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1. Is the following identity true ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad \overset{?}{=} \quad \int_0^\infty \frac{x!}{x^x} dx$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$
2. Does the integral converge ?

3. Does it possess a closed form, or some other alternative expression ?

4. If the integral may prove to be divergent when the two sums start at n = 1, could anyone check the case when the two sums start at n = 2 or 3 ?

• Thank you !

NOTE: If the position of $x$ and $n$ in the numerator of each sum were reversed, and both sums were to start at n = 0, we would have the following identity:

$$\int_0^\infty \frac{E(x)}{e^x} dx \quad = \quad \sum_{n=0}^\infty \frac{n!}{n^n}$$

where $\lim_{n \to 0} n^n = 1,$ and

$$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad and \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

1. Does the following integral converge ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$

 
2. Does it possess a closed form, or some other alternative expression ?

1.Does the following definite integral converge ?

2.Does it have a closed form, or some other alternative expression ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad = \quad ?$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$
3.If the integral may prove to be divergent when the two sums start at n = 1, could anyone check the case when the two sums start at n = 2 or 3 ?

• Thank you !

NOTE: If the position of $x$ and $n$ in the numerator of each sum were reversed, and the second sum were to start at n = 0, we would have the following identity:

$$\int_0^\infty \frac{E(x)}{e^x} dx \quad = \quad \sum_{n=0}^\infty \frac{n!}{n^n}$$

where $\lim_{n \to 0} n^n = 1,$ and

$$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad and \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

1. Is the following identity true ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad \overset{?}{=} \quad \int_0^\infty \frac{x!}{x^x} dx$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$
2. Does the integral converge ?

3. Does it possess a closed form, or some other alternative expression ?

4. If the integral may prove to be divergent when the two sums start at n = 1, could anyone check the case when the two sums start at n = 2 or 3 ?

• Thank you !

NOTE: If the position of $x$ and $n$ in the numerator of each sum were reversed, and both sums were to start at n = 0, we would have the following identity:

$$\int_0^\infty \frac{E(x)}{e^x} dx \quad = \quad \sum_{n=0}^\infty \frac{n!}{n^n}$$

where $\lim_{n \to 0} n^n = 1,$ and

$$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad and \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

1.Does the following definite integral converge ?

2.Does it have a closed form, or some other alternative expression ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad = \quad ?$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$
3.If the integral may prove to be divergent when the two sums start at n = 1, could anyone check the case when the two sums start at n = 2 or 3 ?

• Thank you !

1.Does the following definite integral converge ?

2.Does it have a closed form, or some other alternative expression ?

$$\int_0^\infty \frac{b(x)}{B(x)} dx \quad = \quad ?$$

where

$$b(x) = \sum_{n=1}^\infty \frac{n^x}{n^n} \qquad and \qquad B(x) = \sum_{n=1}^\infty \frac{n^x}{n!}$$
3.If the integral may prove to be divergent when the two sums start at n = 1, could anyone check the case when the two sums start at n = 2 or 3 ?

• Thank you !

NOTE: If the position of $x$ and $n$ in the numerator of each sum were reversed, and the second sum were to start at n = 0, we would have the following identity:

$$\int_0^\infty \frac{E(x)}{e^x} dx \quad = \quad \sum_{n=0}^\infty \frac{n!}{n^n}$$

where $\lim_{n \to 0} n^n = 1,$ and

$$E(x) = \sum_{n=0}^\infty \frac{x^n}{n^n} \qquad and \qquad e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$