can expand $ \frac{e^{-z}}{1 + e^{-2z}} $ in a power series around $ e^{-2z} $:

$$ \frac{e^{-z}}{1 + e^{-2z}} = e^{-z} (1 - e^{-2z} + e^{-4z} - e^{-6z} + \dots) $$

Now, integrating term by term:

$$ \int \mathrm{d}z (z + y)^{2 k +1} e^{-z} e^{-2nz} $$

For each term, you'll have:

$$ \int_{-y}^{0} \mathrm{d}z (z + y)^{2 k +1} e^{-(1+2n)z} $$

This integral will yield polynomials of $y$ when expanded, albeit the calculations may become intensive for large $ k $.

However, an analytical closed-form solution for the integral as a function of $ y $ and $ k $ without the series expansion may not be readily obtainable due to its complexity.

Answer 2

Starting form @Arham Mehta's answer $$\int_{-y}^{0} (z + y)^{2 k +1} e^{-(1+2n)z}\,\mathrm{d}z =$$ $$\frac{e^{(2 n+1) y}}{(2 n+1)^{2( k+1)} }\Big(\Gamma (2 (k+1))-\Gamma (2( k+1),(2 n+1) y)\Big)$$