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A collection of definite integrals

Several well known definite integrals can be found in the book Table of integrals series and products by I.S. Gradshteyn and I.M. Ryzhik. In this article, I present a few definite integrals which I could not find the this book. This work on definite integral is inspired by Ramanujan’s legendary work in definite integrals. I must say that finding beautiful integrals is definitely addictive.

Entry 1. If G denotes the Catalan constant and \gamma denotes the Euler–Mascheroni constant then

\displaystyle{\int_{0}^{\infty}e^{-x^4}\ln^2 x dx = \frac{\Gamma(\frac{1}{4})}{256}\{32G+4\gamma^2+5\pi^2+36\ln^2 2}


Entry 2. If Im(a) >0, b>0 then

\displaystyle{\int_{0}^{\infty} i^{2ax^b}dx = \frac{1}{b}\Gamma\Big(\frac{1}{b}\Big)\Big(\frac{i}{a \pi}\Big)^{1/b}}.

Entry 3. If Re(a) >0, b>1, c>0, d>0 then

\displaystyle{\int_{0}^{\infty} \frac{e^{-ax^b}}{1+dx^b} dx = \frac{e^{a/d}}{d^{1/b}}\Gamma\Big(\frac{b+1}{b}\Big)\Gamma\Big(\frac{b-1}{b}, \frac{a}{d}\Big)}.

Entry 4.

\displaystyle{\int_{0}^{\infty} \frac{e^{-ax^{-b}}}{1+dx^b} dx = \frac{e^{ad}}{d^{1/b}}\Gamma\Big(\frac{b-1}{b}\Big)\Gamma\Big(\frac{1}{b}, \frac{a}{d}\Big)}.

Entry 5.

\displaystyle{\int_{0}^{\infty} \frac{e^{-ax^b}}{1+dx^{-b}} dx = \frac{e^{ad} d^{1/b}}{b^2}\Gamma\Big(\frac{1}{b}\Big)\Gamma\Big(-\frac{1}{b}, ad\Big)}.

Entry 6. If a > 0, b > 0 then

\displaystyle{\int_{0}^{\infty} \ln\Big(1 + \frac{a^2}{x^2}\Big) \cos (bx) dx = \frac{\pi - \pi e^{-ab}}{b}}.

Entry 7. If G denotes the Catalan constant then

\displaystyle{\int_{0}^{1} \Big(\frac{\tan ^{-1} x}{x}\Big)^2dx = G - \frac{\pi^2}{16} + \frac{\pi \ln 2}{4}}.

Entry 8.

\displaystyle{\int_{0}^{1} (\tan ^{-1} x)^2dx = -G + \frac{\pi^2}{16} + \frac{\pi \ln 2}{4}}.


Zeta, gamma and the Stirling numbers

Ramanujan’s work on definite integrals is legendary. In the words of Hardy, “… he could of course evaluate any evaluable definite integral.” Today the great days of definite integrals and beautiful identities are long gone. Nevertheless Ramanujan’s work continues to inspire such beauties. In this article, I present a few definite integrals inspired by the works of Ramanujan. (I have not seen these above integral in mathematical literature. If anybody knows otherwise, please give me the reference.)

The integral formula

\displaystyle{\int_{0}^{\infty} x^{t-1} e^{- ax} (\ln x)^r dx = \frac{\partial^r}{\partial t^r}\{a^{-t} \Gamma (t,a)\}}

where r=0,1,... , Re(a) >0, Re(t) > 0 is well known and can be found in [1]. However this form does not help much in the sense that it does not make it any easier to evaluate the n-th order partial derivative in the RHS. In this article we shall consider definite integrals of the type

\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- ax^m} (\ln x)^r dx}

where m, n and r are positive integers and a \ge 1. We shall see that it is possible to explicitly represent evaluate such definite integrals in terms of the Euler–Mascheroni constant \gamma, the Riemann zeta function \zeta (s) and the Stirling number of the first kind s_{n,k}. As usual let H_n denote the harmonic number defined as H_n = \sum_{r=1}^{n} \frac{1}{r}. We define I_n = \sum_{r=2}^{n} \frac{H_{r-1}}{r} and J_n = \sum_{r=3}^{n} \frac{I_{r-1}}{r!}. We have the following definite integrals.

Entry 1.

\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- x^{m}} (\ln x)^3 dx}

\displaystyle{=\frac{(n-1)!}{m^4} \{-2\zeta(3) + 3\zeta(2)H_{n-1} - 3\gamma \zeta(2)}

\displaystyle{- 6\gamma I_{n-1}+ 3\gamma ^2 H_{n-1} - \gamma ^3 + 6J_{n-1}\}}

\displaystyle{=\frac{(-1)^n}{m^4}[\{2\zeta (3) + 3\gamma \zeta(2) - \gamma ^3\} s_{n,1}+ 3\{\zeta(2)+ \gamma ^2\} s_{n,2} + 6\gamma s_{n,3}+6 s_{n,4}]}

Entry 2.

\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- ax^{m}} (\ln x)^2 dx}

\displaystyle{=\frac{(n-1)!}{m^3 a^n}\{\zeta(2) - 2\gamma H_{n-1} + \gamma ^2 + 2I_{n-1}+ \ln^2 a - 2H_{n-1}\ln a + 2\gamma \ln a\}}

=\frac{(-1)^{n-1}}{m^4}[\{\zeta(2)+(\gamma + \ln a)^2 \}s_{n,1}+2(\gamma + \ln a)s_{n,2}+ 2s_{n,3}].

Entry 3.

\displaystyle{\int_{0}^{\infty} x^{mn-1} e^{- a x^{m}} \ln x dx = \frac{(n-1)!}{m^2 a^n} \{H_{n-1} - \gamma - \ln a\}.}

Corollary. We have the following formula for the Stirling numbers of the first kind.

\displaystyle{s_{n,2} = n!H_n} = n!\sum_{r=1}^{n}\frac{1}{r}

\displaystyle{s_{n,3} = -n!I_n} = -n!\sum_{r=2}^{n}\frac{H_{r-1}}{r}

\displaystyle{s_{n,4} = n!J_n} = n!\sum_{r=3}^{n}\frac{I_{r-1}}{r!}.


[1] I.S. Gradshteyn and I.M. Ryzhik – Table of integrals series and products, (2007)