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I've been trying to evaluate this integral for a while now my friend used complex analysis to evaluate this but he got a wrong result, i tried using real methods but i've been stuck. One can probably use differentiating under the integral sign in the following ways, $$I\left(a\right)=\int _0^{\infty }\frac{\ln \left(x\right)\sin \left(ax\right)}{x^2+1}\:dx$$or$$I\left(a\right)=\int _0^{\infty \:}\frac{x^a\sin \left(x\right)}{x^2+1}\:dx$$ But that seems really complicated, i really have no idea how to proceed, please help me.

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  • $\begingroup$ I very much doubt this integral will be able to be expressed in terms of elementary functions. $\endgroup$ – Giraffes4thewin Jul 1 '20 at 23:02
  • $\begingroup$ What is the original integral that you are dealing with? $\endgroup$ – Luis Sierra Jul 2 '20 at 0:28
  • $\begingroup$ @LuisSierra did you not read the title, its $$\int _0^{\infty }\frac{\ln \left(x\right)\sin \left(x\right)}{x^2+1}\:dx$$. $\endgroup$ – user805212 Jul 2 '20 at 2:12
  • $\begingroup$ It is a good idea to have your post (and title) self contained. $\endgroup$ – Asaf Karagila Jul 2 '20 at 7:57
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Here's a solid start - let me know if you need for me to expand upon the material provided:

Yes employ the second, but with a slight extension $$ I(a, t) = \int_0^\infty \frac{x^a \sin(xt)}{x^2 + 1}\:dx $$

Then if the integral under inspection is $J$, then by the Dominated Convergence Theorem and Leibniz's Integral Rule, we find that $$ J = \frac{\partial I}{\partial a}\bigg|_{(a,t) =(0, 1)} $$ Thus, we need to resolve $I(a,t)$. To do so, we employ Fubini's Theorem and take the Laplace Transform with respect to $t$: \begin{align} \mathscr{L}\left[I(a,t) \right] &= \mathscr{L}\left[\int_0^\infty \frac{x^a \sin(xt)}{x^2 + 1}\:dx \right] = \int_0^\infty \frac{x^a \mathscr{L}\left[\sin(xt)\right]}{x^2 + 1}\:dx = \int_0^\infty x^a \cdot \frac{x}{s^2 + x^2} \cdot \frac{1}{x^2 + 1}\:dx \\ &= \int_0^\infty \frac{x^{a + 1}}{\left(s^2 + x^2\right)\left(x^2 + 1\right)}\:dx = \int_0^\infty x^{a + 1}\left[\frac{1}{s^2 - 1}\left(\frac{1}{x^2 + 1}- \frac{1}{s^2 + x^2} \right)\right]\:dx \\ &= \frac{1}{s^2 - 1}\left[\int_0^\infty \frac{x^{a + 1}}{x^2 + 1} \:dx - \int_0^\infty \frac{x^{a + 1}}{s^2 + x^2}\:dx \right] = \frac{1}{s^2 - 1}\left[I_1 - I_2\right] \end{align} You will observe that both $I_1$ and $I_2$ take the form: $$ H(b,k,n) = \int_0^\infty \frac{x^k}{x^n + b}\:dx = \frac{1}{n} b^{1 - \frac{k + 1}{n}} \Gamma\left(1 - \frac{k + 1}{n} \right)\Gamma\left( \frac{k + 1}{n} \right) $$ Where $\Gamma(x)$ is the Gamma Function.

Thus we observe that: \begin{align} \mathscr{L}\left[I(a,t) \right] &= \frac{1}{s^2 - 1}\bigg[H\left(1, a+1, 2\right) - H\left(s^2, a+1, 2\right)\bigg] \\ &= \frac{1}{s^2 - 1}\bigg[\frac{1}{2} \cdot 1^{\frac{a + 1 + 1}{2} - 1}\Gamma\left(1 - \frac{a + 1 + 1}{2} \right)\Gamma\left( \frac{a + 1 + 1}{2} \right) - \frac{1}{2} \cdot \left(s^2\right)^{\frac{a + 1 + 1}{2} - 1}\Gamma\left(1 - \frac{a + 1 + 1}{2} \right)\Gamma\left( \frac{a + 1 + 1}{2} \right) \bigg] \\ &= \frac{1}{2\left(s^2 - 1\right)}\Gamma\left(1 - \frac{a + 2}{2} \right)\Gamma\left( \frac{a + 2}{2} \right)\bigg[1 - s^{a} \bigg] \end{align} Here as $a$ is to be evaluated at $0$, we may employ Euler's Reflection Formula on the Gamma terms to yield: \begin{align} \mathscr{L}\left[I(a,t) \right] &=\frac{1}{2\left(s^2 - 1\right)}\pi\operatorname{cosec}\left(\pi \cdot \frac{a + 2}{2}\right)\bigg[1 - s^{a} \bigg]\\ &=\frac{\pi}{2\left(s^2 - 1\right)}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\bigg[1 - s^{a} \bigg] \end{align} We now take the Inverse Laplace Transform: \begin{align} I(a,t) &= \mathscr{L}^{-1}\left[ \frac{\pi}{2\left(s^2 - 1\right)}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\bigg[1 - s^{a} \bigg]\right] \\ &= \frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\left[\mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} \right] - \mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} s^{a}\right] \right] \\ &= \frac{\pi}{2}\operatorname{cosec}\left(\frac{\pi}{2} \left( a + 2\right)\right)\left[\sinh(t) - \mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} s^{a}\right] \right] \end{align} For the remaining inversion we employ Convolution: $$ \mathscr{L}^{-1}\left[ F(s)G(s) \right] = \int_0^t f(\tau)g(t - \tau) d\tau $$ Here let

$$ G(s) = \frac{1}{s^2 - 1} \longrightarrow g(t) = \sinh(t) $$ And so $$ F(s) = s^{a} \longrightarrow f(t) = \frac{t^{-(a + 1)}}{\Gamma(-a)} $$

And so, \begin{align} &\mathscr{L}^{-1}\left[\frac{1}{s^2 - 1} s^{a}\right] = \int_0^t \frac{\tau^{-(a + 1)}}{\Gamma(-a)} \sinh(t - \tau)\:d\tau = \frac{1}{\Gamma(-a)} \int_0^t \tau^{-(a + 1)}\sinh(t - \tau)\:d\tau \\ &= \frac{1}{\Gamma(-a)} \int_0^t \tau^{-(a + 1)}\bigg[\sinh(t)\cosh(\tau) - \cosh(t)\sinh(\tau) \bigg]\:d\tau \\ &=\frac{1}{\Gamma(-a)} \left[ \sinh(t)\int_0^t \tau^{-(a + 1)}\cosh(\tau)\:d\tau - \cosh(t)\int_0^t \tau^{-(a + 1)}\sinh(\tau)\:d\tau \right] \end{align}

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  • $\begingroup$ Does $\mathscr{L}\left[I(a,t) \right]$ invert nicely? $\endgroup$ – NoName Jul 3 '20 at 20:21
  • $\begingroup$ @NoName If you break the final bracket apart, then the first term simple involves the inverse of $\frac{1}{s^2 - 1}$ which is simple. The second component which has $s^{a + 1}\frac{1}{s^2 - 1}$ requires convolution. I will expand upon my solution if requested. $\endgroup$ – David Galea Jul 6 '20 at 1:07
  • $\begingroup$ @DavidGalea yes please, id like you to show the full solution, i will checkmark it once done $\endgroup$ – user805212 Jul 8 '20 at 6:04

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