A ridiculously awesome integral with a smooth solution development

[00:00:02]Oh, hey there. Didn't see you.

[00:00:06]How's it going? It's me, Kamal, the integrals guy on YouTube once again. And today we have yet another absolute beast of an integral. We have the integral from 0 to infinity of arc tangent sin alpha / x + cosine alpha dx over 1 + x^2 with alpha being a real number between 0 and my very cursive pi /2.

[00:00:30]Okay, cool.

[00:00:33]Now, how do we approach this thing? The first thing that does come to mind given all the parameterized integrals we' solved on this channel is to apply a Fineman's trick approach that is differentiating under the integral sign.

[00:00:48]But that begs the question of whether or not we can take the derivative operator and the integral operator and switch them up. In other words, is the switch up even mathematically valid?

[00:00:59]And to answer that question, we're going to look at dominated convergence for this integral.

[00:01:05]So take a look at x. It's a positive real number on our interval of integration.

[00:01:11]And that means x + cossine alpha with alpha being bound over here is going to be a positive real number as well.

[00:01:20]And sin alpha itself is going to be a positive real number. So that means we have the arc tangent of something positive. And we know that the arc tangent of something positive is going to be less than pi / 2. So we have r tangent of sin of alpha over x + cossine alpha less than p<unk> / 2 on this interval on this domain for both x and alpha that is and this implies that the target integral I is less than pi / 2 * the integral from 0 to infinity of whatever is left and that is just a dx over 1 + x^2 term and we know exactly what that converges 2 via an arc tangent anti-derivative that is just p<unk> / 2 which results in a p<unk>^2 over 4 term and voila via dominate convergence we do know that this integral converges for all values of alpha over here and we can in fact switch up the order of operators when we differentiate. So let me just copy this down here. Thank you technology and we'll apply a derivative operator terribly sorry about the derivative operator with respect to alpha to the whole thing. So that results in I prime of alpha on the left hand side and on the right we have the integral from 0 to infinity of now the partial derivative because of the liveness rule of arc tangent of sin alpha over x + cossine alpha dx over 1 + x^2 and because we're differentiating partially with respect to alpha that means this 1 / 1 + x^2 term is just a constant. Now for the arc tangent derivative I will need some more writing space. So I have 1 / 1 + sin^ square alpha over x + cossine of alpha squared. And now for the chain rule I have to apply the quotient rule. So we have x + cossine alpha time derivative of s is cosine minus s of alpha.

[00:03:28]Derivative of cosine is s. Okay cool.

[00:03:33]And of course we need the square of the denominator in the denominator.

[00:03:39]So it looks like we have some nice cancellation taking place. And I'll write this as 1 over 1 + x^2 * 1 / on expansion x^2 + cosine square alpha + 2x cosine of alpha + sin square alpha. And the cosine square and the sin square do combine to give us a 1, which is cool.

[00:04:02]And of course we have this term in the denominator that should cancel out cancel out with this thingamabob over there. So the result is well looking a lot more cleaner. And we have up top here x * cosine alpha plus cosine square alpha.

[00:04:20]Terribly sorry about that. Plus a sin square alpha term. And again they combine to give us a one. And that's pretty much it. Although I might as well just write this out given that I've written this bar anyway. So there's the denominator again. And of course now we have the complete cancellation and we can finally rest.

[00:04:48]Well, not yet. There's still plenty of work to do. So all of this implies that I prime of alpha has this cool looking form. That is the integral terribly sorry about that. The integral from 0 to infinity of x * cossine alpha + 1 / x^2 + 1 * x^2 + 2x * cosine alpha + 1 dx.

[00:05:15]Okay, cool. And now we take the simplified picture and make it more complicated because well, why the hell not? I mean, I do have this factor down here in the denominator that does contain an x cosine alpha term. So, I'm tempted to expand using 1/2.

[00:05:33]So, let's just write it in that fashion and see where it gets us. So, I'm going to write this as 2x cosine alpha, but I could also use an x squared term, so why not? And I could use a one as well. So, I'm going to expand using 1/2 as well.

[00:05:49]And what I've done here essentially is just add a couple zeros. And 1/2 of the zero is half of x^2. So that's another half x squ here. And then I need a negative 1/2 term. And where did the one go? It's right here by the law of conservation of ones or some weird scientific mumbo jumbo. So we have x^2 + 1 in the denominator. And we have another quadratic that is x^2 + 2x cossine of alpha + 1 dx.

[00:06:22]Okay, cool. And now we can separate out terms and see what cancellations we get.

[00:06:29]And for some reason I've forgotten how to write the number two. Anyway, so the first term should be the integral from 0 to infinity times 1/2 that is of canceling out these two quadratics involved in the cosine yields dx over 1 + x^2 which is nice because we know exactly what this thing equals. And of course I can factor out the 1/2 and write this as 12 of x^2. So we have another negative sign 12 integral 0 to infinity of dx over the cooler looking quadratic x^2 + 2x cosine alpha + 1 and I need some more writing space because then we have a 1 / quart cortic polinomial term which is dope and that is the integral on which we will focus the majority of our efforts. We have integral of dx / x^2 + 1 * x^2 + 2x cosine alpha + 1.

[00:07:26]Okay, cool. Now the first of these integrals is of course trivial. That is 12. So we have a pi over 4. But then we have a - 1/2 of i sub1 which itself is a pretty fun integral to evaluate. But I will skip over it entirely because we sort of get that integral for free anyway from the integral I sub 2. I sub 2 does look a little bit daunting, but to be honest, the integrant just needs a partial fraction decomposition. So we have 1 /x^2 + 1 * x^2 + 2x cosine of alpha. For some reason, I'm just not able to write cosine of alpha today.

[00:08:03]Anyway, so we have a x + b over x^2 + 1 + a c x + d term over x^2 + terribly sorry about that 2x cosine alpha + 1. So this implies that 1 is equal to a x + b * x^2 + 2x cosine of alpha + 1 + c x + d * x^2 + 1.

[00:08:33]Okay, cool. And from here, it's simply a matter of expanding the multiplication and comparing the coefficients of the linearly independent basis vectors x to the 0, x 1, x^2, and x cub. And then you get the following results. So a here as for my notes is -1 / 2 cossine alpha. B here is zero conveniently. C is 1 /2 cosine of alpha. and d here is again conveniently equal to 1. So this implies that the target integral I sub 2 which is supposed to be ax + b. Now b is just a zero and I am left with a -1 / 2 cosine of alpha integral 0 to infinity x dx over x^2 + 1 and I will preemptively expand by a factor of two and balance that out with a four here. So I can invoke a logarithmic.

[00:09:28]I can invoke an integration via the logarithm.

[00:09:32]And what else do I have on my plate?

[00:09:35]Well, I have the cx plus d. And d is again equal to 1. So I could just write this as + 1 / 2 cosine alpha integral 0 to infinity x and a + 2 cosine alpha term. And that should suffice. And I'm left with x^2 + 2x cosine of alpha + 1 dx. That looks kind of familiar, does it not?

[00:10:01]And it's exactly this familiarity that we're going to make use of via some more algebraic manipulation. So let me just pick a different color. It's been a while since I used the yellow color. So let's take this term.

[00:10:17]Specifically, I'm talking about the integral and write this as integral 0 to infinity. And I could use a 2x over here to invoke the derivative. So that's exactly what I'll do. I have 12 of 2x + 2 cossine alpha. But that means I only have one cosine alpha. So this is where the other cosine alpha term went. And I have this thing in the denominator, our quadratic term that has been following us quite a bit.

[00:10:50]So this yields a 1/2 of the integral from 0 to infinity of 2x + 2 cossine alpha over x^2 + 2x cosine alpha + 1 dx.

[00:11:04]And again, I'm going to need some writing space for the other integral. So that is just a cosine of alpha. And that cosine of alpha term in the uh as a coefficient is actually super useful.

[00:11:17]And I again need some more writing space. So we have dx over x^2 + terribly sorry about that plus 2 cosine no it's not 2 cosine it's 2x cosine alpha + 1 which reminder is actually your i sub1 term.

[00:11:40]Okay, cool. And we have a two cosine alpha term outside. So this would be plus 1/2 I sub1 and the target integral had a - 1/2 of I sub1 which is dope.

[00:11:50]Which means again we have some really cool cancellations taking place. So the result here is just going to be terribly sorry about that. 12 of the logarithm of x^2 + 2x cosine alpha + 1 with the limits being 0 and infinity which I shall not evaluate right now anyway. I plan on evaluating it with this term up here.

[00:12:14]So all of this implies that terribly sorry about that. All of this implies that I sub 2 is just a 1 over 4 cosine of alpha. Now I can just factor that out. So I have 1 over4 again terribly sorry about that 1 over 4 cosine of alpha term times the integral from 0 to infinity of 2x over 1 + x^2 dx plus all of this junk which I will just write as one combined logarithm.

[00:12:47]So we should have log of x^2 + 2x cossine alpha + 1 over what exactly?

[00:12:55]Okay, there's a negative sign. So that means we have over x^2 + 1, which is again pretty convenient given rates of growth.

[00:13:05]And we are left with a + 1 / 2 cosine alpha* cosine alpha is just 1/2. And this is of course our target integral. our other target integral I sub 1.

[00:13:20]Okay, cool. So this implies that I sub 2 is just equal to 1 / 4 cossine of alpha time the logarithm of okay wait I can actually take this x^2 + 1 term separately and then I have a 1 + 2x cosine alpha over x^2 + 1 which actually shows the reasoning very clearly that in the limit as just let me write this out. So in the limit as x tends to zero we do have a log one term and in the limit as x tends to infinity this whole term again goes to zero because the quadratic will go fast will grow faster than the linear term than the linear monomial. So again we have a log one term. So we have a 0 minus 0 term which is just a big fat zero. So let me just crash this down to zero. And this implies that I sub 2 is actually 12 of I sub1. But returning back to the target integral I. Recall that the target integral was just equal to<unk> / 4 - 12 of I sub1. And then I sub 2 is just 12 of I sub1. So there's some lovely cancellation. And wait a second.

[00:14:35]This was not the target integral. This was the derivative of the target integral function. And this implies that the integral function I of alpha is just p<unk> / 4 * alpha plus a constant of integration that I will call b because we all forget the plus c and d examination. So we might as well give it a shot. Okay. So now what exactly is an initial value condition that we can make use of? Recall the target integral which had a sign alpha up top. So notice that as alpha tends to zero from the right I of alpha will also tend to zero from the right. Okay cool. So plugging in in fact why is there why is alpha supposed to be greater than zero?

[00:15:27]Of course alpha can be equal to zero as well. Terribly sorry about that. I believe someone may have pointed that out in the comment section already.

[00:15:37]But the limit is of course correctly written. It's supposed to approach zero from the right. And I'm just being super pedantic right now.

[00:15:46]Okay, cool. So at alpha equal to 0, we have I of 0 equal to 0, which implies that our constant of integration is zero. And this implies that I of alpha just turns out to be<unk> over4 * alpha.

[00:16:02]And recall that this is valid for alpha bound between p<unk> / 2 and zero.

[00:16:11]But what about the case of alpha between pi / 2 and pi?

[00:16:18]So, that is a little homework problem for all of you watching at home [snorts] and in the park and at school.

[00:16:29]I hope you enjoyed the video. Be sure to like and subscribe. Thank you. See you next time.