Can we evaluate the integral using Feynman’s, Frullani’s or double integral without Beta functions?

Question

Background

When I first met the integral

$$I=\int_0^{\infty} \frac{\tan ^{-1}(a x)-\tan ^{-1}(x)}{x} d x,$$

I tried to use Feynman’s Technique by considering the integral with parameter $a$ $$ I(a)=\int_0^{\infty} \frac{\tan ^{-1}(a x)-\tan ^{-1}(x)}{x} d x, \quad \textrm{ where } a>0 $$ Differentiating $I(a)$ gives $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{1}{1+a^2 x^2} dx\\ & =\frac{1}{a}\left[\tan ^{-1}(a x)\right]_0^{\infty} \\ & =\frac{\pi}{2 a} \end{aligned} $$ Integrating $I^{\prime}(a) $ from $a=1$ to $e$ gives $$ \begin{aligned} I(e)-I(1) & =\int_0^e \frac{\pi}{2 x} d x =\frac{\pi}{2}[\ln x]_1^e \\I=I(e)& =\frac{\pi}{2} \end{aligned} $$

Then I discovered that $I$ can be also evaluated as a double integral, Frullani’s integral etc., I started to try similarly but failure to evaluate the following general integral $$I(a)=\int_0^{\infty} \frac{\tan ^{-1}(e x)-\tan ^{-1}(x)}{x^a} d x,$$ where $0<a<1.$

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Investigation by Beta function

In order to convert $I$ into a Beta function, I use integration by parts on $\frac{1}{x^a}$ as below: $$ \begin{aligned} I(a) & =\frac{1}{-a+1} \int_0^{\infty}\left(\tan ^{-1}(e x)-\tan ^{-1} x\right) d\left(x^{-a+1}\right) \\ & =\frac{1}{-a+1}\left[x^{-a+1}\left(\tan ^{-1}(e x)-\tan ^{-1} x\right)\right]_0^{\infty} -\frac{1}{-a+1} \int_0^{\infty}\left(\frac{x^{-a+1} e}{1+e^2 x^2}-\frac{x^{-a+1}}{1+x^2}\right) d x\\&= \frac{e}{1-a} \int_0^{\infty} \frac{x^{1-a}}{1+e^2 x^2} d x+\frac{1}{1-a} \int_0^{\infty} \frac{x^{1-a}}{1+x^2} d x\\& \stackrel{ex\mapsto x}{=} -\frac{1}{(1-a) e^{1-a}} \int_0^{\infty} \frac{x^{1-a}}{1+x^2} d x+\frac{1}{1-a} \int_0^{\infty} \frac{x^{1-a}}{1+x^2} d x\\&= \frac{1}{1-a}\left(1-\frac{1}{e^{1-a}}\right) \int_0^{\infty} \frac{x^{1-a}}{1+x^2} d x \end{aligned} $$ Using the result in the post,we have $$ \boxed{I(a)=\frac{\pi(1-e^{a-1})}{2(1-a)}\csc \left(\frac{\pi a}{2}\right)} $$

For examples,

$$ \begin{aligned} I\left(\frac{1}{2}\right) & =\frac{\pi}{2 \times \frac{1}{2}} \csc \left(\frac{\pi}{4}\right)\left(1-\frac{1}{\sqrt{e}}\right) =\pi \sqrt{2}\left(1-\frac{1}{\sqrt{e}}\right) \\ I\left(\frac{1}{3}\right) & =\frac{3 \pi}{4} \csc \left(\frac{\pi}{6}\right)\left(1-\frac{1}{e^{\frac{2}{3}}}\right) =\frac{3 \pi}{2}\left(1-\frac{1}{e^{\frac{2}{3}}}\right) \end{aligned} $$

Fortunately, the original integral $I=\lim _{a \rightarrow 1} I(a)=\frac{\pi}{2}.$

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My Question: Can we evaluate $I(a)$ using Feynman’s, Frullani’s or double integral without Beta functions?

Answer 1

Consider

$$\vartheta(a,b,c) = \int_{0}^{\infty} \frac{\arctan(bx)-\arctan(cx)}{x^a}dx$$ Do the substitution $x =\sqrt{w}$

$$\vartheta(a,b,c) = \frac{1}{2}\int_{0}^{\infty} \frac{\arctan(b\sqrt{w})-\arctan(c\sqrt{w})}{w^{\frac{1}{2}+\frac{a}{2}}} dw= \frac{1}{2}\int_{0}^{\infty} w^{\frac{2-a}{2}-1}\frac{\arctan(b\sqrt{w})-\arctan(c\sqrt{w})}{\sqrt{w}}dw$$

Now note that,

$$ \frac{\arctan(b\sqrt{w}) -\arctan(c\sqrt{w})}{\sqrt{w}} = \sum_{n=0}^{\infty} \frac{(b^{2n+1}-c^{2n+1})(-w)^{n}}{2n+1} = \sum_{n=0}^{\infty} \frac{n!(b^{2n+1}-c^{2n+1})}{2n+1}\frac{(-w)^{n}}{n!}$$

Denote $\displaystyle \varphi(n) = \frac{n!(b^{2n+1}-c^{2n+1})}{2n+1}$.

Applying the Ramanujan master theorem \begin{align*} \vartheta(a,b,c) =& \frac{1}{2}\int_{0}^{\infty} w^{\frac{2-a}{2}-1}\frac{\arctan(b\sqrt{w})-\arctan(c\sqrt{w})}{\sqrt{w}}dw\\ = &\frac{1}{2} \Gamma\left(\frac{2-a}{2}\right)\varphi\left(\frac{a-2}{2}\right)\\ = & \frac{1}{2} \Gamma\left(1-\frac{a}{2}\right)\Gamma\left(\frac{a}{2}\right)\frac{(b^{a-1}-c^{a-1})}{(a-1)}\\ = & \frac{\pi (b^{a-1}-c^{a-1})}{2\sin\left(\frac{\pi a}{2}\right)(a-1)} \end{align*}

Hence $$\boxed{ \int_{0}^{\infty} \frac{\arctan(bx)-\arctan(cx)}{x^a}dx = \frac{\pi (b^{a-1}-c^{a-1})}{2\sin\left(\frac{\pi a}{2}\right)(a-1)} \quad 0<a<1, \; c,b>0 }$$

Answer 2

Let's take

\begin{align} F(a,b) & = \int_0^{\infty} \frac{\tan ^{-1}(b x)-\tan ^{-1}(x)}{x^a} dx\\ \dfrac{\partial F}{\partial b} & = \int_0^{\infty} \frac{x^{1-a}}{1+b^2x^2} dx\\ & = b^{a-2} \int_0^{\infty} \frac{u^{1-a}}{1+u^2} du \hspace{1cm} [u = bx]\\ & = \dfrac \pi 2 b^{a-2} \sec (\dfrac \pi 2 (1-a)) \hspace{0.7cm} [\text{for |1-a|<1 }] \hspace{4mm}(*)\\ & = \dfrac \pi 2 b^{a-2} \csc (\dfrac {\pi a} 2 )\\ \text{Then, } I(a) & = F(a,e)-F(a,1) = \int_1^e \dfrac \pi 2 b^{a-2} \csc (\dfrac {\pi a} 2 ) db \\ & = \dfrac \pi 2 \dfrac{e^{a-1}-1}{a-1} \csc (\dfrac{\pi a}2) \end{align}

which gives $\lim\limits_{a \rightarrow 1} I(a) = \dfrac \pi 2$

* Using the result in the post

Hope this helps you!

Answer 3

The antiderivative exist in terms of the Gaussian hypergeometric function, $$I(a,b)=\int \frac{\tan ^{-1}(b x)-\tan ^{-1}(x)}{x^a} \,dx$$ $$I(a,b)=\frac{x^{1-a}}{(2-a) (1-a)}\, A$$ where $$A=(2-a) \left(\tan ^{-1}(b x)-\tan ^{-1}(x)\right)+$$ $$x \left(\, _2F_1\left(1,\frac{2-a}{2};\frac{4-a}{2};-x^2\right)-b\, \, _2F_1\left(1,\frac{2-a}{2};\frac{4-a}{2};-b^2 x^2\right)\right)$$ Then, for the definite integral $$J(a,b)=\int_0^\infty \frac{\tan ^{-1}(b x)-\tan ^{-1}(x)}{x^a} \,dx$$ assuming $b>0$ and $0<a<1$ $$J(a,b)=\frac \pi 2\,\frac{1-b^{a-1}}{1-a}\,\csc \left(\frac{\pi a}{2}\right)$$

Answer 4

Try to avoid using Beta function

First of all, splitting the integral interval of $ I(a)=\int_0^{\infty} \frac{x^a}{1+x^2} d x$ into two and then using $x\mapsto\frac{1}{x} $ yields $$ \int_0^{\infty} \frac{x^a}{1+x^2} d x=\int_0^1 \frac{x^a+x^{-a}}{1+x^2} d x $$ Expanding the integrands into the series yields $$ \begin{aligned} I(a) & =\sum_{n=0}^{\infty}\left(-1\right)^n \int_0^1\left(x^a+x^{-a}\right) x^{2 n} d x \\ & =\sum_{n=0}^{\infty}(-1)^n\left[\frac{x^{2 n+a+1}}{2 n+a+1}+\frac{x^{2 n-a+1}}{2 n-a+1}\right]_0^1\\&=\frac{1}{2} \sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{n+\frac{1+a}{2}}+\frac{1}{n+\frac{1-a}{2}}\right)\\\\&= \frac{1}{2}\left[\psi\left(\frac{a+3}{4}\right)-\psi\left(\frac{1-a}{4}\right)+ \psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right]\\&= \frac{1}{2}\left[\psi\left(\frac{a+3}{4}\right) -\psi\left(\frac{1-a}{4}\right) + \psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right]\\&= \frac{1}{2}\left[\pi \cot \left(\pi\left(\frac{1-a}{4}\right)\right)+\pi \cot \left(\pi\left(\frac{1+a}{4}\right) \right)\right]\\&= \frac \pi 2 \sec \left(\frac{\pi a}{2}\right) \end{aligned} $$ where the second last step uses the identity $\psi(1-x)-\psi(x)=\pi \cot (\pi x)$.

Answer 5

The result can be generalized as @Bertrand87‘s and proved by using double integral. $$ \begin{aligned} I & =\int_0^{\infty} \frac{\tan ^{-1}(b x)-\tan ^{-1}(c x)}{x^a} d x \\ & =\int_0^{\infty} \frac{1}{x^{a-1}} \int_b^c \frac{1}{1+y^2 x^2} d y d x \\ & =\int_c^b \int_0^{\infty} \frac{1}{\left(1+y^2 x^2\right) x^{a-1}} d x d y \\ & =\int_c^b \int_0^{\infty} \frac{x^{1-a}}{1+y^2 x^2} d x d y \end{aligned} $$ Letting $z=xy$ and using my post gives $$ \begin{aligned} & \int_0^{\infty} \frac{\left(\frac{z}{y}\right)^{1-a}}{1+z^2} \frac{d z}{y}=y^{a-2} \int_0^{\infty} \frac{z^{1-a}}{1+z^2}dz=\frac{\pi y^{a-2}}{2} \csc \left(\frac{\pi a}{2}\right) \end{aligned} $$ Hence $$ \begin{aligned} I & =\frac{\pi}{2} \csc \left(\frac{\pi a}{2}\right) \int_c^b y^{a-2}dy=\frac{\pi \left(b^{a-1}-c^{a-1}\right)}{2(a-1)} \csc \left(\frac{\pi a}{2}\right) \end{aligned} $$ where $0<a<1, \; c,b>0$.