Integrating a Tricky Cosine Function: Need Help with Substitution

[Dreadfort]

Homework Statement

Solve the integral:
[/B]
$\int_{-\infty}^\infty {\frac {\cos(x)}{x^2 + 1}} \, dx$

Homework Equations

The Attempt at a Solution

I'm a bit stuck here, so what to substitute for tan \theta so as to compute the integral? Help me out here please

 

[phinds]

That's going to be hard since your attachment is unreadable.

 

[Dreadfort]

That's going to be hard since your attachment is unreadable.

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

 

[Nguyen Son]

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

 

[Dreadfort]

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

Not really, I haven't. Could you tell me how to use it, I'm going to try it then

 

[Arman777]

I made a mistake but I am thinking :)

 

[Dreadfort]

there's much more easy way

So how to approach it ?

 

[Nguyen Son]

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?

 

[Dreadfort]

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 I think that'll be better for now, till the time I learn to use Complex variable residues and all that

 

[Nguyen Son]

With complex variable and residues knowledge, I can give you the answer is $\frac{\pi}{e}$, but we should try another way to obtain it

 

[Dreadfort]

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

Yes mate

 

[Arman777]

You can try integration by part to find the solution. It works If I didnt make any mistake. Take as $u=cosx$ and $dx\frac {1} {1+x^2}=dw$ and you ll see that you can find a solution.

 

[Arman777]

I just wonder you solved or not ?

 

[Dreadfort]

I just wonder you solved or not ?

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?

 

[Arman777]

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

Its kind of hard to understand.

Lets call the integral that we are trying to find I. In the end ehat did you get ?

 

[Dreadfort]

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 back I again. So I am going wrong somewhere

 

[Arman777]

I got back I again. So I am going wrong somewhere

I got something like, I=2a-I where a=sinarctanx or it was something like that simply a=uw

 

[Dreadfort]

I got something like, I=2a-I where a=sinarctanx or it was something like that simply a=uw

Solution's given \pi /e

 

[Arman777]

Solution's given \pi /e

Well If I=a then you ll find the solution after the steps.
Its just the part of the solution.

 

[Dreadfort]

Well If I=a then you ll find the solution after the steps.
Its just the part of the solution.

Ok I am working it out

 

[Dick]

Ok I am working it out

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 $F'(1)$.

 

[Mark44]

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

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).

Using these begin/end tags, what you wrote above looks like

I substituted $x = \tan (\theta)$ and the limits to $-\frac {\pi} {2}$ to $+\frac {\pi} {2}. {\sec (\theta)}^2$ cancels out leaving behind $\int {\cos{\tan(\theta)}} \, dx$.

 

[Dreadfort]

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 $F'(1)$.

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

 

[epenguin]

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

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]

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).

I am doing BSc Physics and this is in my first year Mathematics module. The topic for this assignment was Improper Integrals

 

[Ray Vickson]

I am doing BSc Physics and this is in my first year Mathematics module. The topic for this assignment was Improper Integrals

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!

 

[Dreadfort]

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!

Thanks mate I'll be working it out now

 

[epenguin]

I think you were closer to an answer earlier on.

Expecting you to use techniques much more advanced than your course level would be odd.

I even said I thought this was about definite integration. If indefinite integration does not seem to be giving right answers, looking carefully at the definite integrals you might zero in.