1

We use Laplace Transform to solve $\int _{0}^{\infty}\frac{\sin x}{x}dx$ as shown below: $\DeclareMathOperator{\arccot}{arccot}$ We use the property $L(\frac{\sin x}{x})=\int _{s}^{\infty}\frac{1}{s^{\prime2}+1}ds^\prime=\frac{\pi}{2}-\arctan(s)=\arccot(s)$.

Now we use the laplace transform definition $\int _{0}^{\infty}e^{-sx}\frac{\sin x}{x}dx=\arccot(s)$ and put $s=0$, getting:

$\int _{0}^{\infty}\frac{\sin x}{x}dx=\arccot(0)=\frac{\pi}{2}$

I would like to know why this is the case (if possible in a intuitive way) and why this function is not solvable (the integration continues indefinitely) using normal (Integration by parts) integration methods.

Edit: The edits are in bold (in the original question).

Edit 2: I thank everyone for your answers and comments. I would request you to put up some visual/graphic representation (if possible) to explain this so that I can have an intuitive understanding. For example, The definite integral gives the area under the curve right ? So what changes when we solve it using Laplace (or other methods) and why is it not possible to find the area of the curve using integration by parts?

J.G.
  • 115,835
paulplusx
  • 1,636
  • There are a lot of methods to attack this integral .... Laplace is one of the them . – Tolaso Aug 14 '18 at 14:48
  • @Tolaso Yes, exactly but why is it solvable when we use such methods. – paulplusx Aug 14 '18 at 14:50
  • You can also evaluate it using contour integration in the complex plane. – ncmathsadist Aug 14 '18 at 14:51
  • See https://math.stackexchange.com/q/891812/27978 for some ways. – copper.hat Aug 14 '18 at 14:53
  • 1
    Most elementary functions don't have elementary antiderivatives. Look here. In many cases, it happens to be possible to the value for some particular interval of integration, often when the interval becomes infinite. In some cases, this happens when some complicated term in an estimate goes to $0$ at infinity. I think the question "why this is the case" is really too broad to have an answer. – saulspatz Aug 14 '18 at 14:58
  • What do you mean by 'solvable' and 'the normal way'? – copper.hat Aug 14 '18 at 15:00
  • Define "normal." – Mark Viola Aug 14 '18 at 15:05
  • @MarkViola "integration by parts" I mean. – paulplusx Aug 14 '18 at 15:49
  • @copper.hat I have edited the question. – paulplusx Aug 14 '18 at 15:55
  • @ncmathsadist That's the kind of intuition I am looking for. Can you please elaborate it with references? – paulplusx Aug 14 '18 at 15:59
  • I am not 100% sure I understand what you are looking for. It is solvable as an improper Riemann or Lebesgue integral using integration by parts. It does not exists as a proper Riemann integral because the domain is not a compact interval. It does not exist as a Lebesgue integral because it is not absolutely integrable. – copper.hat Aug 14 '18 at 16:36
  • @copper.hat To put it simply I want to understand intuitively why we can't calculate the area under the curve (for the given function) when we use integration by parts (without any special condition) and what happens/changes (again intuitively/graphically) when when we use Laplace (or Riemann, etc. but lets stick to Laplace for the scope of this question) that we are then able to evaluate it. I am really sorry for being so vague, I think I don't have enough understanding to explain my own question :-( – paulplusx Aug 14 '18 at 16:42

4 Answers4

3

By Louville's Thoerem, the sinc function has no elementary anti-derivative and so the improper integral has to be calculated through other methods such as Laplace Transform or through complex analytical methods.

aleden
  • 4,007
  • Thank you. All the answers and comments suggest that it's quite difficult (if not impossible) to graph and get an intuitive understanding for my problem. Hence, I am accepting this answer because it explains part of my question and puts me on the right track to do the research I require to obtain that intuitive understanding. If I get that understanding I'll post an answer myself. – paulplusx Aug 15 '18 at 13:34
1

As said by aleden, the case of the indefinite integrals has been settled by Liouville and others (Risch), and the antiderivative of $\dfrac{\sin x}x$ is proven to have no closed-form expression.

But the case of definite integrals, some of which are sporadically solvable (by various methods such as residues) when their definite counterparts are not, remains black magic.

  • I would like to understand the concept visually/graphically. Is it possible ? – paulplusx Aug 14 '18 at 16:19
  • @paulplusx: which concept ??? –  Aug 14 '18 at 17:05
  • concept : why the area cannot be evaluated for the given curve and what changes when we use Laplace (or other methods - I don't want to be broad though ) which makes the area computable. – paulplusx Aug 14 '18 at 17:13
  • @paulplusx: I have absolutely no idea how one could represent that graphically. Especially a negative result. –  Aug 14 '18 at 17:55
0

You can solve it using Feynman's technique. But it boils down to providing the same calculation steps, with an intuitive explanation. See this video for example

Andrei
  • 37,370
0

There are loads of ways to evaluate this so-called Dirichlet integral. Just to expand on Andrei's answer, Feynman's trick gets $$\int_0^\infty\frac{\sin x dx}{x}=\int_0^\infty dy\int_0^\infty\sin x \exp -xy dx=\int_0^\infty\frac{dy}{1+y^2}=\frac{\pi}{2}.$$That first $=$ sign is where we use the trick, but I've concealed the details. Let's spell them out:$$\int_0^\infty\frac{\sin x dx}{x}=\left.\int_{0}^{\infty}\frac{\sin x\exp-xydx}{x}\right|_{y=0}=-\int_0^\infty\partial_y\int_{0}^{\infty}\frac{\sin x\exp-xydx}{x},$$since we're trying to evaluate at $y=0$ a function that vanishes at $y=\infty$. But I didn't hide all that just so no one line of LaTeX would be too long. You see, Feynman's trick isn't even needed to obtain the double-integral expression; you can simply use $\frac{1}{x}=\int_0^\infty e^{-xy}dy$. After that, the result of the argument is the same either way; but I prefer this use of $\frac{\Gamma(s)}{x^s}=\int_0^\infty y^{s-1}\exp -xy dy$ because it's probably easier than a double-layered Feynman's trick for the next power of sinc.

J.G.
  • 115,835
  • Thank you for your answer. I still would like a graphical/intuitive method (if possible) to have a proper understanding. – paulplusx Aug 14 '18 at 16:52
  • @paulplusx I suppose you can imagine a surface of height $\sin x\exp -xy$. Integrating over $y$ first means slicing the volume under the surface with close-together constant-$y$ cuts, producing pieces of volume $\sin x dx$. If we cut along perpendicular constant-$x$ lines instead, each piece has volume $\tfrac{dy}{1+y^2}$. – J.G. Aug 14 '18 at 17:40
  • Could you please reframe your answer to fit this concept : why the area cannot be evaluated for the given curve and what changes when we use Laplace (or other methods - I don't want to be broad though ) which makes the area computable. – paulplusx Aug 14 '18 at 18:24
  • @paulplusx I don't think there's any easy explanation why IBP in particular isn't helpful; every technique has integrals it won't help you with. In theory you could do it "by parts" by converting it to $\int_0^\infty\frac{1-\cos x}{x^2}dx$, which is equivalent to evaluating $\int_0^\infty\frac{\sin^2 x}{x^2}dx$, which you can then evaluate by techniques such as those discussed herein. Whether you consider that "cheating" is up to you. – J.G. Aug 14 '18 at 18:29