Closed form of $\displaystyle \int_{1}^{x}\cos\left(\pi\left(n+1\right)\right)e^{\frac{x}{n}}dn$

Question

The graph of $$ \int_{1}^{x}\cos\left(\pi\left(n+1\right)\right)e^{\frac{x}{n}}dn $$ looks really similar to $\Gamma(x)$ plus some sinusoidal function. Wolfram Alpha couldn't give an answer. Is there some way to evaluate this integral? Maybe in terms of the gamma function?

Answer 1

Note that $$ \int_1^x {\cos (\pi (t + 1)){\rm e}^{x/t} {\rm d}t} = - \int_1^x {\cos (\pi t){\rm e}^{x/t} {\rm d}t} = - \int_{1/x}^1 {\frac{1}{{u^2 }}\cos \left( {\frac{\pi }{u}} \right){\rm e}^{xu} {\rm d}u} . $$ Now assume that $x\ge 2$. We split the integral at $u=\frac{1}{2}$. Then $$ \left| {\int_{1/x}^{1/2} {\frac{1}{{u^2 }}\cos \left( {\frac{\pi }{u}} \right){\rm e}^{xu} {\rm d}u} } \right| \le \frac{1}{2}x^2 {\rm e}^{x/2} . $$ By Watson's lemma \begin{align*} - \int_{1/2}^1 {\frac{1}{{u^2 }}\cos \left( {\frac{\pi }{u}} \right){\rm e}^{xu} {\rm d}u} & = - {\rm e}^x \int_0^{1/2} {\frac{1}{{(1 - t)^2 }}\cos \left( {\frac{\pi }{{1 - t}}} \right){\rm e}^{ - xt} {\rm d}t} \\ & \sim \frac{{{\rm e}^x }}{x}\left( {1 + \frac{2}{x} + \frac{{6 - \pi ^2 }}{{x^2 }} + \frac{{12(2 - \pi ^2 )}}{{x^3 }} + \ldots } \right) \end{align*} as $x\to +\infty$. Accordingly, $$ \int_1^x {\cos (\pi (t + 1)){\rm e}^{x/t} {\rm d}t} \sim \frac{{{\rm e}^x }}{x}\left( {1 + \frac{2}{x} + \frac{{6 - \pi ^2 }}{{x^2 }} + \frac{{12(2 - \pi ^2 )}}{{x^3 }} + \ldots } \right) $$ as $x\to +\infty$. If you compare this with Stirling's formula for $\Gamma(x)$, you can see that your integral grows much slower than $\Gamma(x)$. Note also that the oscillatory contribution does not affect the value of the integral for large $x$.

Answer 2

We use a cosine series:

$$-\int_1^x \cos(\pi n)e^\frac xn dn=\sum_{k=0}^\infty \int_1^\infty\frac{(-1)^{k+1} \pi^{2k} n^{2k}e^\frac xn }{(2k)!}dn$$

Using $n\to \frac 1n$ implies the En function E$_v(z)$ and “generalized” incomplete gamma function $\Gamma(a,x,y)$ integral representations:

$$\int_1^x \cos(\pi n)e^\frac xn dn =x\sum_{k=0}^\infty\frac{(-1)^k (\pi n)^{2k}\Gamma\left(-2k-1,-x,-\frac x n\right)}{(2k)!}=\sum_{k=0}^\infty\frac{(-1)^k\pi^{2k}\left(n^{2k+1}\text E_{2k+2}\left(-\frac xn\right)-\text E_{2k+2}(-x)\right)}{(2k)!}$$

Shown here is the gamma result. This series does not imply a closed form.

Answer 3

Notice that $\cos(\pi(n+1)) = -\cos(\pi n)$.

Don't know if this could help but as $x\to +\infty$ we have an asymptotic estimation for the integral:

$$\int_1^x e^{x/n} \cos(\pi n)\ \text{d}n \sim -x \text{Ei}(x)-\text{Ei}(1) x \cos (\pi x)+e^x+e x \cos (\pi x)$$

Where Ei is the Exponential Integral function. There is a relation between the Exponential Integral function and the incomplete Gamma Function:

$$E_n(x) = \int_1^{+\infty} \dfrac{e^{-xt}}{t^n}\ \text{d}t \equiv x^{n-1} \Gamma(1-n, x)$$

There is a connection between Ei and $E_n$ too, but you will find more interesting the whole reading, like here as a start: https://en.wikipedia.org/wiki/Exponential_integral#Generalization

Answer 4

Maple also doesn't find a closed-form antiderivative. This is the "pure transcendental" case, where Maple (and I think Mathematica/Wolfram Alpha too) should have a full implementation of the Risch algorithm, so I think we can be confident that there is no elementary antiderivative. It's not impossible that there could be a closed-form antiderivative involving non-elementary functions such as $\Gamma$, but since neither Maple nor Wolfram Alpha find one , I think it's unlikely.