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Consider $$\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx$$

I thought about making some substitution like : $x^{y}$(it's easy to calculate $$\int_{0}^{1}\frac{x^{b}-x^{a}}{\log(x)}dx = \int_{a}^{b}dy\int_{0}^{1}x^{y}dx=\log(b-1)-\log(a-1)$$).

But I couldn't find substitution like that. Any hints?

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  • $\begingroup$ Have you considered splitting the integral and considering something like $$I(\alpha) = \int_{0}^{1} \frac{x^{\alpha}}{\ln x} \sin(\alpha \ln x) dx \implies I'(\alpha) = \int_{0}^{1} x^{\alpha} \cos(\alpha \ln x) dx$$? $\endgroup$ – mattos Nov 15 '17 at 8:30
  • $\begingroup$ @Mattos I guess there should be $x^{a}$, not $\alpha$? $\endgroup$ – openspace Nov 15 '17 at 8:59
  • $\begingroup$ Yes, it was supposed to be $x^{a}$. Though you could also try considering $$\int_{0}^{1} \frac{x^{\alpha}}{\ln x} \sin(\ln x) dx$$ instead. $\endgroup$ – mattos Nov 15 '17 at 9:24
  • $\begingroup$ Observing this $\frac{d}{dt}\left(\frac{x^t}{\ln x}\right) =x^t$ it becomes easy see below $\endgroup$ – Guy Fsone Nov 15 '17 at 14:49
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Your integral consists of two similar part and we only do one of them, namely: \begin{align} \int_0^1 \frac{x^a}{\ln(x)}\sin(\ln(x)) dx \end{align} Set $x=e^{-u}$ so that $dx = -e^{-u}du$, and we get: \begin{align} \int^\infty_0 \frac{e^{-(a+1)u}}{u}\sin(u)du \end{align} Now consider: \begin{align} I(z) = \int^\infty_0 \frac{e^{-zu}}{u}\sin(u)du \end{align} This is well defined as a real number for $z>0$, otherwise the integral won't converge. We also have: \begin{align} I'(z) = -\int^\infty_0 e^{-zu}\sin(u) du \end{align} I assume you can evaluate that, two times partial integration and you get: \begin{align} I'(z) = -\frac{1}{z^2+1} \end{align} Integrating this gives us $I(z)$ namely: \begin{align} I(z) = -\arctan(z) + C \end{align} Now we would actually determine $C$. But that is not needed, since our original integral is a diference of $I(b+1)$ and $I(a+1)$ so that $C$ will be canceled. So: \begin{align} \int_0^1 \frac{x^b-x^a}{\ln(x)}\sin(\ln(x)) dx =I(b+1) - I(a+1) = \arctan(a+1) - \arctan(b+1) \end{align} And we have (indirectly) seen that is only convergent if $a>-1, b>-1$.

Edit: If you really want to determine the value of $C$: Note that we know the following: \begin{align} \lim_{z\to \infty} I(z) = -\frac{\pi}{2}+C \end{align} And we also know: \begin{align} \lim_{z\to \infty} I(z) = \lim_{z\to \infty} \int^\infty_0 \frac{e^{-zu}} {u}\sin(u)du \stackrel{\text{DCT}}{=} \int^\infty_0 \lim_{z\to \infty} \frac{e^{-zu}} {u}\sin(u)du = 0 \end{align} Where DCT means Dominated Convergence Theorem. Now we can solve for $C$ and finally we get: \begin{align} I(z) = -\arctan(z) +\frac{\pi}{2} \end{align}

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It suffices to evaluate $$\int_0^1 \frac{x^a}{\ln x}\sin(\ln x) dx = \int_0^\infty \frac{e^{-(a+1)x}}{x}\sin x dx = \int_0^\infty {e^{-(a+1)x}}\sin x d(\ln x)$$

Integration by parts shows it suffices to evaluate integrals like $$\int_0^\infty {e^{-(a+1-i)x}}\ln x dx$$

Using the following formula, valid for $\Re(z)> 0$, enables you to finish the problem: $$\int_0^\infty e^{-zx}\ln x dx = -\frac{\gamma+\ln z}{z}$$ where $\gamma$ is this constant.

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Observes that: $$\color{red}{\frac{d}{dt}\left(\frac{x^t}{\ln x}\right) =x^t}$$ we have,

\begin{align}\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx &= \int_{0}^{1} \left[\frac{x^{t}}{\log(x)}\right]_a^b\sin(\log(x))dx \\ &=\int_a^b\int_{0}^{1} x^t\sin(\log(x))dx dt\\ & \overset{\color{red}{u =-\ln x}}{=}- \int_a^b\int_{0}^{\infty} e^{-ut-u}\sin(u)du dt\\ &= -\int_a^b Im\left(\int_{0}^{\infty} e^{(-t-1+i)u}du \right)dt\\ &=-\int_a^b Im\left[\frac{1}{-t-1+i}e^{(-t-1+i)u} \right]_0^\infty dt \\ &=- \int_a^b \frac{dt}{(t+1)^2+1} \\ &=\color{blue}{\arctan (a+1)-\arctan (b+1)}\end{align}

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    $\begingroup$ You made a mistake on the third line. You have $dx = e^{-u}du$ $\endgroup$ – Dylan Nov 15 '17 at 18:33
  • $\begingroup$ @Dylan you are right. Thanks I corrected it $\endgroup$ – Guy Fsone Nov 15 '17 at 19:09
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    $\begingroup$ From the substitution $u = - \ln x$ the $\sin(\ln(x))$ term becomes $\sin (-u) = - \sin(u)$ meaning the final answer should be $\arctan(a + 1) - \arctan (b + 1)$. $\endgroup$ – omegadot Nov 16 '17 at 1:21
  • $\begingroup$ Thanks you are right $\endgroup$ – Guy Fsone Nov 16 '17 at 6:12
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We can use a Frullani Integral approach, but in $\mathbb{C}$, it requires Cauchy's Integral Theorem instead of direct cancellation. $$ \newcommand{\Im}{\operatorname{Im}} \begin{align} \int_0^1\frac{x^b-x^a}{\log(x)}\sin(\log(x))\,\mathrm{d}x &=\int_0^\infty\frac{e^{-(b+1)x}-e^{-(a+1)x}}{x}\sin(x)\,\mathrm{d}x\tag1\\ &=\Im\left(\int_0^\infty\frac{e^{-(b+1-i)x}-e^{-(a+1-i)x}}{x}\,\mathrm{d}x\right)\tag2\\ &=\lim_{\epsilon\to0}\Im\left(\int_\epsilon^{1/\epsilon}\frac{e^{-(b+1-i)x}-e^{-(a+1-i)x}}{x}\,\mathrm{d}x\right)\tag3\\ &=\lim_{\epsilon\to0}\Im\left(\int_{(b+1-i)\epsilon}^{(b+1-i)/\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{(a+1-i)\epsilon}^{(a+1-i)/\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\right)\tag4\\ &=\lim_{\epsilon\to0}\Im\left(\int_{(b+1-i)\epsilon}^{(a+1-i)\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x-\int_{(b+1-i)/\epsilon}^{(a+1-i)/\epsilon}\frac{e^{-x}}{x}\,\mathrm{d}x\right)\tag5\\ &=\Im\left(\log\left(\frac{a+1-i}{b+1-i}\right)\right)\tag6\\[6pt] &=\tan^{-1}(a+1)-\tan^{-1}(b+1)\tag7 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto e^{-x}$
$(2)$: $\sin(x)=\Im\left(e^{ix}\right)$
$(3)$: write improper integral as a limit
$(4)$: separate integrals and substitute $x\mapsto\frac{x}{b+1-i}$ and $x\mapsto\frac{x}{a+1-i}$
$(5)$: apply Cauchy's Integral Theorem
$(6)$: as $\epsilon\to0$, $e^{-x}\to1$ in the first integral and $e^{-x}\to0$ in the second
$(7)$: $\Im(\log(x+iy))=\tan^{-1}(y/x)$

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