A shortcut for evaluating some log integrals from products and limits

TL;DR

This paper presents a simple, elementary method for exactly evaluating certain definite integrals involving logarithms of sine, cosine, and tangent functions, using manipulations of sums and Riemann sums, offering an alternative to traditional techniques.

Contribution

The paper introduces a novel elementary approach to evaluate specific log-trigonometric integrals and related integrals involving the gamma function, avoiding complex primitive searches.

Findings

Exact evaluations of key log-trigonometric integrals.

Method simplifies calculations without primitive search.

Application to gamma function integral.

Abstract

In this short paper, I introduce an elementary method for exactly evaluating the definite integrals $\, \int_0^{\pi}{\ln{(\sin{\theta})}\,d\theta}$, $\int_0^{\pi/2}{\ln{(\sin{\theta})}\,d\theta}$, $\int_0^{\pi/2}{\ln{(\cos{\theta})}\,d\theta}$, and $\int_0^{\pi/2}{\ln{(\tan{\theta})}\,d\theta} \,$ in finite terms. The method consists in to manipulate the sums obtained from the logarithm of certain products of trigonometric functions at rational multiples of $\pi$, putting them in the form of Riemann sums. As this method does not involve any search for primitives, it represents a good alternative to more involved integration techniques. As a bonus, I show how to apply the method for easily evaluating $\,\int_0^1{\ln{\Gamma(x)} \, d x}$.

Equations

$$\frac{i}{2}{\mathrm{Li}}_2\left(e^{2ix}\right)+i\frac{x^2}{2}+x\ln \left(\sin x\right)-x\ln \left(1-e^{2ix}\right),$$

$$F(x)=\sum _{j=0}^N(-1{)}^j(\genfrac{}{}{0ex}{}{2N+1}{2j+1})x^{N-j}=(2N+1)\prod _{n=1}^N\left(x-{\cot }^2{\theta }_n\right),$$

$$\prod _{n=1}^N{\cot }^2{\theta }_n=(-1{)}^N\frac{(-1{)}^N}{\left(\genfrac{}{}{0ex}{}{2N+1}{1}\right)}=\frac{1}{2N+1}.$$

$$\prod _{n=1}^N\tan \left(\frac{n\pi }{2N+1}\right)=\sqrt{2N+1}.$$

$$\prod _{n=1}^{N-1}\sin \left(\frac{\pi n}{N}\right)=\frac{N}{2^{N-1}}.$$

$$\prod _{n=1}^{N-1}\sin \left(\frac{\pi n}{N}\right)=\prod _{n=1}^{\lfloor \frac{N}{2}\rfloor }\sin \left(\frac{\pi n}{N}\right)\sin \left(\frac{\pi (N-n)}{N}\right)=\prod _{n=1}^{\lfloor \frac{N}{2}\rfloor }\sin \left(\frac{\pi n}{N}\right)\sin \left(\pi -\frac{\pi n}{N}\right),$$

$$\prod _{n=1}^{\lfloor \frac{N}{2}\rfloor }{\sin }^2\left(\frac{\pi n}{N}\right)=\frac{N}{2^{N-1}}.$$

$$\prod _{n=1}^{\lfloor \frac{N}{2}\rfloor }{\cos }^2\left(\frac{\pi }{2}-\frac{\pi n}{N}\right)=\frac{N}{2^{N-1}}.$$

$$\sum _{n=1}^{N-1}\ln \left[\sin \left(\pi \frac{n}{N}\right)\right]=\ln N-(N-1)\ln 2.$$

$$\sum _{n=1}^{N-1}\frac{\ln \left[\sin \left(\pi \frac{n}{N}\right)\right]}{N-1}=\frac{\ln N}{N-1}-\ln 2.$$

$$\sum _{n=1}^{N-1}\ln \left[\sin \left(\pi x_n\right)\right]\Delta x=\frac{\ln N}{N-1}-\ln 2.$$

$$\underset{N\to \infty }{\lim }\sum _{n=1}^{N-1}\ln \left[\sin \left(\pi x_n\right)\right]\Delta x=-\ln 2,$$

$${\int }_{0^{+}}^{1^{-}}\ln \sin \left(\pi x\right)dx=-\ln 2.$$

$${\int }_0^{\pi }\ln \sin \theta d\theta =-\pi \ln 2.$$

$$2\sum _{n=1}^{\lfloor N/2\rfloor }\ln \left[\sin \left(\pi x_n\right)\right]\Delta x=\frac{\ln N}{N-1}-\ln 2.$$

$$2{\int }_0^{\frac{1}{2}}\ln \sin \left(\pi x\right)dx=-\ln 2.$$

$${\int }_0^{\pi /2}\ln \sin \theta d\theta =-\frac{\pi }{2}\ln 2.$$

$$2\sum _{n=1}^{\lfloor N/2\rfloor }\ln \left[\cos \left(\frac{\pi }{2}-\pi x_n\right)\right]\Delta x=\frac{\ln N}{N-1}-\ln 2.$$

$$2{\int }_0^{\frac{1}{2}}\ln \cos \left(\frac{\pi }{2}-\pi x\right)dx=-\ln 2.$$

$${\int }_0^{\pi /2}\ln \cos \theta d\theta =-\frac{\pi }{2}\ln 2.$$

$$\sum _{n=1}^N\ln \left[\tan \left(\pi \frac{n}{2N+1}\right)\right]=\frac{1}{2}\ln (2N+1).$$

$$\sum _{n=1}^{(M-1)/2}\ln \left[\tan \left(\pi x_n\right)\right]\Delta x=\frac{\ln M}{2M},$$

$$\underset{M\to \infty }{\lim }\sum _{n=1}^{(M-1)/2}\ln \left[\tan \left(\pi x_n\right)\right]\Delta x=\underset{M\to \infty }{\lim }\frac{\ln M}{2M}=\frac{1}{2}\underset{M\to \infty }{\lim }\frac{1}{M}=0,$$

$${\int }_0^{\frac{1}{2}}\ln \tan \left(\pi x\right)dx=0.$$

$${\int }_0^{\pi /2}\ln \tan \theta d\theta =0.$$

$$(N-1)\ln \pi -\sum _{n=1}^{N-1}\ln \left[\sin \left(\pi \frac{n}{N}\right)\right]=(N-1)\ln \pi -\ln N+(N-1)\ln 2.$$

$$\sum _{n=1}^{N-1}\ln \left[\frac{\pi }{\sin \left(\pi x_n\right)}\right]=(N-1)\ln (2\pi )-\ln N.$$

$$\sum _{n=1}^{N-1}\frac{\ln \Gamma (x_n)+\ln \Gamma (1-x_n)}{N-1}=\ln (2\pi )-\frac{\ln N}{N-1}.$$

$${\int }_0^1\left[\ln \Gamma (x)+\ln \Gamma (1-x)\right]dx=\ln (2\pi ).$$

$$2{\int }_0^1\ln \Gamma (x)dx=\ln (2\pi ),$$

Full text

††journal: :  Int. J. Math. Educ. Sci. Technol.

1 Introduction

The simplicity of the functions $\,\ln{(\sin{\theta})}$, $\ln{(\cos{\theta})}$, and $\ln{(\tan{\theta})}$, as well as the fact they are continuous (even differentiable) except at some isolated points, suggests that the evaluation of the definite integrals $\int_{0}^{\pi}{\ln{(\sin{\theta})}\,d\theta}$, $\int_{0}^{\pi/2}{\ln{(\sin{\theta})}\,d\theta}$, $\int_{0}^{\pi/2}{\ln{(\cos{\theta})}\,d\theta}$, and $\int_{0}^{\pi/2}{\ln{(\tan{\theta})}\,d\theta}$ via the Fundamental Theorem of Calculus should be a straightforward task,111 These integrals are, in fact, improper because each integrand has at least one infinite discontinuity in the integration interval. which is not true.222 Note that it is not possible to express the corresponding indefinite integrals in a finite, closed-form expression involving only elementary functions. The usual methods for the analytic evaluation of such integrals involve advanced integration techniques, such as to expand the integrand in a power series and then integrate it term-by-term (see Sec. 11.9 of Ref. [1]), or the evaluation of a suitable contour integral on the complex plane in view to apply the Cauchy’s residue theorem (see Secs. 4.1 and 4.2 of Ref. [2]). However, these methods present some disadvantages when applied to definite integrals of “log-trig” functions. The expansion of the integrand in a Maclaurin series, e.g., is not possible for $\ln{(\sin{\theta})}$ and $\ln{(\tan{\theta})}$. Though this series expansion is possible for $\ln{(\cos{\theta})}$, it is difficult to determine a closed-form expression for the general term and then to recognize the number it represents. The evaluation of a contour integral on the complex plane has the inconveniences of requiring the choice of a suitable integration path, usually a difficult task, and yielding logarithms of complex (non-real) numbers and/or non-elementary transcendental functions (e.g., dilogarithm, elliptic, and hypergeometric functions), which often makes it obscure the final result since these functions are either unknown or badly-known for most students. These inconveniences are just what one finds by appealing to mathematical softwares. For instance, Maple (release 13) and Mathematica (release 7) both return the following ‘stodgy’ result for $\int{\ln{(\sin{x})}\,dx}$:

[TABLE]

where $\mathrm{Li}_{2}(x):=\sum_{n=1}^{\infty}{\dfrac{x^{n}}{n^{2}}}$ is the dilogarithm function.

In this work, the integrals $\int_{0}^{\pi}{\ln{(\sin{\theta})}\,d\theta}$, $\int_{0}^{\pi/2}{\ln{(\sin{\theta})}\,d\theta}$, $\int_{0}^{\pi/2}{\ln{(\cos{\theta})}\,d\theta}$, and $\int_{0}^{\pi/2}{\ln{(\tan{\theta})}\,d\theta}$ are easily evaluated by applying the natural logarithm to certain products of trigonometric functions at rational multiples of $\pi$, which yields sums that can be written in the form of Riemann sums. The closed-form expressions emerge when we take suitable limits (as the number of terms tends to infinity), without any search for primitives. I also show how the method can be used for easily evaluating $\,\int_{0}^{1}{\ln{\Gamma(x)}\,dx}$.

2 Some products of trigonometric functions

Let me present the products of trigonometric functions at rational multiples of $\pi$ that will serve as the basis for my method.

In Appendix A.3 of Ref. [3], in presenting an elementary proof for the Euler result $\,\sum_{n=1}^{\infty}{{\,1/n^{2}}}={\,\pi^{2}/6}$, the authors prove an identity involving $\cot^{2}{\theta}$, $\theta$ being a rational multiple of $\pi$. By applying the well-known rule for the product of roots of a polynomial equation $\,F(x)=0\,$ to

[TABLE]

where $\theta_{n}={\,n\pi/(2N+1)}$, $n=1,\ldots,N$, it is shown that

[TABLE]

By extracting the square-root of the inverse of each side, one finds

[TABLE]

In that appendix, we also find a proof for the following identity:

[TABLE]

From the symmetry relation $\sin{(\pi/2+\alpha)}=\sin{(\pi/2-\alpha)}$, it is easy to deduce that

[TABLE]

valid for all $N>1$, which implies that

[TABLE]

For a product of cosines, note that, $\forall\;\alpha\in\left[0,{\,\pi/2}\right]$, $\sin{\alpha}=\cos{({\,\pi/2}-\alpha)}$. By taking $\alpha={\,\pi n/N}$ and applying this trigonometric identity on Eq. (3), one readily finds

[TABLE]

These are the trigonometric products needed for evaluating the integrals we are interested in here.

3 Evaluation of definite integrals from products

The general idea underlying my method is to take the logarithm of a product of positive terms, convert it into a sum of logarithms and then to put this sum in the form of a Riemann sum with equally-spaced subintervals, whose limit as the number of terms tends to infinity is a definite integral. Let us apply this procedure to the products of trigonometric functions in Eqs. (1)–(4).

For instance, by taking the logarithm of each side of Eq. (2), one has

[TABLE]

By dividing both sides by $N-1$, one gets

[TABLE]

Now, let us define $x_{n}={\,n/N}$ and $\Delta x={\,1/(N-1)}$. Equation (6) then reads

[TABLE]

Clearly, the sum at the left-hand side has the form of a Riemann sum in which the grid points $x_{n}$ are equally spaced by $\Delta x$. By taking the limit as $N\rightarrow\infty$ on both sides of this equation and noting that $\lim_{N\rightarrow\infty}{\dfrac{\ln{N}}{N-1}}=0$, which follows from L’Hopital rule, one has

[TABLE]

which means that

[TABLE]

The change of variable $\theta=\pi\,x$ promptly yields

[TABLE]

When the above procedure is applied to Eq. (3) one finds that

[TABLE]

The limit as $N\rightarrow\infty$ yields

[TABLE]

The change of variable $\theta=\pi\,x$ yields

[TABLE]

Now, from the product of cosines in Eq. (4) one has

[TABLE]

The limit as $N\rightarrow\infty$ yields

[TABLE]

The change of variable $\theta=\frac{\pi}{2}-\pi\,x$ yields

[TABLE]

From the product of tangents in Eq. (1), one has

[TABLE]

By substituting $2N+1=M$ (hence $M$ is an odd positive integer) and then dividing both sides by $M$, one has

[TABLE]

where $x_{n}={\,n/M}$ and $\Delta x={\,1/M}$. The limit as $M\rightarrow\infty$ yields

[TABLE]

which means that

[TABLE]

The change of variable $\theta=\pi\,x$ yields

[TABLE]

Surprisingly, this method also works in evaluating the ‘impossible’ integral $\int_{0}^{1}{\ln{\Gamma(x)}\,dx}$, where $\Gamma(x)$ is the Euler gamma function.333Here, the word impossible certainly reflects the opinion of most students. For this, let us add $(N-1)\,\ln{\pi}$ on both sides of Eq. (5), which yields

[TABLE]

This promptly simplifies to

[TABLE]

Now, let us make use of the reflection property $\Gamma{(x)}\cdot\Gamma{(1-x)}={\,\pi/\sin{(\pi\,x)}}$, valid for all $x\not\in\mathbb{Z}$. By substituting this in Eq. (14) and then dividing both sides by $N-1$, one finds that

[TABLE]

By taking the limit as $N\rightarrow\infty$, one finds

[TABLE]

Now, note that $\int_{0}^{1}{\left[\ln{\Gamma{(x)}}+\ln{\Gamma{(1-x)}}\right]\>dx}=\int_{0}^{1}{\ln{\Gamma{(x)}}\>dx}+\int_{0}^{1}{\ln{\Gamma{(1-x)}}\>dx}$. By substituting $y=1-x$ in the latter integral, one finds

[TABLE]

which means that

[TABLE]

This nice evaluation appears on the cover of the gripping book Irresistible integrals, by Boros and Moll [4].

I left for the reader the less obvious task of using the identity

[TABLE]

valid for all positive integer $N$ and all $\theta\in\mathbb{R}$, for evaluating the definite integral

[TABLE]

The proof of the trigonometric identity in Eq. (17) is proposed as an exercise in Ref. [3] (see its Appendix A.3, Ex. 6).