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A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \big( \frac{x}{m} \big)!\,\big( \frac{x-1}{m} \big)!\dots \big( \frac{x-m+1}{m} \big)!\, \,$$

For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:

$$\prod_{k=1}^{m}\big(\frac{k}{m} \big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$

Take a logarithm; divide by $m$ and let $m\to\infty$: one finds

$$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$

or, as a multiplicative integral

$$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$

Now the question: How to evaluate the above integral by means of standard integral calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.

A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \Big( \frac{x}{m} \Big)!\,\Big( \frac{x-1}{m} \Big)!\dots \Big( \frac{x-m+1}{m} \Big)!\, \,$$

For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:

$$\prod_{k=1}^{m}\Big(\frac{k}{m} \Big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$

Take a logarithm; divide by $m$ and let $m\to\infty$: one finds

$$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$

or, as a multiplicative integral

$$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$

Now the question: How to evaluate the above integral by means of standard integral calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.

A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \big( \frac{x}{m} \big)!\,\big( \frac{x-1}{m} \big)!\dots \big( \frac{x-m+1}{m} \big)!\, \,$$

For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:

$$\prod_{k=1}^{m}\big(\frac{k}{m} \big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$

Take a logarithm; divide by $m$ and let $m\to\infty$: one finds

$$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$

or, as a multiplicative integral

$$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$

Now the question: How to evaluate the above integral by means of standard integral calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.

A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \big( \frac{x}{m} \big)!\,\big( \frac{x-1}{m} \big)!\dots \big( \frac{x-m+1}{m} \big)!\, \,$$

For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:

$$\prod_{k=1}^{m}\big(\frac{k}{m} \big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$

Take a logarithm; divide by $m$ and let $m\to\infty$: one finds

$$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$

or, as a multiplicative integral

$$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$

Now the question: How to evaluate the above integral by means of standard integral calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.

A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

$$x!=(2\pi)^{-\frac{m-1}{2}}\\ m^{x+\frac{1}{2}}\\ \big( \frac{x}{m} \big)!\\,\big( \frac{x-1}{m} \big)!\dots \big( \frac{x-m+1}{m} \big)!\\,\\,$$

For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:

$$\prod_{k=1}^{m}\big(\frac{k}{m} \big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$

Take a logarithm; divide by $m$ and let $m\to\infty$: one finds

$$\int_0^1\log(x!)\\, dx=\frac{1}{2}\log(2\pi )-1,$$

or, as a multiplicative integral

$$\prod_0^1 (x!\\, dx)=\frac{\sqrt{2\pi}}{e}.$$

Now the question: How to evaluate the above integral by means of standard integral calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.

A recent question on the notion and notation of multiplicative integrals ( What is the standard notation for a multiplicative integral? ) induced me to play with the Riemann products of the Gamma function, in order to evaluate the multiplicative integral of $\Gamma(x)$, exploiting the multiplicative formula. I will, however, put the question mainly in terms of a standard integral; and I will also use the factorial function $x!=\Gamma(x+1)$ instead (that seems to be more appreciated here). Consider the multiplicative formula for $x!$:

$$x!=(2\pi)^{-\frac{m-1}{2}}\, m^{x+\frac{1}{2}}\, \big( \frac{x}{m} \big)!\,\big( \frac{x-1}{m} \big)!\dots \big( \frac{x-m+1}{m} \big)!\, \,$$

For $x=m\in\mathbb{N}$ we get, using the Stirling asymptotics for $m!$:

$$\prod_{k=1}^{m}\big(\frac{k}{m} \big)!\sim (2\pi)^{\frac{m}{2}}e^{-m} $$

Take a logarithm; divide by $m$ and let $m\to\infty$: one finds

$$\int_0^1\log(x!)\, dx=\frac{1}{2}\log(2\pi )-1,$$

or, as a multiplicative integral

$$\prod_0^1 (x!\, dx)=\frac{\sqrt{2\pi}}{e}.$$

Now the question: How to evaluate the above integral by means of standard integral calculus?

I guess it's feasible, but how? Otherwise, it would be a remarkable case of an integral that one can only (edit: or say "more easily") evaluate directly from the definition of Riemann sums, like one does e.g. with $x^2$ in introductory calculus courses.