How to solve $I_1=\int_0^\infty\frac{(x^3e^{-2x}-x^2e^{-3x}+1)\sin x}{x}dx$ from $I$ with differentiation under integral sign

Question

I was solving problems on differentiation under integral sign :

I encountered this problem : Evaluate this integral$$I=\int_0^\infty\frac{(e^{-ax}-e^{-4bx})sin(x)}{x}dx$$ and then use it to solve the integral $$I_1=\int_0^\infty\frac{(x^3e^{-2x}-x^2e^{-3x}+1)sin(x)}{x}dx$$

I evaluated the integral I , then I divided integral I1 to 3 integrals $$I_1=\int_0^\infty x^2e^{-2x}sin(x)dx+\int_0^\infty-x e^{-3x}sin(x)dx+\int_0^\infty \frac{sin(x)}{x}dx$$ I can solve the first 2 integrals (by differentiating the integral as I clarify below).. My problem is the third integral of $I_1$. How can I deduce $$\int_0^\infty \frac{\sin x}{x}dx $$ from the integral I ?

Here is my solution to evaluate I , then the first 2 parts of the integral I1: ( I know the below solution is correct .. But how to complete to evaluate the last part of I1) $$\frac{\partial I}{\partial a}=\int_0^\infty - e^{-ax} sin(x)=\frac{-1}{1+a^2}$$ $$I=\int\frac{-1}{1+a^2}da+f(b)$$ $$I=-\tan^{-1}(a)+f(b)$$ $$f(b)=\tan^{-1}(4b)$$ $$I=-\tan^{-1}(a)+\tan^{-1}(4b)$$ $$\frac{\partial^2 I}{\partial a^2}=\int_0^\infty x e^{-ax} sin(x)=\frac{d}{da}(\frac{-1}{1+a^2})=\frac{2a}{(1+a^2)^2}$$ $$\frac{\partial^3 I}{\partial a^3}=\int_0^\infty -x^2 e^{-ax} sin(x)=\frac{d}{da}(\frac{2a}{(1+a^2)^2})=\frac{-8a^3+2a^2-8a+2}{(a^2+1)^3}$$ $$\int_0^\infty x^2e^{-2x}sin(x)dx=-\frac{\partial^3 I}{\partial a^3}|_{a=2}$$ $$\int_0^\infty-x e^{-3x}sin(x)dx=-\frac{\partial^2 I}{\partial a^2}|_{a=3}$$

Answer 1

By the complex version of Frullani's theorem, for any $a,b>0$ we have

$$ \int_{0}^{+\infty}\frac{(e^{-ax}-e^{-4bx})\sin x}{x}\,dx = \text{Im}\log\left(\frac{4b-i}{a-i}\right)=\arctan(4b)-\arctan(a). $$ This can be proved also by computing the Laplace transform of $\mathbb{1}_{(0,a)}(x)$, turning the LHS into $$ \int_{0}^{+\infty}\frac{\mathbb{1}_{(0,4b)}(s)-\mathbb{1}_{(0,a)}(s)}{s^2+1}\,ds $$ which is an elementary integral.

Answer 2

Well, for the first integral we have:

$$\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b}\right):=\int_0^\infty\frac{\exp\left(\text{a}\cdot x\right)-\exp\left(\text{b}\cdot x\right)}{x}\cdot\sin\left(\text{n}\cdot x\right)\space\text{d}x\tag1$$

Using Laplace transform, we can write:

$$\text{F}_{\space\text{n}}\left(\text{s}\right):=\mathscr{L}_x\left[\frac{\exp\left(\text{a}\cdot x\right)-\exp\left(\text{b}\cdot x\right)}{x}\cdot\sin\left(\text{n}\cdot x\right)\right]_{\left(\text{s}\right)}\tag2$$

Using the 'frequency-domain integration' property of the Laplace transform:

$$\text{F}_{\space\text{n}}\left(\text{s}\right)=\int_\text{s}^\infty\mathscr{L}_x\left[\left(\exp\left(\text{a}\cdot x\right)-\exp\left(\text{b}\cdot x\right)\right)\cdot\sin\left(\text{n}\cdot x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma=$$ $$\int_\text{s}^\infty\mathscr{L}_x\left[\exp\left(\text{a}\cdot x\right)\cdot\sin\left(\text{n}\cdot x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma-\int_\text{s}^\infty\mathscr{L}_x\left[\exp\left(\text{b}\cdot x\right)\cdot\sin\left(\text{n}\cdot x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag3$$

Using the 'frequency shifting' property of the Laplace transform:

$$\text{F}_{\space\text{n}}\left(\text{s}\right)=\int_\text{s}^\infty\mathscr{L}_x\left[\sin\left(\text{n}\cdot x\right)\right]_{\left(\sigma-\text{a}\right)}\space\text{d}\sigma-\int_\text{s}^\infty\mathscr{L}_x\left[\sin\left(\text{n}\cdot x\right)\right]_{\left(\sigma-\text{b}\right)}\space\text{d}\sigma\tag4$$

Using 'table of selected Laplace transforms':

$$\text{F}_{\space\text{n}}\left(\text{s}\right)=\int_\text{s}^\infty\frac{\text{n}}{\text{n}^2+\left(\sigma-\text{a}\right)^2}\space\text{d}\sigma-\int_\text{s}^\infty\frac{\text{n}}{\text{n}^2+\left(\sigma-\text{b}\right)^2}\space\text{d}\sigma\tag5$$

Now, when $\text{s}=0$ and $\text{n}=1$, we get your integral:

$$\text{F}_{\space1}\left(0\right)=\pi+\arctan\left(\text{a}\right)-\arctan\left(\text{b}\right)\tag6$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{0}^{\infty}{\pars{\expo{-ax} - \expo{-4bx}}\sin\pars{x} \over x}\,\dd x = \int_{0}^{\infty}\pars{\expo{-ax} - \expo{-4bx}}\ \overbrace{{1 \over 2}\int_{-1}^{1}\expo{\ic k x}\,\dd k} ^{\ds{{\sin\pars{x} \over x}}}\ \,\dd x \\[5mm] & = {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\bracks{\expo{-\pars{a - \ic k}x} - \expo{-\pars{4b - \ic k}x}}\dd x\,\dd k = {1 \over 2}\int_{-1}^{1}\pars{{1 \over a - \ic k} - {1 \over 4b - \ic k}}\dd k \\[5mm] & = \int_{0}^{1}\bracks{{a \over k^{2} + a^{2}} - {4b \over k^{2} + \pars{4b}^{2}}}\dd k = \mrm{sgn}\pars{a}\int_{0}^{1/\verts{a}}{\dd k \over k^{2} + 1} - \mrm{sgn}\pars{4b}\int_{0}^{1/\pars{4\verts{b}}}{\dd k \over k^{2} + 1} \\[5mm] & = \arctan\pars{1 \over a} - \arctan\pars{1 \over 4b} = \bbx{\mrm{arccot}\pars{a} - \mrm{arccot}\pars{4b}} \end{align}

Answer 4

You already established

$$\frac{dI}{da}=\int_0^\infty - e^{-ax} \sin(x)=-\frac{1}{1+a^2}$$ which is used below in evaluating \begin{align} \int_0^\infty \dfrac{\sin x}{x}\ dx =& \int_0^{\infty }\int _0^{\infty}e^{-ax }\sin x \>dx\:da\\ =&\> -\int_0^\infty\frac{dI}{da}da = \int_0^\infty \frac 1{1+a^2}da = \dfrac{\pi}{2} \end{align}