6

Proposed:

$$\int_{0}^{\pi/2}e^{-{\pi\over 2}\tan t}\mathrm dt=\operatorname{Ci}\left({\pi\over 2}\right)\tag1$$

Where $\operatorname{Ci}$ is the Cosine Integral

$$\operatorname{Ci}(x)=-\int_{x}^{\infty}{\cos t\over t}\mathrm dt\tag2$$

My try:

Recall (I doubt it would be any useful) $$e^{\tan t}=1+t+{t^2\over 2}+{t^3\over 2}+{3t^3\over 8}+\cdots\tag3$$

$u={\pi\over 2}\tan t\implies {\pi\over 2}\sec ^2 t$ then $(1)$ becomes

$$2\pi\int_{0}^{\infty}e^{-u}\cdot{\mathrm du\over 4u^2+\pi^2}\tag4$$

Recall $$\int{\mathrm du\over u^2+a^2}={1\over a}\tan^{-1}{u\over a}\tag5$$

probably $(4)$ we may apply integration by parts?

How does one prove $(1)$?

Gonçalo
  • 15,869
gymbvghjkgkjkhgfkl
  • 8,528

2 Answers2

5

Define $$ F(\alpha)=\int_0^{\pi/2}e^{-\alpha\tan x}\,dx,\quad\alpha>0. $$ Differentiating $F$ two times we get \begin{align} F''(\alpha)&=\int_0^{\pi/2}e^{-\alpha\tan x}\tan^2x\,dx=-F(\alpha)+\int_0^{\pi/2}e^{-\alpha\tan x}(\underbrace{1+\tan^2x}_{\tan'(x)})\,dx=\\ &=-F(\alpha)+\int_0^{+\infty}e^{-\alpha t}\,dt=-F(\alpha)+\frac{1}{\alpha}. \end{align} The resulting differential equation $$ F''+F=\frac{1}{\alpha},\quad\alpha>0, $$ can be solved e.g. by variation of parameters $$ F(\alpha)=A(\alpha)\cos\alpha+B(\alpha)\sin\alpha $$ that gives $$ \begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} A'\\B' \end{bmatrix}=\begin{bmatrix} 0\\1/\alpha \end{bmatrix}\quad\Rightarrow\quad \begin{bmatrix} A'\\B' \end{bmatrix}=\begin{bmatrix} -\frac{\sin\alpha}{\alpha}\\\frac{\cos\alpha}{\alpha} \end{bmatrix}. $$ Together with the initial conditions $$ F(+\infty)=F'(+\infty)=0 $$ it gives the solution $$ F(\alpha)=-\cos\alpha\int_{+\infty}^\alpha\frac{\sin t}{t}\,dt+\sin\alpha\int_{+\infty}^\alpha\frac{\cos t}{t}\,dt. $$ Now $\alpha=\frac{\pi}{2}$ gives the result.

A.Γ.
  • 30,381
1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\pi/2}\expo{-\pi\tan\pars{t}/2}\,\dd t & \,\,\,\stackrel{x\ =\ \tan\pars{t}}{=}\,\,\, \int_{0}^{\infty}{\expo{-\pi x/2} \over x^{2} + 1}\,\dd x = \Im\int_{0}^{\infty}{\expo{-\pi x/2} \over x - \ic}\,\dd x \\[5mm] & = \Im\int_{-\ic}^{\infty - \ic}{\expo{-\pi\ic/2}\expo{-\pi x/2} \over x}\,\dd x = -\,\Re\int_{-\ic}^{\infty - \ic}{\expo{-\pi x/2} \over x}\,\dd x \\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} \Re\int_{\infty}^{\epsilon}{\expo{-\pi x/2} \over x}\,\dd x + \Re\int_{0}^{-\pi/2} {\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} + \Re\int_{-\epsilon}^{-1}{\expo{-\pi\ic y/2} \over \ic y}\,\ic\,\dd y \\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} -\int_{\pi\epsilon/2}^{\infty}{\expo{-x} \over x}\,\dd x - \int^{\pi\epsilon/2}_{1}{\cos\pars{x} \over x}\,\dd y \\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\, \int^{\infty}_{\pi\epsilon/2}{\cos\pars{x} \over x}\,\dd y - \int_{\pi\epsilon/2}^{\infty}{\expo{-x} \over x}\,\dd x - \int^{\infty}_{\pi/2}{\cos\pars{x} \over x}\,\dd y \\[5mm] & = -\,\mrm{Ci}\pars{{\pi \over 2}\,\epsilon} - \,\mrm{Ei}\pars{{\pi \over 2}\,\epsilon} + \,\mrm{Ci}\pars{\pi \over 2} \label{1}\tag{1} \end{align}

$\ds{\mrm{Ei}}$ is the Exponential Integral Function. Note that, as $\ds{z \to 0}$, $\ds{\,\mrm{Ci}\pars{z} \sim \gamma + \ln\pars{z} + \,\mrm{O}\pars{z^{2}}}$ and $\ds{\,\mrm{Ei}\pars{z} \sim -\gamma - \ln\pars{z} + \,\mrm{O}\pars{z^{1}}}$

such that \eqref{1} becomes $$ \bbx{\int_{0}^{\pi/2}\expo{-\pi\tan\pars{t}/2}\,\dd t = \,\mrm{Ci}\pars{\pi \over 2}} $$

Felix Marin
  • 94,079