Improper integral with exponential function

Abstract

In this paper, we explore the improper integral with exponential function f = x^x is approached to infinite series, and also prove the convergence of these series. An improper integral converges if the limit defining it exists. We use Maple code to calculate the infinite series. The application of improper integral appear in several domain in science. As an application in this paper, three examples are given to illustrate the effectiveness of our main result.

Keywords

Improper integral exponential function infinite series

1 Introduction

In mathematical analysis, an improper integral provides an excellent tool in various scientific and mathematical fields due to high profile accuracy and usability. Improper integral has made great advances in the past years, especially the Laplace transform, and the Fourier transform. Also, many special functions like Beta and Gamma are defined using improper integrals, which appear in a lot of problems and applications in probability and statistics [13]. For example, the electric potential created by a charged sphere of radius R for r ≥ R is given by: $V (r) = - \int_{\infty}^{r} E (r) dr$ , where E is the electric field (modulus) generated by the sphere and r is the distance from the center of the sphere. This also has a very simple physical meaning since the integral (when the minus sign is considered) represents the work to be done by somebody (or something) to bring the considered amount of electrical charge (responsible for creating the electric field E) from ∞ to r. In addition, architects use integration in determining the amount of the necessary material to construct curved shape constructions (e.g. dome over a sports arena) and also to measure the weight of that structure. Integrals are used to improve the architecture not only of buildings but also of important infrastructures such as bridges. Besides, integrals are used in the field of Electrical Engineering to determine the exact length of power cable needed to connect two substations, which are miles away from each other. Exponential functions are best applied in real world problems especially when the causation factor itself propagates against another factor. Its omnipresence in pure and applied mathematics has led the famous mathematician Rudin [13]. Exponential functions have many different applications, such as predicting population growth, computation in astronomy, economics, finance and banking, radioactive decay, and heat reduction of objects in the natural environment, especially modeling the above phenomena. For more details of exponential function and its applications, we refer the readers to see [1 , 10–12]. In the past many scholars have studied the improper integral with exponential functions. In [7], Euler, proved the necessary and sufficient condition for the convergence of undersetn → ∞ lim xⁿ. In [5], Baker and Rippon discussed the convergence of infinite exponential for the case a is a complex number. In [6], Macdonnell investigate some critical points on the hyper power function f = x^{x
^{x
^⋯}} also some properties of the function f (x) = undersetn → ∞ lim xⁿ is introduced. Also some scholars studied and discussed a different topics, considering the following explanation: The serial operations as production, power and ultra power respectively. As we know, due to algebraic properties of power, it can be extended from natural to rational numbers (and then other numbers) but this cannot be done for ultra power, because it does not have any useful algebraic properties except aⁿ = a^{a
^{a
^⋯}}. Perhaps this is the reason that ultra power is not extended so far [4, 7]. The rest of the contents of the paper is organized as follows. In Section 2, we deduce some propositions associated by convergence. In Section 3, we introduce main theorem. At the end, we give some examples to illustrate the feasibility of our proposed theoretical results.

2 Some propositions of convergence

In this section we deduce some propositions associated with convergence, which help us to prove the main theorem.

Proposition 1.

The series $\sum_{m \geq 1} \frac{1}{m^{m}}$ is convergent.

The improper integral $\int_{0}^{1} \frac{dx}{x^{x}}$ is convergent.

Proof 1. The convergence of the series can be obtained as follows: For m ≥ 2, we have m^m ≥ m² then, $0 \leq \frac{1}{m^{m}} \leq \frac{1}{m^{2}}$ (1) since, $\sum_{m \geq 1} \frac{1}{m^{2}}$ converges by Riemann criteria then, by comparison then, $\sum_{m \geq 1} \frac{1}{m^{m}}$ is a convergent series.

Proof 2. Convergence of $\int_{0}^{1} \frac{dx}{x^{x}} .$ We deal an improper integral with a singularity at x = 0. Let $f (x) = \frac{1}{x^{x}} = e^{- ln (x^{x})} = e^{- x ln (x)}, \forall x \in (0, 1] .$ f is a continuous function on [a, 1] ∀ a > 0 so, $\int_{a}^{1} f (x) dx exists .$ Now, we have $lim_{x \to 0^{+}} f (x) = 1 .$ Hence, we define f (0) =1 therefore, f can be define on [0, 1] by adding f (0) =1 . So, we have a continuous function defined on [0, 1] so, f is integrable function that lead to $\int_{0}^{1} f (x) dx$ is convergent. Which complete the proof.

Proposition 2.For all n, m ∈ N, there exist u_n,m such that $u_{m, n} = \int_{0}^{1} x^{n} {ln}^{m} (x) dx, \forall n, m \in N .$

Proof. Let f (x) = xⁿ ln ^m (x) is a continuous function on [0, 1] , then, $\sqrt{x} f (x) ⟶ 0$ , as x ⟶ 0, thus $f (x) = 0 (\frac{1}{\sqrt{x}}) .$ Since $\int_{0}^{1} \frac{1}{\sqrt{x}} dx$ exist that is $\frac{1}{\sqrt{x}}$ is integrable on [0, 1] . Then, by comparison $\int_{0}^{1} f (x) dx$ exist. Finally, u_n,m exist.

Proposition 3. For all n, m ∈ N . We get, $u_{n, m} = \frac{(- 1)^{m} m!}{(n + 1)^{m + 1}} .$

Proof. Assume n ≥ 1, using integration by parts, we have $\begin{matrix} u_{n, m} & = \int_{0}^{1} x^{m} \ln^{m} (x) \\ = {[\frac{1}{n + 1} x^{n + 1} \ln^{m} x]}_{0}^{1} \\ - \int_{0}^{1} (\frac{1}{n + 1} c^{m} . m {ln}^{m - 1} x) dx \\ = 0 - \frac{m}{n + 1} u_{n, m - 1} \end{matrix}$ from (1) we deduce that $u_{n, m} = - \frac{m}{n + 1} u_{n, m - 1}$ (2) to achieve the solution we will use (Telescopy).

Remark 1. Telescopy sum are finite sum in which pair of consecutive cancel other, leaving only the initial and final sum terms. Let a_n be a sequence of numbers. Then, $\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} = lim_{N \to \infty} \sum_{n = 1}^{N} (\frac{1}{n} - \frac{1}{n + 1})$ and $\prod_{n = 1}^{\infty} (1 - \frac{1}{n^{2}}) = \prod_{n = 2}^{\infty} \frac{(n - 1) (n + 1)}{n^{2}}$ $= lim_{N \to \infty} \prod_{n = 1}^{N} \frac{n - 1}{n} - \times \prod_{n = 2}^{N} \frac{n + 1}{n} .$ $\sum_{n = 1}^{N} (a_{n} - a_{n - 1}) = a_{N} - a_{0},$ if a_n ⟶ 0 then, we get $\sum_{n = 1}^{\infty} (a_{n} - a_{n - 1}) = - a_{0} .$ Similarly, in product we obtain $\prod_{n = 1}^{N} \frac{a_{n - 1}}{a_{n}} = \frac{a_{0}}{a_{N}}, if a_{n} ⟶ 0$ then, we get $\prod_{n = 1}^{\infty} \frac{a_{n - 1}}{a_{n}} = a_{0} .$ Now, recall that, if f is non negative a continuous function on the [a, b] therefore, we get $\int_{a}^{b} f (x) dx = 0,$ then, f = 0. In our case f (x) = - xⁿ ln ^m (x) is non negative continuous function on [0, 1], so if $\int_{0}^{1} f (x) dx = 0,$ then, f (x) =0, which impossible then, we deduce that $\int_{0}^{1} f (x) dx \neq 0,$ and u_n,m ≠ 0. From (2) we have $\frac{u_{n, m}}{u_{n, m - 1}} = - \frac{m}{n + 1} . (u_{n, m - 1} \neq 0) .$ Then, we get $\prod_{k = 1}^{m} \frac{u_{n, k}}{u_{n, k - 1}} = \prod_{k = 1}^{m} \frac{- k}{n + 1} = \frac{\prod_{k = 1}^{m} - k}{\prod_{k = 1}^{m} (n + 1)} = \frac{(- 1)^{m} m!}{(n + 1)^{m}}$ $\frac{\prod_{k = 1}^{m} u_{n, k}}{\prod_{k = 1}^{m} u_{n, k - 1}} = \frac{(- 1)^{m} m!}{(n + 1)^{m} .}$ Since $\frac{\prod_{k = 1}^{m} u_{n, k}}{\prod_{k = 0}^{m - 1} u_{n, k}} = \frac{(- 1)^{m} m!}{(n + 1)^{m}}$ then, we get $\frac{u_{n, m}}{u_{n, 0}} = \frac{(- 1)^{m} m!}{(n + 1)^{m}}$ and $u_{n, m} = \frac{u_{n, 0} (- 1)^{m} m!}{(n + 1)^{m}} .$ By definition $u_{n, 0} = \int_{0}^{1} x^{n} dx = \frac{1}{n + 1} .$ Hence, we obtain $u_{n, m} = \frac{(- 1)^{m} m!}{(n + 1)^{m + 1}},$ which completes the proof.

Proposition 4. For all n ∈ N we get, $\int_{0}^{1} \frac{1}{x^{n}} dx = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{n!} u_{n, n} .$

Proof. Let $\frac{1}{x^{x}} = x^{- x} = e^{ln x^{- x}} = e^{- x ln x},$ using the power series of exponential u, where u ∈ R, $e^{u} = \sum_{n = 0}^{\infty} \frac{u^{n}}{n!} \cdot$ Then, we get $\begin{matrix} \int_{0}^{1} \frac{1}{x^{x}} dx & = & \int_{0}^{1} e^{- x ln x} dx \\ = & \int_{0}^{1} \sum_{n = 0}^{\infty} \frac{(- x ln x)^{n}}{n!} \\ = & \int_{0}^{1} \sum_{n = 0}^{\infty} \frac{1}{n!} x^{n} (- ln x)^{n} dx \cdot \end{matrix}$ Now, we use Dominated Convergence Theorem (DCT)[3] we can invert integral and summation. Now, let N ≥ 1 then, we obtain $\sum_{n = 0}^{N} \frac{x^{n} (- ln x)^{n}}{n!} \leq \sum_{n = 0}^{\infty} \frac{x^{n} (- ln x)^{n}}{n!} \leq \frac{1}{x^{x}},$ where $\frac{1}{x^{x}}$ is integrable function on [0, 1] by (DCT), we can invert integral and summation, we get, $\begin{matrix} \int_{0}^{1} \frac{dx}{x^{x}} & = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{n!} \int_{0}^{1} x^{n} {ln}^{n} xdx \\ = \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{n!} u_{n, n} . \end{matrix}$ Finally, we deduce that, $\int_{0}^{1} \frac{dx}{x^{x}} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} u_{n, n}}{n!} .$

Proposition 5. For all m ∈ N, and exponential function $\frac{1}{x^{x}}$ we get the following: $\int_{0}^{1} \frac{1}{x^{x}} = \sum_{m = 1}^{\infty} \frac{1}{m^{m}} .$

Proof. Using Proposition (4), we have $\int_{0}^{1} \frac{dx}{x^{x}} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} u_{n, n}}{n!} .$ Then, we obtain $\int_{0}^{1} \frac{dx}{x^{x}} = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (- 1)^{n} n!}{n! (n + 1)^{n + 1}} = \sum_{m = 0}^{\infty} \frac{1}{(n + 1)^{n + 1}} = \sum_{m = 1}^{\infty} \frac{1}{m^{m}} .$ (We can change the index n, using m = n + 1). Which completes the proof.

3 Main theorem

Theorem 1. For any α, c ∈ R, the improper integral $\int_{0}^{1} x^{x} dx$ satisfies the following: $\begin{matrix} \int_{0}^{1} x^{{cx}^{α}} dx & = & \sum_{N = 0}^{\infty} \frac{c^{N} (- 1)^{N}}{(1 + α N)^{N + 1}} = 1 - \frac{c}{(1 + α)^{2}} \\ + & \frac{c^{2}}{(1 + 2 α)^{3}} - \frac{c^{3}}{(1 + 3 α)^{4}} + \dots \end{matrix}$ Proof. Let x^{cx
^α} = (e^ln(x)) ^{cx
^α} = e^{cx^αln(x)}, and $e^{y} = \sum_{N = 0}^{\infty} = \frac{y^{N}}{N!}$ then, we get $\begin{matrix} x^{{cx}^{α}} & = e^{{cx}^{α} \ln (x)} = \sum_{N = 0}^{\infty} \frac{{cx}^{α} \ln (x))^{N}}{N!} \\ = \sum_{N = 0}^{\infty} \frac{c^{N} x^{α N} (\ln (x))^{N}}{N!} \cdot \end{matrix}$ Now, let $I = \int_{0}^{1} x^{α N} (\ln (x))^{N} dx$ and $u = ln (x) \Rightarrow x = e^{u}, dx = e^{u} du$ $u (0^{+}) = \ln (0^{+}) = \infty, u (1) = 0 .$ This gives $\begin{matrix} \int_{0}^{1} x^{α N} (\ln (x))^{N} dx & = \int_{\infty}^{0} e^{α N u} u^{N} e^{u} du \\ = \int_{\infty}^{0} e^{u (1 + α N)} u^{N} du \cdot \end{matrix}$ Let $v = - u (1 + α N), v (\infty) = \infty, v (0) = 0$ $dv = - (1 + α N) du \cdot$ Therefore, we get $\begin{matrix} \int_{0}^{1} x^{α N} (\ln (x))^{N} dx & = \int_{\infty}^{0} e^{u} (1 + α N) u^{N} du \\ = \int_{\infty}^{0} e^{(- v)} (\frac{- v}{1 + α N})^{N} \frac{dv}{- (1 + α N)} \\ = \frac{(- 1)^{N}}{(1 + α N)^{N + 1}} \int_{0}^{\infty} e^{- v} v^{N} dv \\ = \frac{(- 1)^{N}}{(1 + α N)^{N + 1}} Γ (N + 1) \\ = \frac{(- 1)^{N}!}{(1 + α N)^{N + 1}} \cdot \end{matrix}$ Finally, we deduce that $\begin{matrix} \int_{0}^{1} x^{{cx}^{α}} dx & = \sum_{N = 0}^{\infty} \frac{c^{N}}{N!} \int_{0}^{1} x^{α N} (\ln (x))^{N} dx \\ = \sum_{N = 0}^{\infty} \frac{c^{N}}{N!} \frac{(- 1)^{N} N!}{(1 + α N)^{N + 1}} \end{matrix}$ $\int_{0}^{1} x^{{cx}^{α}} dx = \sum_{N = 0}^{\infty} \frac{c^{N} (- 1)^{N}}{(1 + α N)^{N + 1}}$ $\begin{matrix} \int_{0}^{1} x^{{cx}^{α}} dx & = & \sum_{N = 0}^{\infty} \frac{c^{N} (- 1)^{N}}{(1 + α N)^{N + 1}} = 1 - \frac{c}{(1 + α)^{2}} \\ + & \frac{c^{2}}{(1 + 2 α)^{3}} - \frac{c^{3}}{(1 + 3 α)^{4}} + \dots \cdot \end{matrix}$ Which completes the proof.

4 Examples

Example 1. Evaluate the improper integral $\int_{0}^{1} x^{x} dx .$

Solution: Let $\int_{0}^{1} x^{x} dx = \int_{0}^{1} e^{ln (x^{x})} dx = \int_{0}^{1} e^{x ln (x)} dx$

Note that: $\begin{matrix} e^{z} & = \sum_{N = 0}^{\infty} \frac{z^{N}}{N!} \\ = \int_{0}^{1} \sum_{N = 0}^{\infty} \frac{(xlnx)^{N}}{N!} dx \\ = \sum_{N = 0}^{\infty} \int_{0}^{1} \frac{(xlnx)^{N}}{N!} dx \\ = \sum_{N = 0}^{\infty} \frac{1}{N!} \int_{0}^{1} x^{N} (\ln (x))^{N} dx, \end{matrix}$ this gives, $\int_{0}^{1} x^{N} (\ln (x))^{N} dx .$ Let u = - ln(x) ⇒ x = e^-u $du = - \frac{1}{x} dx \Rightarrow dx = - xdu = - e^{- u} du$ and u (0) = - ln(0⁾ = - (-∞) = ∞ u (1) = - ln(1) =0 . $\begin{matrix} \int_{0}^{1} x^{N} (ln (x))^{N} dx & = & \int_{\infty}^{0} (e^{- u})^{N} (- u)^{N} (- e^{- u})^{1} du \\ = & \int_{0}^{\infty} (e^{- u})^{N + 1} (- 1)^{N} u^{N} du \\ = & (- 1)^{N} \int_{0}^{\infty} e^{- (N + 1) u} u^{N} du . \end{matrix}$ Let $t = (N + 1) u \Rightarrow u = \frac{t}{N + 1}$ $dt = (N + 1) du \Rightarrow du = \frac{dt}{N + 1}$ we obtain that $(- 1)^{N} \int_{0}^{\infty} e^{- (N + 1) u} u^{N} du =$ $(- 1)^{N} \int_{0}^{\infty} e^{- t} \frac{t^{N}}{(N + 1)^{N}} \frac{dt}{(N + 1)}$ $= (- 1)^{N} \int_{0}^{\infty} e^{t} t^{N} dt \cdot$ Where N = (N + 1) -1 . Recall that $Γ (z) = \int_{0}^{\infty} e^{t} t^{z - 1} dt .$ Now, $\begin{matrix} (- 1)^{N} \int_{\infty}^{0} e^{- t} t^{N} dt & = & \frac{(- 1)^{N}}{(N + 1)^{N + 1}} Γ (N + 1) \\ = & \frac{(- 1)^{N}}{(N + 1)^{N + 1}} N! \end{matrix}$ $\begin{matrix} \int_{0}^{1} x^{x} & = \sum_{N = 0}^{\infty} \frac{1}{N!} \int_{0}^{1} x^{N} (\ln (x))^{N} dx \\ = \sum_{N = 0}^{\infty} \frac{1}{N!} \frac{(- 1)^{N}}{(N + 1)^{N + 1}} N! \\ = \sum_{N = 0}^{\infty} \frac{(- 1)^{N}}{(N + 1)^{N + 1}} = \frac{1}{1^{1}} - \frac{1}{2^{2}} + \frac{1}{3^{3}} + \dots \end{matrix}$ $\begin{matrix} \int_{0}^{1} x^{x} & = \sum_{N = 1}^{\infty} \frac{(- 1)^{N + 1}}{N^{N}} \\ = \frac{1}{1^{1}} - \frac{1}{2^{2}} + \frac{1}{3^{3}} + \frac{1}{4^{4}} \dots . \end{matrix}$ Now, to evaluate the improper integral $\int_{0}^{1} x^{x} dx,$ we proceed as follows to evaluate the above integral using numerical series, using the following technique: Let $f (x) = x^{- x}, x \geq 1,$ which is defined on the (0, 1) , we differentiate both sides w.r.to x $\begin{matrix} \frac{d}{dx} ln (y) & = & \frac{d}{dx} (- x ln (x)) \\ \frac{y^{'}}{y} & = & - (x . \frac{1}{x} + ln (x)) \\ y^{'} & = & - x^{- x} (1 + ln (x)), x \geq 0, \end{matrix}$ for x ≥ 1, we have ln(x) ≥ ln(1) =0, so (1 + ln(x)) ≥1, thus using x^-x is non negative function thus, lead to $- (1 + ln (x)) \leq - 1 < 0$ $- x^{x} (1 + ln (x)) < 0$ so, f′ < 0, ∀ x ≥ 1 therefore, f is decreasing function. Furthermore f is non negative function so, we are able to apply alternating series criteria, more precisely $\sum_{n = 1}^{\infty} (- 1)^{n - 1} f (n)$ is convergent, that leads to $\sum_{n = 1}^{\infty} (- 1)^{n - 1} \frac{1}{n^{n}} < \infty$ is convergent so that, the original integral becomes $\int_{0}^{1} x^{x} dx = 0.7834305107 is convergent .$ The number 0.7834305107 calculated using maple code.

Example 2. Evaluate the improper integral $\int_{0}^{1} x^{- x} dx .$

Solution: Let $\begin{matrix} \int_{0}^{1} x^{- x} dx & = \int_{0}^{1} e^{- xlnx} dx \\ = \int_{0}^{1} \sum_{N = 0}^{\infty} \frac{(- xln (x))^{N}}{N!} dx \\ = \sum_{N = 0}^{\infty} \frac{(- 1)^{N}}{N!} \int_{0}^{1} x^{N} (\ln (x))^{N} dx \end{matrix}$ $\begin{matrix} \int_{0}^{1} x^{- x} dx & = \sum_{N = 0}^{\infty} \frac{(- 1)^{N}}{N!} \frac{(- 1)^{N} N!}{(N + 1)^{N + 1}} \\ = \sum_{N = 0}^{\infty} \frac{1}{(N + 1)^{N + 1}} \\ = \sum_{N = 1}^{\infty} = \frac{1}{1^{1}} + \frac{1}{2^{2}} + \frac{1}{3^{3}} + . . . . \end{matrix}$ Now, to evaluate the improper integral $\int_{0}^{1} x^{- x} dx$ we proceed as follows to evaluate the above integral using the following technique: Let $2 < 3 \Rightarrow 2^{3} < 3^{3}$ now, we have $\frac{1}{2^{3}} > \frac{1}{3^{3}},$ let k ≥ 2 such that 2 ≤ k then, we get $2^{k} \leq k^{k}$ $\frac{1}{2^{k}} \geq \frac{1}{k^{k}},$ we deduce that $\sum_{k = 2}^{\infty} \frac{1}{2^{k}} \geq \sum_{k = 2}^{\infty} \frac{1}{k^{k}}$ therefore, we get $\frac{1}{2} \geq \sum_{k = 2}^{\infty} \frac{1}{k^{k}} = \sum_{k = 1}^{\infty} \frac{1}{k^{k}} - 1$

$1 + \frac{1}{2} \geq \sum_{k = 1}^{\infty} \frac{1}{k^{k}}$

$\frac{3}{2} \geq \sum_{k = 1}^{\infty} \frac{1}{k^{k}}$ now, let $S_{n} = \sum_{k = 1}^{n} \frac{1}{k^{k}}$ we show that the partial sum of the series S_n is convergent so, the series is convergent. Recall that:

First: $\begin{matrix} S_{n + 1} - S_{n} & = & \sum_{k = 1}^{n + 1} \frac{1}{k^{k}} - \sum_{k = 1}^{n} \frac{1}{k^{k}} \\ = & \frac{1}{(n + 1)^{n + 1}} > 0 \end{matrix}$ so, the sequence S_n is increasing sequence and has an upper bound (bounded above) so, the sequence S_n is convergent and so, $\sum_{n = 1}^{\infty} \frac{1}{n^{n}} < \infty is finite \cdot$

Second: We conclude that the improper integral namely, $\int_{0}^{1} x^{- x} dx = 1.291285997 is convergent .$ The number 1.291285997 calculated using maple code.

Example 3. Evaluate the improper integral $\int_{0}^{1} x^{\sqrt{x}} dx .$

Solution: See Theorem (1), when α = 1; $c = \frac{1}{2}$ then, we obtain $\begin{matrix} \int_{0}^{1} x^{\frac{1}{2} x} dx & = \int_{0}^{1} (\sqrt{x})^{x} \\ = \sum_{N = 0}^{\infty} \frac{(\frac{1}{2})^{N} (- 1)^{N}}{(1 + N)^{N + 1}} \\ = 1 - \frac{\frac{1}{2}}{2^{2}} + \frac{\frac{1}{4}}{3^{3}} - \frac{\frac{1}{8}}{4^{4}} + \dots \\ = 0.8837903267 . \end{matrix}$ Furthermore let $α = \frac{1}{2}$ ; c = 1, then, $\begin{matrix} \int_{0}^{1} x^{\sqrt{x}} dx & = & \sum_{N = 0}^{\infty} \frac{(- 1)^{N}}{(1 + \frac{N}{2})^{N + 1}} \\ = & 1 - \frac{1}{3^{2}} + \frac{1}{5^{3}} - \frac{1}{7^{4}} + \dots \\ = & 0.6585823541 . \end{matrix}$

5 Discussion

From the results and analysis of the section 3 and 4, the authors observed that an improper integral converges if the limit defining it exists. If f is a non-negative function which is unbounded in a domain A, then the improper integral of f is defined by truncating f at some cutoff M, integrating the resulting function, and then taking the limit as M tends to infinity. The new proposed approach useful to use in probability and statistics, which is an important tool in certain areas of the social, managerial and life sciences. We expect to study the quantum white noise case which is now attractive in mathematical physics area. As applicable examples of improper integral is:

•NASA has studied improper integral and fuzzy control system for automated space docking: simulations show that a fuzzy control system can greatly reduce fuel consumption

• The Aumann fuzzy improper integral and its application use to solve fuzzy integro-differential equations. (see [2]).

• The concept of fuzzy integral as developed and has been used in the problems of optimization with inexact constraints.

• Integrals over infinite intervals arise naturally in physics and in statistics as an example: The work needed to move the probe a distance dr against the gravitational force $F = G \frac{mM}{r^{2}}$ of Earth is Fdr. M denotes the mass of Earth, m is the mass of the probe, r is the distance between the centers of mass and G is the gravitational constant. Summing up the $\int F dr = \int G \frac{mM}{r^{2}} dr$ . The lower bound should be R, the radius of the Earth, while the upper bound would need to be infinity. Thus, to complete this example we will need to compute an integral over an infinite interval. (Use improper integral).

• Application of improper integral involving marginal profit, by using an improper integral to find profit.

• If you have a probability density function that can get any number, then for normalization and to calculate moments you need to calculate improper integrals.

• We might want to know how a system behaves after it relaxed to equilibrium or to a steady state, i.e. After an infinite time. This may entail an integral where the time is the limit of the integration.

6 Conclusion

In this research we confirm the approach, evaluated the integral and series using Maple code and we proved the convergent of series mathematically. We get that the both sides are equal approximation, this confirms that both sides of the following equation: $\begin{matrix} \int_{0}^{1} x^{{cx}^{α}} dx & = & \sum_{N = 0}^{\infty} \frac{c^{N} (- 1)^{N}}{(1 + α N)^{N + 1}} = 1 - \frac{c}{(1 + α)^{2}} \\ + & \frac{c^{2}}{(1 + 2 α)^{3}} - \frac{c^{3}}{(1 + 3 α)^{4}} + \dots \end{matrix}$ ∀ α, c ∈ R .

Footnotes

Acknowledgments

The authors would like to thank the Deanship of Scientific Research at Umm Al-Qura University for supporting this work by Grant code: 22UQU4310382DSR01.

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