# Does my definition of an exponential-like function make any sense?

The definition of the exponential function is based on an infinite series $$e^x = \sum_{k=0}^{\infty}\frac{x^k}{k!}$$ To make things more complicated, we could replace the factorial with the Gamma function $$e^x = \sum_{k=0}^{\infty}\frac{x^k}{\Gamma(k+1)}$$ The advantage of the Gamma function is that it is defined also for continuous values, so we are tempted to replace the infinite sum with an integral $$\epsilon^x = \int_{k=0}^{\infty}\frac{x^k}{\Gamma(k+1)}dk$$ It turns out that this definition is somewhat similar to the exponential function, but not quite the same. The constant $$\epsilon$$ is computed by Wolfram Alpha as $$\epsilon = \int_{k=0}^{\infty}\frac{1^k}{\Gamma(k+1)}dk = 2.2665345...$$ but from the integral we have $$\epsilon^0 = \int_{k=0}^{\infty}\frac{0^k}{\Gamma(k+1)}dk = 0$$ which seems to be in contradiction to the rule $$a^0 = 1$$ for every $$a\neq 0$$.

So is my integral definition for $$\epsilon^x$$ consistent at all? How does the integral behave for $$x \rightarrow \pm\infty$$ ? Is there any literature about this kind of integral and it's relation to the definition of the exponential function?

• I think the first question we should ask, is for which $x$ your 'semi-exponent', so to speak, is convergent. – Jakobian Feb 7 at 22:28
• Of course that final integral is $0$; the integrand is $0$ everywhere except (possibly, depending on convention) at one point! – Theo Bendit Feb 7 at 22:37
• Your definition is flawed because you are assuming that the integral has the form $\epsilon ^{x}$ for some constant $\epsilon$. – Kavi Rama Murthy Feb 7 at 23:25
• Wait, so you're defining some function $f(x)$. You wrote it as $\epsilon^x$ but that's just the notation you picked. As you noticed, your $f$ is not the same as the function $x \mapsto e^x$. Is there a reason to be surprised that $f(0)\ne 1$ then? What I'm saying is it is not in any way a contradiction to the rule $a^0 = 1$. – I want to make games Feb 8 at 0:47

Your integral $$\ \int_{k=0}^{\infty}\frac{x^k}{\Gamma(k+1)}dk\$$ is perfectly well-defined for $$\ x \ge 0\$$. For $$\ x < 0\$$ you run into the problem that there is no unique definition for $$\ x^k\$$ when $$\ k\$$ is not an integer. In general, there are several possible values, $$\ \vert x\vert^k\,e^{(2n\pi+1)k i}\$$, to choose for it, and these becone infinite in number if $$\ k\$$ is irrational. Here, $$\ n\$$ can be any integer, but if you simply choose one particular value for it ($$\ n=0\$$, for instance), then the integral will be well-defined for $$\ x < 0\$$ as well, although it will typically be complex-valued in that case.
You have no grounds, however, for assuming that the integral will have the value $$\ \ \left(\int_{k=0}^{\infty}\Gamma(k+1)^{-1}dk\right)^x\$$, as a couple of commenters have already pointed out, and, in fact, it doesn't, except when $$\ x=1\$$, so the apparent inconsistency which puzzles you has resulted from this assumption, rather than from any problem with the definition of the integral itself.
You can show that the integral is well-defined for $$\ x \ge 0\$$, and obtain bounds for it, by observing that $$\frac{x^k}{\Gamma(k+1)}\$$ is bounded and continuous for $$k\ge0\$$, and for $$\ 0\le i\le k\le i+1$$ we have $$\ i\,! \le \Gamma\left(k+1\right) \le \left(i+1\right)\,!\$$ and $$\ x^i \le x^k \le x^{i+1}\$$ if $$\ x \ge 1\$$, or $$\ x^{i+1} \le x^k \le x^i\$$ if $$x < 1\$$. Therefore $$\frac{x^i}{\left(i+1\right)\,!} \le \frac{x^k}{\Gamma\left(k+1\right)} \le \frac{x^{i+1}}{i\,!}\$$ if $$\ x \ge 1\$$, and $$\frac{x^{i+1}}{\left(i+1\right)\,!} \le \frac{x^k}{\Gamma\left(k+1\right)} \le \frac{x^i}{i\,!}\$$ if $$x < 1$$.
Thus, for $$\ x\ge 1\$$, $$\sum_{i=0}^r \frac{x^i}{\left(i+1\right)\,!}\le \ \int_{k=0}^{r+1}\frac{x^k}{\Gamma(k+1)}dk\ \le \sum_{i=0}^r \frac{x^{i+1}}{i\,!}\le x\,e^x\ \ ,$$ so the integral converges as $$\ r\rightarrow\infty\$$, is bounded below by $$\ \frac{e^x-1}{x}\$$ and above by $$\ x\,e^x\$$.
On the other hand, for $$\ x < 1\$$, $$\sum_{i=0}^r \frac{x^{i+1}}{\left(i+1\right)\,!}\le \ \int_{k=0}^{r+1}\frac{x^k}{\Gamma(k+1)}dk\ \le \sum_{i=0}^r \frac{x^i}{i\,!}\le e^x\ \ ,$$ so the integral again converges as $$\ r\rightarrow\infty\$$, is bounded below by $$\ e^x-1\$$ and above by $$\ e^x\$$.