0
$\begingroup$

I am having troubles evaluating this integral: $$\int_{-\infty}^{\infty} \frac{\cos x-1}{x^2(x^2+4)}dx$$ Well, actually, I'm quite sured about the numerical part: the value of this integral is likely to be $-\frac{(e^{-2}+1)\pi}{8}$. However, I am having troubles when proving it.

Firstly, by a given hint, I constructed a new function which is: $$f(z)=\frac{e^{iz}-1-iz}{z^2(z^2+4)}$$ Then, I constructed a contour $\gamma =\gamma_1\cup\gamma_2\cup\gamma_r\cup\gamma_R$, with:$$\gamma_1=\{x+0i:-R\le x\le-r\}$$ $$\gamma_2=\{x+0i:r\le x \le R\}$$ $$ \gamma_r = \{re^{i\theta}:\theta\in[\pi,0]\}$$$$ \gamma_R = \{Re^{i\theta}:\theta\in[0, \pi]\}$$ Then, by residue theorem, I evaluate the integral $\int_{\gamma}f(z)dz$ to be equal to $-\frac{(e^{-2}+1)\pi}{8}$. Then, what's left is to prove that $\int_{\gamma_1}f(z)dz$ and $\int_{\gamma_2}f(z)dz$ are both equal to zero as $r\to0$ and $R\to\infty$ and that's what's troubling me.

I have no idea how to prove that the integral: $$\lim_{r\to0}\int_{\gamma_r}\frac{e^{iz}-1-iz}{z^2(z^2+4)}dz$$where clearly, $z=re^{i\theta}$, is equal to $0$, and also, for the case where the contour is $\gamma_{R}$.

Please help me! Thanks a lot.

$\endgroup$
  • 1
    $\begingroup$ HINT: $$\frac{e^{iz}-1-iz}{z^2(z^2+4)}$$ has a removable singularity at $z=0$, hence it is bounded locally at $z=0$. This proves that the limit is zero as $r \to 0$. $\endgroup$ – Crostul Mar 15 at 16:44
  • $\begingroup$ Look at page 102 (section "Third type") in books.google.com/…. Also, it's not clear to me why not work with the function $$ f(z) = {(e^{iz} + e^{-iz})/2 - 1 \over z(z^2+1)}. $$ $\endgroup$ – avs Mar 15 at 16:47
  • 1
    $\begingroup$ @avs I have no idea why there is a strange $iz$ term but I was instructed to do so on the paper. Also, thanks for the book! I'm reading it to try to find some hints. $\endgroup$ – Jamie Carr Mar 15 at 16:54
  • $\begingroup$ @Crostul Thanks for the hint. What about the case when $R\to\infty$? $\endgroup$ – Jamie Carr Mar 15 at 16:56
1
$\begingroup$

Appreciate you are seeking a solution via Residues. Here is a solution using Feynman's Trick coupled with Laplace Transforms (which is Residue in disguise). Here your integral is: \begin{equation} I = \int_{-\infty}^{\infty} \frac{\cos(x) - 1}{x^2\left(x^2 + 1\right)}\:dx\nonumber \end{equation} Here we introduce the function: \begin{equation} J(t) = \int_{-\infty}^{\infty} \frac{\cos(tx) - 1}{x^2\left(x^2 + 1\right)}\:dx\nonumber \end{equation} Here we see that $J(0) = 0$ and $\lim_{t \rightarrow 1^+} J(t) = I$. We begin by taking the derivative with respect to $t$. To do so, we need to employ Leibniz's Integral Rule: \begin{equation} J'(t) = \int_{-\infty}^{\infty} \frac{\frac{\partial}{\partial t}\left[\cos(tx) - 1\right]}{x^2\left(x^2 + 1\right)}\:dx = \int_{-\infty}^{\infty} \frac{-x\sin(tx)}{x^2\left(x^2 + 1\right)}\:dx = \int_{-\infty}^{\infty} \frac{-\sin(tx)}{x\left(x^2 + 1\right)}\:dx \end{equation} We note that $J'(0) = 0$. Differentiating with respect to $t$ again, yields: \begin{equation} J''(t) = \int_{-\infty}^{\infty} \frac{-x\cos(tx)}{x\left(x^2 + 1\right)}\:dx = - \int_{-\infty}^{\infty} \frac{\cos(tx)}{x^2 + 1}\:dx \end{equation} We proceed by taking the Laplace Transform with respect to $t$. To do so, we must employ Fubini's Theorem: \begin{align} \mathscr{L}\left[J''(t)\right] &=- \int_{\infty}^{\infty}\frac{\mathscr{L}\left[\cos(tx)\right]}{x^2 + 1}\:dx \nonumber \\ s^2\mathscr{L}\left[J(t)\right] - sJ(0) - sJ'(0) &= -\int_{\infty}^{\infty} \frac{s}{s^2 + x^2} \cdot \frac{1}{x^2 + 1}\:dx \nonumber \\ s^2\mathscr{L}\left[J(t)\right] &= -s \int_{\infty}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 1\right)} \:dx \nonumber \\ -s\mathscr{L}\left[J(t)\right] &= \frac{1}{s^2 - 1}\int_{\infty}^{\infty}\left[\frac{1}{x^2 + 1} - \frac{}{s^2 + x^2} \right]\:dx = \frac{1}{s^2 - 1}\left[\arctan(x) - \frac{1}{s}\arctan\left(\frac{x}{s} \right) \right]_{-\infty}^{\infty} \nonumber \\ &= \frac{1}{s^2 - 1}\left[\pi - \frac{\pi}{s} \right] = \frac{\pi}{s^2 - 1}\cdot\frac{s - 1}{s} = \frac{\pi}{s\left(s + 1\right)} \end{align} And so, \begin{equation} \mathscr{L}\left[J(t)\right] = -\frac{\pi}{s^2\left(s + 1\right)} \Longrightarrow J(t) = \mathscr{L}^{-1}\left[-\frac{\pi}{s^2\left(s + 1\right)} \right] = -\pi\left( t + e^{-t} - 1\right) \end{equation} We now may resolve $I$ by taking the limit as above: \begin{equation} I = \int_{-\infty}^{\infty} \frac{\cos(x) - 1}{x^2\left(x^2 + 1\right)}\:dx = -\pi\left(1 + e^{-1} - 1\right) = -\pi e^{-1}\nonumber \end{equation}

$\endgroup$
  • $\begingroup$ Hi, thanks for your detailed answer! It must take some time and I appreciate your work. The problem is, I checked the value of this integral with WolframAlpha which gave me a number approximately equal to -0.445845 and my answer, which is $-\frac{(e^{-2}+1)\pi}{8}$, is also equal to that number. However, $-\pi e^{-1}$ is approximately equal to -1.55727. Could you please double-check your calculations so that I can learn more about it? Many thanks $\endgroup$ – Jamie Carr Mar 16 at 21:37
  • $\begingroup$ @JamieCarr - Yes, I will look into that now. $\endgroup$ – user619699 Mar 17 at 2:28
  • $\begingroup$ @JamieCarr - Are you sure you put it into wolframalpha correctly? I just put in and got the $-\pi e^{-1}$ result: wolframalpha.com/input/?i=%E2%88%AB(cos(x)+-+1)%2F(x%5E2*(x%5E2+%2B+1))+from+-%E2%88%9E+to+%E2%88%9E $\endgroup$ – user619699 Mar 17 at 2:31
0
$\begingroup$

Hi I am new to the site and was looking to comment a small fix, but am not immediately able to, so anyways I am just going to write it out here (hopefully no one takes offense to that, sorry in advance).

So one thing I saw in the explanation of your work is that you said "Then, what's left is to prove that ∫γ1f(z)dz and ∫γ2f(z)dz are both equal to zero as r→0 and R→∞ and that's what's troubling me.", but those two parts are combined to form the orginal integral that you were trying to solve.

So really you are just trying to prove that the two semicircles both go to zero. And in response to the end of your question when you want to know how to show that the γr integral becomes zero, think about the fact that the smaller semi-circle is bounded by the absolute value of that smaller semi-circle being less than or equal to ((pi)r)/(r^2-4), which when taking the limit as r goes to zero clearly becomes zero.

Sorry for not having much knowledge or understanding of formatting the math but hopefully you could still understand some of what I wrote, and if I made any mistakes on something please tell me so I can fix it.

$\endgroup$
  • $\begingroup$ Thanks for your answer! No worries about the format, if you want to learn about it, check out this link: math.meta.stackexchange.com/questions/5020/…. Can you tell me more about the upper-bound, $\frac{\pi r}{r^2-4}$? I'm kind of confused about how you derive this. Thank you! $\endgroup$ – Jamie Carr Mar 16 at 21:44
0
$\begingroup$

Oh my bad I didn't even think to explain that part, but for example, if we have a slightly different integral such as the integral of (e^ix)/x from negative infinity to positive infinity we can rewrite it as f(z)=(e^iz)/z, which has a simple pole at the origin but is analytic everywhere else. So let's say we used the same contour as you did for your original integral (so the smaller semi-circle within the larger semi-circle) for this one.

Then the sum of all the individual contours that define this closed contour would be equal to zero, because there are no singularities within the closed contour in order for us to use residue theory with.

Now this is somewhat obvious but your original integral has a singularity within the closed contour, and as such the sum of all the individual contours will equal 2(pi)i*(sum of the residues).

So now it is just a matter of proving that your two semi-circle integrals go to zero, and to start with the larger one from -R to R, we can make use of Jordan's Lemma.

And actually, I am going to give a different example for this in order to show that the larger semi-circle goes to zero, and let you try to figure out the smaller semi-circle with some of the intuition you might pick up here(though if you are still stumped on the smaller circle then please comment and I will try to explain it). So the integral I will use is (cosx)/(x^2+1), where we rewrite it as f(z)=(e^iz)/(z^2+1) and use a semi-circular contour made up of two integrals, the semi-circle and the real axis. So we want to prove that the semi-circle integral goes to zero and that the real axis integral is the original integral(which I won't do here since you already know that). So we can rewrite (absolute value f(z) is |f(z)|, still not good with the math formatting, wanted to clarify) |f(z)| as |f(x+iy)| which is equal to (|e^ix*e^-y|)/|z^2+1| which is equal to (e^-y)/|z^2+1|, and in the upper half plane (y is greater than or equal to 0) we get |f(z)|=1/|z^2+1|.

Now with this what we want to do is bound the absolute value of the integral of the larger semi-circle (the circle from -R to R, with radius R) with a function that will include the singularity within our contour, for this example that singularity is i, so the function will become (pi)R/(R^2-1) where R is greater than 1. Now as R goes to infinity the new function we just made will go to zero and then, as a result, the semi-circle integral will go to zero as well because of the inequality.

All this being said hopefully I was able to help you out a little with this problem.

$\endgroup$

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.