- #1
Dreadfort
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Homework Statement
Solve the integral:
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##\int_{-\infty}^\infty {\frac {\cos(x)}{x^2 + 1}} \, dx##
Homework Equations
The Attempt at a Solution
Attachments
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So what I did, I substituted x = \tan (\theta) and the limits to -\frac {\pi} {2} to +\frac {\pi} {2}. {\sec (\theta)}^2 cancels out leaving behind \int {\cos{\tan(\theta)}} \, dx. I'm stuck after thatphinds said:That's going to be hard since your attachment is unreadable.
Not really, I haven't. Could you tell me how to use it, I'm going to try it thenNguyen Son said:I think you can't continue with ##\cos(\tan \theta)##. I have a question, have you learn about Complex variable, Residues? We can use them to solve some improper integral like this
So how to approach it ?Arman777 said:Theres much more easy way
Yes I think that'll be better for now, till the time I learn to use Complex variable residues and all thatNguyen Son said:Hmmm, so I think we should try another way, because you haven't learned complex variable and residues. I think your teacher will not happy/satisfy if you use Calculus II knowledge to solve Calculus I problem, right?
Yes mateNguyen Son said:By complex variable and residues knowledge, I can give you the answer is ##\frac{\pi}{e}##, but we should try another way to obtain it
Its kind of hard to understand.Dreadfort said:I substituted u and w like you said but in the end I got back the same integral again. Dunno where am going wrong. Am I supposed to substitute u in terms of w?
View attachment 221206
I got back I again. So I am going wrong somewhereArman777 said:Its kind of hard to understand.
Lets call the integral that we are trying to find I. In the end ehat did you get ?
I got something like, I=2a-I where a=sinarctanx or it was something like that simply a=uwDreadfort said:I got back I again. So I am going wrong somewhere
Solution's given \pi /eArman777 said:I got something like, I=2a-I where a=sinarctanx or it was something like that simply a=uw
Well If I=a then you ll find the solution after the steps.Dreadfort said:Solution's given \pi /e
Ok I am working it outArman777 said:Well If I=a then you ll find the solution after the steps.
Its just the part of the solution.
Dreadfort said:Ok I am working it out
Please take a look at out LaTeX tutorial (https://www.physicsforums.com/help/latexhelp/). It looks like you already understand a lot of it, but at this site you have to use a pair of # characters at each end of your expression (for inline math expressions) or a pair of $ characters (for standalone math expressions).Dreadfort said:So what I did, I substituted x = \tan (\theta) and the limits to -\frac {\pi} {2} to +\frac {\pi} {2}. {\sec (\theta)}^2 cancels out leaving behind \int {\cos{\tan(\theta)}} \, dx. I'm stuck after that
I substituted ##x = \tan (\theta)## and the limits to ##-\frac {\pi} {2}## to ##+\frac {\pi} {2}. {\sec (\theta)}^2## cancels out leaving behind ##\int {\cos{\tan(\theta)}} \, dx##.
So its a Calculus II problem then, don't know why its given in our assignment. I'll brush up on contour integration to solve this oneDick said:I would have a little more to say if you had posted something more legible, but I wouldn't break your neck pushing the integration by parts. It's really not going to work. The problem is a perfectly straightforward contour integral. The only other way I know of to do it is to use a Feynman parametric integral approach. I.e. write ##F(y)=\int_{-\infty}^{\infty} \frac{\sin(xy)}{x (1+x^2)} dx## and try to find a differential equation for ##F(y)## that you can solve. Once you've done that your integral is ##F'(1)##.
Dreadfort said:So its a Calculus II problem then, don't know why its given in our assignment. I'll brush up on contour integration to solve this one
I am doing BSc Physics and this is in my first year Mathematics module. The topic for this assignment was Improper Integralsepenguin said:Booth actually is the subject matter your course is currently covering? Usually assignment questions about what is currently being covered.
I'll bet it's about aspects of definite integration come, indefinite integration of this looking too difficul if possible at all my method of your level (or mine).
Dreadfort said:I am doing BSc Physics and this is in my first year Mathematics module. The topic for this assignment was Improper Integrals
Thanks mate I'll be working it out nowRay Vickson said:I will outline another approach that does not use contour integration, but is still way beyond most Calculus II material.
A fairly standard trick when dealing with such problems is to introduce a function like
$$F(k) = \int_{-\infty}^{\infty} \frac{\cos(kx)}{x^2+1} \, dx, $$
and to note that you want to compute ##F(1)##. We can try differentiating twice ##F(k)## under the integral sign to get back something close to ##F(k)## again, and so get a differential equation to solve. However, in this case, differentiating once or twice leads to a divergent integral, so to get around that we introduce an additional "mollifier" to ensure convergence. In other words, we look at
$$G_a(k) = \int_{-\infty}^{\infty} \frac{e^{-ax^2} \cos(kx)}{x^2+1} \, dx,$$
where ##a>0##. At the very end we will take the limit ##a \to 0.##
Differentiating twice wrt ##k## under the integral sign gives
$$\frac{d^2}{dk^2} G_a(k) = -\int_{-\infty}^{\infty} \frac{e^{-ax^2} \, x^2 \,\cos(kx)}{x^2+1} \, dx.$$
Writing ##-x^2 = -(x^2+1) +1## in the numerator, we obtain
$$\frac{d^2}{dk^2} G_a(k) = - \int_{-\infty}^{\infty} \cos(kx) e^{-a x^2} \, dx + G_a(k)$$
The integral on the right is do-able, so the differential equation for ##G_a## is
$$\frac{d^2}{dk^2} G_a(k) = G_a(k) - \frac{\sqrt{\pi}}{\sqrt{a}} e^{-k^2/(4a)}.$$
This can be solved in terms of the so-called "error function", and the resulting solution contains two constants of integration. If we note that ##G_a(k) = G_a(-k)## for all real ##k##, we can express one of the integration constants in terms of the other. Now we also have that ##G_a(0) = \int e^{-ax^2}/(x^2+1) \, dx,## which is do-able in terms of the error function. We can also get ##G_a(0)## by putting ##k=0## in the solution of the differential equation. Equating these two quantities fixes the constant of integration, so now we have an explicit formula for ##G_a(k)## that no longer contains any unknown constants. Now we can put ##k=1## and then take the limit as ##a \to 0##.
We get the answer ##\pi/e##, which could have been obtained in 3 or 4 lines of easy work using contour integration!
An integral is a mathematical concept that represents the area under a curve on a graph. It is used to find the exact value of a quantity that is changing continuously over a given interval.
Integrals can be challenging to solve, especially for complex functions. It requires a good understanding of mathematical concepts and techniques, and sometimes, it can be difficult to know where to start. Therefore, seeking help can make the process easier and more efficient.
Some common techniques for solving integrals include substitution, integration by parts, and using trigonometric identities. It is also useful to know the basic properties of integrals, such as linearity and the fundamental theorem of calculus.
One can check if their integral solution is correct by differentiating the solution and comparing it to the original function. If they are the same, then the solution is correct. Additionally, online integral calculators can be used to verify the solution.
Some tips for solving integrals include breaking down the integral into smaller parts, recognizing patterns, and practicing regularly. It is also helpful to understand the concepts behind the techniques used for solving integrals rather than just memorizing formulas.