0
$\begingroup$

Finding $\displaystyle \int^{\infty}_{0}\bigg(\frac{1-\cos 7x}{x}\bigg)e^{-x}dx$

Plan

$$I =\int^{\infty}_{0}\frac{1}{x}\bigg(1-\frac{(7x)^2}{2!}+\frac{(7x)^4}{4!}+\cdots \bigg)\bigg(1-\frac{x}{1!}+\frac{x^2}{2!}+\cdots \bigg)dx$$

How do i solve it Help me please

$\endgroup$
  • 1
    $\begingroup$ It should be $$\frac{\log (50)}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner Jun 13 at 13:54
  • 1
    $\begingroup$ Expanding in a power series is a good idea, but you should only expand the left term in the integrand, not the exponential. You also messed up the first term after expanding it. $\endgroup$ – Cameron Williams Jun 13 at 13:56
  • $\begingroup$ math.stackexchange.com/questions/1807410/… $\endgroup$ – Olivier Oloa Jun 13 at 14:11
  • $\begingroup$ @Dr. Sonnhard Graubner , you are right, you may see Answer no. 3. $\endgroup$ – Dr Zafar Ahmed DSc Jun 13 at 17:28
4
$\begingroup$

Hint. Assume $a>0$. Set $$ I(a):= \int^{\infty}_{0}\bigg(\frac{1-\cos 7x}{x}\bigg)e^{-ax}dx $$ then, by differentiating with respect to $a$, one gets $$ I'(a)= -\int^{\infty}_{0}\bigg(1-\cos 7x\bigg)e^{-ax}dx=-\frac{1}{a}+\frac{a}{a^2+49} $$ giving $$ I(a)=\frac{1}{2} \log \left(\frac{49}{a^2}+1\right). $$

Hope it helps.

$\endgroup$
  • $\begingroup$ Hi Olivier ! Good to see you and very nice answer. $\endgroup$ – Claude Leibovici Jun 13 at 14:04
  • $\begingroup$ Hi Claude! Thank you. $\endgroup$ – Olivier Oloa Jun 13 at 14:04
0
$\begingroup$

Use $1/x=\int_{0}^{\infty} e^{-tx} dt$, then $$I=\int_{0}^{\infty}\int_{0}^{\infty} (1-\cos 7x) e^{-x(1+t)} dt~ dx=\Re \left(\int_{0}^{\infty}\int_{0}^{\infty}[e^{-x(1+t)}-e^{-x(1+7i+t)}] dx ~ dt \right)=\Re \left(\int_{0}^{\infty} \left( \frac{dt}{1+t}- \frac{dt}{1+7i+t)} \right) \right).= \Re \left( \ln\frac{1+t}{1+7i+t}\right)_{0}^{\infty}= -\Re \left(\ln \left[\frac{1}{1+7i}\right]\right)=\frac{\ln 50}{2}.$$

$\endgroup$
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}\bracks{1 - \cos\pars{7x} \over x}\expo{-x}\dd x} = \Re\int_{0}^{\infty}\bracks{1 - \expo{7\ic x} \over x}\expo{-x}\dd x \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,& -\Re\int_{0}^{\infty}\ln\pars{x}\bracks{-\expo{-x} + \pars{1 - 7\ic}\expo{-\pars{1 - 7\ic}x}}\dd x \\[5mm] = &\ \int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x - \Re\int_{0}^{\pars{1 - 7\ic}\infty}\bracks{\ln\pars{x} - \ln\pars{1 - 7\ic}}\expo{-x}\dd x \\[5mm] = &\ \Re\ln\pars{1 - 7\ic} = \bbx{{1 \over 2}\,\ln\pars{50}} \approx 1.9560 \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.