Another Interesting Putnam Integral.

[Amad27]

Helllo,

I have found another interesting integral from the old Putnam's.

Please ONLY use the "differentiation under the integral sign" rule for this one, after we have done that we can explore other possibilities. I really want to try "differentiation under the integral sign" for this.

Problem:
-----------

$$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+((\tan\left({x}\right))^\sqrt{2}} \,dx $$

So first we need to explore what we can let be $$a$$ there are a few possibilities. $$\int_{0}^{\frac{\pi}{2}}\frac{1}{1+(\tan\left({ax}\right))^\sqrt{2}} \,dx $$

OR

$$\int_{0}^{\frac{\pi}{2}}\frac{1}{1+(\tan\left({x}\right))^a} \,dx $$

So in the end, we have two ways to go,

$$ I(a) = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+(\tan\left({ax}\right))^\sqrt{2}} \,dx $$

$$ J(a) = \int_{0}^{\frac{\pi}{2}}\frac{1}{1+(\tan\left({x}\right))^a} \,dx $$$$I'(a)$$ is too complicated (checked on WolframAlpha) so it is not worth it.

The other option is $$J'(a)$$

$$J'(a) = \int_{0}^{1} \frac{-(\tan\left({x}\right))^a\cdot\ln\left({\tan\left({x}\right)}\right)}{((\tan\left({x}\right))^a+1)^2} \,dx $$

Which can be found easily by a combination of the quotient rule and log rules.

Here comes the problem... It is very tough to find the antiderivative. Just checked on WolframAlpha, I found out that WolframAlpha CANT even compute it! It runs out of "time"

Any other suggestions?

Thanks

 

[Euge]

Let

$\displaystyle J(a) = \int_0^{\pi/2} \frac{dx}{1 + (\tan{x})^a}$.

The answer we seek is $J(\sqrt{2})$. If you use symmetry, you find readily that $J(a) = \frac{\pi}{4}$ for all $a$. But since you want differentiation

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\tan{x})^a \ln(\tan{x})}{[1 + (\tan{x})^a]^2}\, dx$.

Via the u-sub $u = \frac{\pi}{2} - x$, we have

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\cot{u})^a \ln(\cot{u})}{[1 + (\cot{u})^a]^2}\, du$

$\displaystyle = -\int_0^{\pi/2} \frac{(\tan{u})^a \ln(\tan{u})}{[1 + (\tan{u})^a]^2}\, du$

$\displaystyle = -J'(a)$.

Hence $J'(a) = 0$, which implies $J(a)$ is constant. Now

$\displaystyle J(0) = \int_0^{\pi/2} \frac{du}{2} = \frac{\pi}{4}$.

Hence $J(\sqrt{2}) = J(0) = \frac{\pi}{4}$.

 

[Amad27]

Let

$\displaystyle J(a) = \int_0^{\pi/2} \frac{dx}{1 + (\tan{x})^a}$.

The answer we seek is $J(\sqrt{2})$. If you use symmetry, you find readily that $J(a) = \frac{\pi}{4}$ for all $a$. But since you want differentiation

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\tan{x})^a \ln(\tan{x})}{[1 + (\tan{x})^a]^2}\, dx$.

Via the u-sub $u = \frac{\pi}{2} - x$, we have

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\cot{u})^a \ln(\cot{u})}{[1 + (\cot{u})^a]^2}\, du$

$\displaystyle = -\int_0^{\pi/2} \frac{(\tan{u})^a \ln(\tan{u})}{[1 + (\tan{u})^a]^2}\, du$

$\displaystyle = -J'(a)$.

Hence $J'(a) = 0$, which implies $J(a)$ is constant. Now

$\displaystyle J(0) = \int_0^{\pi/2} \frac{du}{2} = \frac{\pi}{4}$.

Hence $J(\sqrt{2}) = J(0) = \frac{\pi}{4}$.

HI,

Let's explore BOTH ways.

The answer we seek is $J(\sqrt{2})$. If you use symmetry, you find readily that $J(a) = \frac{\pi}{4}$ for all $a$. But since you want differentiation

How CAN you use symmetry? Tangent is an odd function, so it is "symmetric" with respect to the x-axis. This is interesting too. Can you give me some advice?

Next, we go to the differentiation.

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\tan{x})^a \ln(\tan{x})}{[1 + (\tan{x})^a]^2}\, dx$.

Via the u-sub $u = \frac{\pi}{2} - x$, we have

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\cot{u})^a \ln(\cot{u})}{[1 + (\cot{u})^a]^2}\, du$

$\displaystyle = -\int_0^{\pi/2} \frac{(\tan{u})^a \ln(\tan{u})}{[1 + (\tan{u})^a]^2}\, du$

$\displaystyle = -J'(a)$.

Hence $J'(a) = 0$, which implies $J(a)$ is constant. Now

$\displaystyle J(0) = \int_0^{\pi/2} \frac{du}{2} = \frac{\pi}{4}$.

Hence $J(\sqrt{2}) = J(0) = \frac{\pi}{4}$.

I understand one part of this,

J'(a) = \int_0^{\pi/2} \frac{(\tan{x})^a \ln(\tan{x})}{[1 + (\tan{x})^a]^2}\, dx$.

Via the u-sub $u = \frac{\pi}{2} - x$, we have

$\displaystyle J'(a) = \int_0^{\pi/2} \frac{(\cot{u})^a \ln(\cot{u})}{[1 + (\cot{u})^a]^2}\, du$

When you substitution the u = $$\frac{\pi}{2} - x$$ you get the cotangents, that makes sense, the NEXT step is confusing. Somehow you have transformed cotangents into tangents again.

$\displaystyle = -\int_0^{\pi/2} \frac{(\tan{u})^a \ln(\tan{u})}{[1 + (\tan{u})^a]^2}\, du$

And also,

Hence $J(\sqrt{2}) = J(0) = \frac{\pi}{4}$.

I'm sorry for acting like a beginner but how did $J(\sqrt{2}) = J(0)$?

Thanks

 

[anemone]

Hey Olok,(Smile)

This is one old problem from the High School Problem Of The Week section (http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-108-april-21st-2014-a-10229.html) and I've included two answers provided by the participants in that problem and I hope you will gain some insights from them.:)

@Euge, I hope you don't mind me posting a link to refer the OP to seeing an old thread of mine in the POTW section, I just thought perhaps the OP wanted also to see other methods to solve for this problem and I'm by no means to hijack this thread from you.

 

[MarkFL]

Hey Olok,(Smile)

This is one old problem from the High School Problem Of The Week section (http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-108-april-21st-2014-a-10229.html) and I've included two answers provided by the participants in that problem and I hope you will gain some insights from them.:)

@Euge, I hope you don't mind me posting a link to refer the OP to seeing an old thread of mine in the POTW section, I just thought perhaps the OP wanted also to see other methods to solve for this problem and I'm by no means to hijack this thread from you.

You know, I thought that problem seemed familiar...:D Thanks for clearing it up for me. (Yes)

 

[Amad27]

Hey Olok,(Smile)

This is one old problem from the High School Problem Of The Week section (http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-108-april-21st-2014-a-10229.html) and I've included two answers provided by the participants in that problem and I hope you will gain some insights from them.:)

@Euge, I hope you don't mind me posting a link to refer the OP to seeing an old thread of mine in the POTW section, I just thought perhaps the OP wanted also to see other methods to solve for this problem and I'm by no means to hijack this thread from you.

Hello,

Thank you, MarkFL's solution was very quick and useful.

The only issue is that it is not obvious. It is rather intuitive to make the $$ u = \frac{\pi}{2} - x$$ substitution.

I TRIED differentiation under the integral sign BECAUSE It is **NOT** intuitive, it is simply a method.

I will be grateful if anyone else can still answer the questions I asked Euge, to just use this method (of differentiation).

 

[Euge]

Hi Olok,

Since you want to explore both approaches I mentioned, let me start with the non-differentiation method first. Multiplying the integrand of $J(a)$ by $\frac{(\cos{x})^a}{(\cos{x})^a}$, we get

$\displaystyle (a) = \int_0^{\pi/2} \frac{(\cos{x})^a}{(\cos{x})^a + (\sin{x})^a}\, dx$.

Via the u-sub $u = \frac{\pi}{2} - x$, we also have

$\displaystyle J(a) = \int_0^{\pi/2} \frac{(\sin{u})^a}{(\cos{u})^a + (\sin{u})^a}\, du$.

Thus

$\displaystyle 2J(a) = \int_0^{\pi/2} \frac{(\cos{x})^a + (\sin{x})^a}{(\cos{x})^a + (\sin{x})^a}\, dx = \frac{\pi}{2}$.

Dividing by two gives $J(a) = \frac{\pi}{4}$.

Now for your other questions. By symmetry, I was referring to the equation

$\int_0^b f(x) \, dx = \int_0^b f(b - x)\, dx$,

where $f$ is integrable on $[0, b]$. This was used in both methods.

You wanted to find out how I transformed cotangents to tangents in one of my steps. This was done using the relations $\cot{u} = \frac{1}{\tan{u}}$ and the log rule $\ln{\frac{1}{t}}= -\ln{t}$.

To answer your last question, the fact that $J$ is constant implies $J(\sqrt{2}) = J(0)$.

 

[Amad27]

Hi Olok,

Since you want to explore both approaches I mentioned, let me start with the non-differentiation method first. Multiplying the integrand of $J(a)$ by $\frac{(\cos{x})^a}{(\cos{x})^a}$, we get

$\displaystyle (a) = \int_0^{\pi/2} \frac{(\cos{x})^a}{(\cos{x})^a + (\sin{x})^a}\, dx$.

Via the u-sub $u = \frac{\pi}{2} - x$, we also have

$\displaystyle J(a) = \int_0^{\pi/2} \frac{(\sin{u})^a}{(\cos{u})^a + (\sin{u})^a}\, du$.

Thus

$\displaystyle 2J(a) = \int_0^{\pi/2} \frac{(\cos{x})^a + (\sin{x})^a}{(\cos{x})^a + (\sin{x})^a}\, dx = \frac{\pi}{2}$.

Dividing by two gives $J(a) = \frac{\pi}{4}$.

Now for your other questions. By symmetry, I was referring to the equation

$\int_0^b f(x) \, dx = \int_0^b f(b - x)\, dx$,

where $f$ is integrable on $[0, b]$. This was used in both methods.

You wanted to find out how I transformed cotangents to tangents in one of my steps. This was done using the relations $\cot{u} = \frac{1}{\tan{u}}$ and the log rule $\ln{\frac{1}{t}}= -\ln{t}$.

To answer your last question, the fact that $J$ is constant implies $J(\sqrt{2}) = J(0)$.

Hi there,

That was all very helpful, I have one simple question left.

Where did you find that

$$ \int_{0}^{b}f(x) \,dx = \int_{0}^{b}f(b-x) \,dx $$

In other words, how do you know this property is true? Where would you find this online?

The graph is symmetric with respect to what though? I mean does this apply for the y-axis?

That is all I need to know. Very nicely done!

 

[Euge]

Hi Olok,

You just use the u-sub $u = b - x$ to get that result. There's no graph involved.

 

[Amad27]

Hi Olok,

You just use the u-sub $u = b - x$ to get that result. There's no graph involved.

Thanks =) We are done.