Challenge Micromass' big integral challenge

[micromass]

#9: \int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx

(I'm not sure if it is a valid solution since I looked up the Γ-function and its formulas ... after despaired on 2016th roots of unity and many ugly looking sheets of paper ...)

For convenience I take $n=2016$ and $m=1916$.
The disturbing denominator can be transformed into an integral:

$$\int_0^\infty e^{(-x^n -1) y} dy = [- \frac{1}{x^n + 1} * e^{(-x^n -1) y } ]_{ y=0 }^{ y=\infty } = \frac{1}{x^n + 1}$$
Thus
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \int_0^\infty x^{m} \int_0^\infty e^{-x^n y} e^{-y} dy dx$$ and with Fubini's theorem
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \int_0^\infty\int_0^\infty x^m e^{-x^n y} e^{-y} dx dy$$.

Substitution $z = x^n y$ with $dx = \frac{1}{ny} x^{1-n} dz$ yields
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \frac{1}{n} \int_0^\infty e^{-y} (\frac{1}{y})^\frac{m+1}{n} dy \int_0^\infty e^{-z} z^\frac{m-n+1}{n} dz = \frac{1}{n} Γ(\frac{n-m-1}{n}) Γ(\frac{m+1}{n}) = \frac{1}{n} Γ(1-\frac{m+1}{n}) Γ(\frac{m+1}{n}) $$
$$\int_0^\infty \frac{x^{1916}}{x^{2016} + 1}dx = \frac{1}{2016} π \frac{1}{\sin{\frac{1917 π}{2016}}} ≈ 0.0101411901 $$

That is an extremely elegant solution! Congratulations! Don't worry about having looked up the $\Gamma$-function and similar formulas, you were allowed to do that.

 

[StudentOfScience]

This is a question regarding 7. This is not the answer.

Spoiler: Attempt at 7

I tried evaluating # 7 by finding an antiderivative and then using the Fundamental Theorem. I thought I found the antiderivative until I checked it on Desmos (I don't know if that is even allowed). I probably made a mistake that messed the whole problem up. Thus, I cannot be sure if the primitive can be represented in closed form with elementary functions. This is a lot to ask, but if anyone is willing to follow my work to see where I possibly went wrong, that would be great. The outline of my solution is as follows.

1. Use weierstrass substitution
2. Simplify
3. Use integration by parts
   1. $u= arccos(\frac{1-s^2}{3-s^2})$
   2. ## du=\frac{4s}{(3-s^2)^2*(1-(\frac{1-s^2}{3-s^2})^2)^\frac{1}{2}} ## (the bolded 2 is an exponent)
   3. $dv=1+s^2$
4. Simplify integrand resulting from the vdu section of integration by parts
5. Substitute: $p=3-s^2$
6. Rewrite integrand in terms of p
7. Simplify integrand
8. Trig Substitution: $p+1/2=\frac{\sec(q)}{2}$
9. Rewrite in terms of q
10. Simplify
11. Split numerator and integrate the remaining three integrands
    1. Note that all three follow the same pattern in how they were integrated; use long division and then integrate
12. Back-substitute (I didn't do this since this would take a very long time and would be considerably tedious)

I did omit details in some parts (like in steps 2, 6, 9, and 11). If you want to see my full work, the link below should take you to a OneNote notebook with the (attempted) solution. If you would like me to put the details for 2, 6, 9, and 11 in, just let me know. Finally, kudos to all of you; you are way above me in integration (I'm only at the calc 2 level). I'm sure you could come up with an elegant solution that is less than a fourth of my work. Thank you for taking the time to read my post.
Notebook link: https://onedrive.live.com/redir?page=view&resid=B4AB9991BD097345!2597&authkey=!AFXWqJAgfdt0erA
Desmos link: https://www.desmos.com/calculator/vnvhapmyy1

 

[StudentOfScience]

I'll actually give #7 a shot (even though I "tried" already). Also, I hope it's valid to check your answer with a graphing calculator. I'm probably wrong anyways, so don't regard my answer as much. This is what I have.

I think $\int\arccos(\frac{cosx}{1+2cosx}) = 2[\arctan(s)]*[\arccos(\frac{1-s^2}{3-s^2})+\arcsin(\frac{1-s^2}{3-s^2})] + C$, where $s= \frac{sinx}{1+cosx}=tan(x/2)$. I checked the antiderivative on the graphing calculator, but it was wrong since the derivative of the primitive didn't coincide with the integrand (see link: https://www.desmos.com/calculator/dpizsq7x7a). I will include work of how I obtained the primitive below. Before I do so, when I changed the limits of integration after the substitution, I got 0 and 1 as the limits. Using the antiderivative to evaluate the original definite integral yielded a value of $\pi^2/4$ This is probably wrong, but at least I tried.

How I found the (wrong) primitive:
1) Use the weierstrass substitution to rewrite the integrand as $2 \int\arccos(\frac{1-s^2}{3-s^2})*\frac{1}{1+s^2}ds$
2) Using integration by parts where $u=\arccos(\frac{1-s^2}{3-s^2})$ let's us write $2 \int\arccos(\frac{1-s^2}{3-s^2})*\frac{1}{1+s^2}ds= 2[\arctan(s)][arccos(\frac{1-s^2}{3-s^2})]-2\int\frac{4s\arctan(s)}{(3-s^2)^2*(1-(\frac{1-s^2}{3-s^2})^2)^\frac{1}{2}}ds$
3)To evaluate the remaining integral, use parts again with $u=\arctan(s)$. This allows us to say $-2\int\frac{4s\arctan(s)}{(3-s^2)^2*(1-(\frac{1-s^2}{3-s^2})^2)^\frac{1}{2}}ds = 2(\arctan(s))*(arcsin(\frac{1-s^2}{3-s^2}))+2\int\frac{arcsin(\frac{1-s^2}{3-s^2}){1+s^2}ds$ For some reason, the LaTeX above is not working, so I will attach an image below:
Screen Shot 2016-04-23 at 2.29.50 PM.png
4) Now, let us solve the integral $2\int\frac{\arcsin(\frac{1-s^2}{3-s^2})}{1+s^2}ds$. Using parts once more where $u=\arcsin(\frac{1-s^2}{3-s^2})$ gives an interesting result. That is, we can combine like integrals (the procedure used to solve the integral $\int\sec^3(x)$). So, combining like integrals, we have that $-2*\int\frac{4s\arctan(s)}{(3-s^2)^2*(1-(\frac{1-s^2}{3-s^2})^2)^\frac{1}{2}}ds =2[\arctan(s)}*[\arcsin(\frac{1-s^2}{3-s^2})$ Again, the LaTeX may not be working, so here is another image (my apologies for attaching many images): Screen Shot 2016-04-23 at 2.44.36 PM.png
5) Thus, combining steps 2 and 3-4, we get the primitive $2[\arctan(s)]*[\arccos(\frac{1-s^2}{3-s^2})+\arcsin(\frac{1-s^2}{3-s^2})] + C$

Thank you all for taking the time to read my work and putting up with me. Anyhow, I'm sure my work is wrong (and I've had 2 tries in addition to using a graphing calculator).

 

[StudentOfScience]

By the way, you can ignore post 32 since I found my error. I have not found my error in post 33 though.

 

[fresh_42]

By the way, you can ignore post 32 since I found my error. I have not found my error in post 33 though.

You're at least close to the correct solution and as far as I can see with some correct ideas. However, it's hard for me to read your text and find eventual errors. The many substitutions needed in a brute force trial IMO require a careful and extended presentation (through formulas and equations rather than text). I still hope to find a "nice" trick instead of ever changing the variable and often return to the original integrand. Plus the substitution I've found to solve a similar integral isn't really nice and easy.

 

[micromass]

I'll post the solution to 7 tomorrow, unless somebody really wants to find this himself.

 

[StudentOfScience]

I'll post the solution to 7 tomorrow, unless somebody really wants to find this himself.

If possible, I would prefer if you wait to post the solution until a bit longer. Others may disagree with me, but right now, I have exams to worry about as well. Thus, I may need a couple of days to find time to return to the problem. Thank you for your consideration.

 

[Samy_A]

Half of my challenge problems are already solved. I did not expect this to happen so soon! The people who found these solutions are truly integral masters!

Anyway, I wish to make a little adertisement now for the book containing these problems. I got these problems (except 6 which is from Apostol) from the beautiful book "Inside interesting integrals" from Paul J. Nahin. So if you want to learn tricky solutions or are up for a challenge. This is the book for you!

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

Judging by the integrals in this thread, this must be a great book.

Ordered it!

 

[Ssnow]

I agree with @Samy_A, I am curious on the techniques in this book, if this challenge was a way to publicize this book it works very well!

 

[Samy_A]

Just received the book.
Of course I first looked for the solution of integral 7.
Indeed very difficult.

A consolation for those how tried it: it was apparently solved by none other than the great G.H. Hardy!

 

[fresh_42]

Just received the book.
Of course I first looked for the solution of integral 7.
Indeed very difficult.

A consolation for those how tried it: it was apparently solved by none other than the great G.H. Hardy!

Now that's a motivation. It's like saying "Ramanujan solved this prime problem" or "Wiles could solve this problem on modular forms" ...

 

[micromass]

Now that's a motivation. It's like saying "Ramanujan solved this prime problem" or "Wiles could solve this problem on modular forms" ...

Gauss found the answer to the sum $1+2+3+4+5+...+1000$.

 

[fresh_42]

Gauss found the answer to the sum $1+2+3+4+5+...+1000$.

... at the age of about 6 ...

 

[micromass]

... at the age of about 6 ...

Details...

 

[fresh_42]

Details...

https://nrich.maths.org/2478

 

[Ssnow]

I ordered the book last week but I am far (I need at least two weeks...). So in this time I can drink tea at five in order to reproduce all Hardy-physical conditions that will permit me to solve the integral ...

 

[fresh_42]

Is anyone interested in the anecdote of Hardy's very special life insurance?

 

[Samy_A]

Of course!

 

[fresh_42]

It is not quite clear whether Hardy believed in God or was just superstitious. However, in any case he believed God will do everything to make his life tough and complicated. One day he was on a journey back home. (I've heard it with Harald Bohr and Copenhagen, but also found Norway on the internet.) Anyway. He had to take a ship and the boat he got didn't look very trustful. Typically, he thought, why me?
So he sent a postcard before boarding - say to Bohr - claiming he has found the proof of Riemann's hypothesis (conjecture).
When afterwards asked why he replied: Well, if the ship sank the proof would have been lost but I would have become the most famous mathematician of my generation. God won't allow this to happen. This way I only had to write Bohr another postcard in which I stated to have made a mistake.

 

[member 587159]

Can someone post the solution of the remaining integral? I have waited long enough ;)

 

[micromass]

OK, I'll post it soon. I just thought some people were still thinking about it.

 

[fresh_42]

I admit I gave up after many attempts on something like $\int q(x)^{-1} \arctan{x} dx$ with $q(x) = (3x^2+1)\sqrt{2x^2-1}$ or similar. (Don't count on it, it's just out of memory.)

 

[micromass]

Alright, here is the first part of the solution to $I = \int_0^{\pi/2} \cos^{-1}\left(\frac{\cos(x)}{1+2\cos(x)}\right)dx$. I will give people the chance to complete the solution. I will post the second half on monday.

Fist, notice that the double-angle formula from trigonometry say that for any $\theta$, we have $\cos(2\theta) = 2\cos^2(\theta)-1$. Call this $(*)$.

If we write $u=\cos(\theta)$ - and so $\theta = \cos^{-1}(u)$ - then $(*)$ says that $\cos(2\theta) = 2u^2 - 1$, from which immediately follows that
\cos^{-1}(2 u ^2 - 1) = \cos^{-1}(\cos(2\theta)) = 2\theta = 2\cos^{-1}(u)

So, since $u$ is simply an arbitrary variable, we can write

\cos^{-1}(2\theta^2 - 1) = 2\cos^{-1}(\theta)

Next, we write $\alpha = 2\theta^2 - 1$, which means that $\theta = \sqrt{\frac{1+\alpha}{2}}$. It follows that

\cos^{-1}(\alpha) = 2\cos^{-1}\left(\sqrt{\frac{1+\alpha}{2}}\right)

We write $\alpha = \frac{\cos(x)}{1+2\cos(x)}$, to get

\cos^{-1}\left(\frac{\cos(x)}{1 + 2\cos(x)}\right) = 2\cos^{-1}\left(\sqrt{ \frac{1 + \frac{\cos(x)}{1 + 2\cos(x)}}{2}}\right) = 2\cos^{-1}\left(\sqrt{\frac{1 + 3\cos(x)}{2 + 4\cos(x)}}\right).

Applying the Pythagorean theorem to a right triangle with acute angle whose cosine is $\sqrt{\frac{1 + 3\cos(x)}{2 + 4\cos(x)}}$, yu'll see that the tangent of that same angle is $\sqrt{\frac{1 + \cos(x)}{1 + 3\cos(x)}}$. Thus

\cos^{-1}\left(\frac{\cos(x)}{1 + 2\cos(x)}\right) = 2\tan^{-1}\left(\sqrt{\frac{1+\cos(x)}{1 + 3\cos(x)}}\right)

And so we have

I = 2\int_0^{\pi/2} \tan^{-1}\left(\sqrt{\frac{1 + \cos(x)}{1 + 3\cos(x)}}\right)dx

Make the chance of variable $x = 2y$, so $dx = 2dy$, to get

I = 4\int_0^{\pi/4} \tan^{-1}\left(\sqrt{\frac{1 + \cos(2y)}{1 + 3 \cos(2y)}}\right)dy

Using $(*)$ again, we can find

\sqrt{\frac{1 + \cos(2y)}{1 + 3\cos(2y)}} = \frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}}

And so

I = 4\int_0^{\pi/4} \tan^{-1}\left(\frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}} \right)dy

Notice that

\int_0^1 \frac{1}{1 + \left(\frac{\cos^2(y)}{2 - 3\sin^2(y)}\right)t^2}dt

is of the form

\int_0^1 \frac{1}{1 + b^2t^2}dt = \frac{1}{b^2}\int_0^1 \frac{1}{\frac{1}{b^2} + t^2}dt = \frac{1}{b^2} [b\tan^{-1}(bt)]_0^1 = \frac{1}{b} \tan^{-1}(b),
where $b = \frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}}$.
Thus

\int_0^1 \frac{1}{1 + \left(\frac{\cos^2(y)}{2 - 3\sin^2(y)}\right)t^2 } dt = \frac{\sqrt{2 - 3\sin^2(y)}}{\cos(y)} = \tan^{-1}\left(\frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}}\right)

That is,

<br /> \begin{eqnarray*}<br /> I &amp; = &amp; 4\int_0^{\pi/4} \frac{\cos(y)}{\sqrt{ 2 - 3\sin^2(y)}} \left(\int_0^1 \frac{1}{1 + \left(\frac{\cos^2(y)}{2 - 3\sin^2(y)}\right)t^2}\right)dy\\<br /> &amp; = &amp; \int_0^{\pi/4} \int_0^1 \frac{4\cos(y)(2 - 3\sin^2(y))}{\sqrt{2 - 3\sin^2(y)}(2 - 3\sin^2(y) + t^2\cos^2(y))}dt dy\\<br /> &amp; = &amp; \int_0^{\pi/4} \int_0^1 \frac{4\cos(y)\sqrt{ 2 - 3\sin^2(y)}}{2 - 3\sin^2(y) + t^2 - t^2\sin^2(y)}dt dy\\<br /> &amp; = &amp; \int_0^{\pi/4} \int_0^1 \frac{4\cos(y) \sqrt{2 - 3\sin^2(y)} }{(t^2 + 2) - (t^2 + 3)\sin^2(y)}dt dy<br /> \end{eqnarray*}<br />

Next, we change the variables $\sin(y) = \sqrt{\frac{2}{3}} \sin(w)$, and so $dy = \sqrt{\frac{2}{3}} \frac{\cos(w)}{\cos(y)}dw$. We have $w=0$ when $y=0$, and when $y = \pi/4$, we have $\sin(\pi/4) = \frac{1}{\sqrt{2}}$. And so $\sin(w) = \sqrt{\frac{3}{2}}\frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}$, which says $w = \pi/3$. So

<br /> \begin{eqnarray*}<br /> I<br /> &amp; = &amp; \int_0^{\pi/3} \int_0^1 \frac{4\cos(y) \sqrt{2 - 3\frac{2}{3}\sin^2(w)}}{(t^2 + 2) - (t^2 + 3) \frac{2}{3} \sin^2(w)}dt \frac{\cos(w)}{\cos(y)} dw \sqrt{\frac{2}{3}}\\<br /> &amp; = &amp; \int_0^{\pi/3} \int_0^1 \frac{4 \sqrt{2 - 2( 1 - \cos^2(w))}}{(t^2 + 2) - (t^2 + 3) \frac{2}{3} ( 1 - \cos^2(w))} dt \cos(w) dw \sqrt{\frac{2}{3}}\\<br /> &amp; = &amp; \int_0^{\pi/3} \int_0^1\frac{4 \sqrt{2} \cos(w) \sqrt{2} \cos(w) }{(t^2 + 2) - (t^2 + 3) \frac{2}{3} ( 1 - \cos^2(w)} dt \frac{1}{\sqrt{3}} dw\\<br /> &amp; = &amp; \int_0^{\pi/3} \int_0^1 \frac{8\sqrt{3}\cos^2(w)}{t^2 + (2t^2 + 6)cos^2(w)} dt dw<br /> \end{eqnarray*}<br />

Our next step is another change of variable to $s = \tan(w)$. Thus as $\tan(w) = \frac{\sin(w)}{\cos(w)}$, we have

\frac{ds}{dw} = \frac{1}{\cos^2(w)}

and so $dw = \cos^2(w) ds$. Since

1 + s^2 = 1 + \tan^2(w) = \frac{1}{\cos^2(w)}

we have

\frac{1}{1 + s^2} = \cos^2(w)

and so

dw = \frac{ds}{ 1+ s^2}

Therefore, since $s=0$ when $w=0$ and $s= \sqrt{3}$, when $w = \pi/3$, we have

<br /> \begin{eqnarray*}<br /> I<br /> &amp; = &amp; \int_0^{\sqrt{3}} \int_0^1 \frac{8\sqrt{3} \frac{1}{1 + s^2}}{t^2 + (2t^2 + 6) \frac{1}{1 + s^2}} dt \frac{ds}{1 + s^2}\\<br /> &amp; = &amp; \int_0^{\sqrt{3}} \int_0^1 \frac{8\sqrt{3}}{t^2(1 + s^2)^2 + (2t^2 + 6)(1 + s^2)} dt ds\\<br /> &amp; = &amp; \int_0^{\sqrt{3}} \int_0^1 \frac{8\sqrt{3}}{(1 + s^2)(t^2s^2 + 3t^2 + 6)} dt ds<br /> \end{eqnarray*}<br />

Let's make a partial fraction expansion:

\frac{1}{(1+ s^2)(t^2 s^2 + 3t^2 + 6)} = \frac{A}{1+s^2} + \frac{B}{t^2 s^2 + 3t^2 + 6}
where it is easy to confirm that
A = \frac{1}{2t^2 + 6}~~\text{and}~~ B = - \frac{t^2}{2t^2 + 6}

and so

<br /> \begin{eqnarray*}<br /> I<br /> &amp; = &amp; \int_0^{\sqrt{3}} \int_0^1 8\sqrt{3} \left(\frac{\frac{1}{2t^2 + 6}}{1 + s^2} - \frac{\frac{t^2}{2t^2 + 6}}{t^2 + 3t^2 + 6}\right)dt ds\\<br /> &amp; = &amp; \int_0^1 \frac{4\sqrt{3}}{t^2 + 3}\left(\int_0^{\sqrt{3}} \frac{ds}{1+s^2} - \int_0^{\sqrt{3}} \frac{ds}{s^2 + 3 + \frac{6}{t^2}}\right) dt<br /> \end{eqnarray*}<br />

The first inner intergral on the right is easy:

[\tan^{-1}(s)]_0^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}
The second inner integral is almost as easy and equals

<br /> \begin{eqnarray*}<br /> \int_0^{\sqrt{3}} \frac{ds}{s^2 + \left(\sqrt{3 + \frac{6}{t^2}}\right)^2}<br /> &amp; = &amp; \frac{1}{\sqrt{3 + \frac{6}{t^2}}} \left. \left(\tan^{-1}\left(\frac{s}{\sqrt{3 + \frac{6}{t^2}}}\right)\right)\right|_0^{\sqrt{3}}\\<br /> &amp; = &amp; \frac{t}{\sqrt{3}\sqrt{t^2 + 2}} \tan^{-1}\left.\left(\frac{st}{\sqrt{3}\sqrt{t^2 + 2}}\right)\right|_0^{\sqrt{3}}\\<br /> &amp; = &amp; \frac{t}{\sqrt{3}\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right)<br /> \end{eqnarray*}<br />

Thus

<br /> \begin{eqnarray*}<br /> I<br /> &amp; = &amp; \int_0^1 \frac{4\sqrt{3}}{t^2 + 3}\left(\frac{\pi}{3} - \frac{t}{\sqrt{3}\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right)\right)dt\\<br /> &amp; = &amp; \frac{4\sqrt{3}\pi}{3}\int_0^1 \frac{dt}{t^2 + (\sqrt{3})^2} - 4\int_0^1 \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) dt\\<br /> &amp; = &amp; \frac{4\sqrt{3}\pi}{3}\left.\left(\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\right)\right|_0^1 - 4\int_0^1 \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) dt<br /> \end{eqnarray*}<br />

Since

\frac{4\sqrt{3}\pi}{3}\left.\left(\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\right)\right|_0^1 = \frac{4\pi}{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{2\pi^2}{9}

we have

I = \frac{2\pi^2}{9} - 4\int_0^1 \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) dt

We can do this by parts, let

u = \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right)

and

dv = \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} dt

Then

\frac{du}{dt} = \frac{1}{(t^2 + 1)\sqrt{t^2 + 2}}

and you can verify that

v = \tan^{-1}(\sqrt{t^2 + 2})

by simply sifferentiating this $v$ and observing we get the above $dv$ back. So, plugging all this into the integration by parts formula, we have

<br /> \begin{eqnarray*}<br /> I<br /> &amp; = &amp; \frac{2\pi^2}{9} - 4\left(\left.\left(\tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right)\tan^{-1}(\sqrt{t^2 + 2})\right)\right|_0^1 - \int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt\right)\\<br /> &amp; = &amp; \frac{2\pi^2}{9} - 4\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\tan^{-1}(\sqrt{3}) - \int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt\right)\\<br /> &amp; = &amp; \frac{2\pi^2}{9} - 4\left(\frac{\pi}{6}\frac{\pi}{3} - \int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt\right)\\<br /> &amp; = &amp; \frac{2\pi^2}{9} - \frac{2\pi^2}{9} + 4\int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt<br /> \end{eqnarray*}<br />

And so

I = 4\int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt

This is the first part. I leave the second part up to you. I will post the solution monday unless somebody doesn't want me to

 

[member 587159]

How was anyone supposed to find that? :P

 

[micromass]

Alright, here is the second part. We let

I(u) = \int_0^1 \frac{\tan^{-1}\left(u\sqrt{2 +x^2}\right)}{(1+x^2)\sqrt{2 + x^2}}dx

Note that we wish to find $I(1)$. Notice that if $u\rightarrow +\infty$, then the argument for the inverse tangent also goes to $+\infty$ for all $x>0$. So since $\tan^{-1}(+\infty) = \frac{\pi}{2}$, we have

I(+\infty) = \frac{\pi}{2} \int_0^1 \frac{dx}{(1+x^2)\sqrt{2+x^2}}

This integral is easy to do if you remember the following standard formula:

\frac{d}{dx} \tan^{-1}(f(x)) = \frac{1}{1+f^2(x)}\frac{df}{dx}

We can use this formula to calculate

\frac{d}{dx} \tan^{-1}\left(\frac{x}{\sqrt{2+x^2}}\right) = \frac{1}{(1+x^2)\sqrt{2 + x^2}}

Thus

\begin{eqnarray*}<br /> I(+\infty) <br /> &amp; = &amp; \frac{\pi}{2}\int_0^1 \frac{d}{dx} \left(\frac{x}{\sqrt{2+x^2}}\right)dx\\<br /> &amp; = &amp; \frac{\pi}{2} \left.\left(\tan^{-1}\left(\frac{x}{\sqrt{2+x^2}}\right)\right)\right|_0^1\\<br /> &amp; = &amp; \frac{\pi}{2}\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(0)\right)\\<br /> &amp; = &amp; \frac{\pi^2}{12}<br /> \end{eqnarray*}

Now we differentiate $I(u)$ with respect to $u$. We get

\frac{dI}{du} = \int_0^1\frac{dx}{(1+x^2)(1 + 2u^2 + u^2 x^2)}

With a partial fraction expansion, this becomes

<br /> \begin{eqnarray*}<br /> \frac{dI}{du} <br /> &amp; = &amp; \int_0^1 \frac{1}{1+u^2} \left(\frac{1}{1+x^2} - \frac{u^2}{1 + 2u^2 + u^2 x^2}\right)dx\\<br /> &amp; = &amp; \frac{1}{1+u^2}\left(\int_0^1 \frac{dx}{1+x^2} - \int_0^1 \frac{dx}{\frac{1 + 2u^2}{u^2} + x^2}\right)<br /> \end{eqnarray*}

These last two integrals are of the form

\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)

And so, doing the integrals, we have

<br /> \begin{eqnarray*}<br /> \frac{dI}{du}<br /> &amp; = &amp; \frac{1}{1+u^2} \left. \left( \tan^{-1}(x) - \frac{u}{\sqrt{1 + 2u^2}} \tan^{-1}\left(\frac{xu}{\sqrt{1 + 2u^2}}\right) \right) \right|_0^1\\<br /> &amp; = &amp; \frac{1}{1+u^2}\left(\frac{\pi}{4} - \frac{u}{\sqrt{1 + 2u^2}}\tan^{-1} \left(\frac{u}{\sqrt{1 + 2u^2}}\right)\right)<br /> \end{eqnarray*}

Now we integrate both sides from $1$ to $+\infty$ with respect to $u$. On the left, we get:

\int_1^{+\infty} \frac{dI}{du} du = \int_1^{+\infty} dI = I(+\infty) - I(1)

On the right, we get

\frac{\pi}{4}\int_1^{+\infty} \frac{du}{1+u^2} - \int_1^{+\infty} \frac{u}{(1+u^2)\sqrt{1 + 2u^2}} \tan^{-1}\left(\frac{u}{\sqrt{1 + 2u^2}}\right)du

The first integral is easy:

\frac{\pi}{4} \int_1^{+\infty} \frac{du}{1+u^2} = \frac{\pi}{4} \left(\tan^{-1}(+\infty) - \tan^{-1}(1)\right) = \frac{\pi}{4}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{\pi^2}{16}

Thus

I(+\infty) - I(1) = \frac{\pi^2}{16} - \int_1^{+\infty} \frac{u}{(1+u^2)\sqrt{1 + 2u^2}}\tan^{-1} \left(\frac{u}{\sqrt{1 + 2u^2}}\right)du

In this final integral, we change variable $t = \frac{1}{u}$, and so $du = - \frac{1}{t^2} dt$ as follows:

<br /> \begin{eqnarray*}<br /> &amp; &amp; \int_1^{+\infty} \frac{u}{(1+u^2)\sqrt{1 + 2u^2}} \tan^{-1}\left(\frac{u}{\sqrt{1 + 2u^2}}\right) du\\<br /> &amp; = &amp; \int_1^0 \frac{\frac{1}{t}}{\left(1 + \frac{1}{t^2}\right)\sqrt{1 + \frac{2}{t^2}}} \tan^{-1}\left(\frac{\frac{1}{t}}{\sqrt{1 + \frac{2}{t^2}}}\right) \left(-\frac{1}{t^2} dt\right)\\<br /> &amp; = &amp; \int_0^1 \frac{\frac{1}{t}}{(t^2 + 1)\frac{\sqrt{t^2 + 2}}{t}} \tan^{-1} \left(\frac{\frac{1}{t}}{\frac{\sqrt{t^2 + 2}}{t}}\right)dt\\<br /> &amp; = &amp; \int_0^1 \frac{1}{(t^2 + 1)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right)dt<br /> \end{eqnarray*}<br />

Now recall the identity $\tan^{-1}(s) + \tan^{-1}\left(\frac{1}{s}\right) = \frac{\pi}{2}$, which becomes instantly obvious if you draw a right triangle with perpendicular sides of lengths $1$ and $s$ and remember that the two acute angles add to $\pi/2$. This says

\tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right) = \frac{\pi}{2} - \tan^{-1}\left(\sqrt{t^2 + 2}\right)

And so we can write

<br /> \begin{eqnarray*}<br /> &amp; &amp; \int_0^1\frac{1}{(t^2 + 1)\sqrt{t^2 + 2}}\tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right)dt\\<br /> &amp; = &amp; \frac{\pi}{2}\int_0^1 \frac{dt}{(t^2 + 1)\sqrt{t^2 + 2}} -\int_0^1 \frac{\tan^{-1}\left(\sqrt{t^2 + 2}\right)}{(t^2 + 1)\sqrt{t^2 + 2}}dt<br /> \end{eqnarray*}

That is, we have

I(+\infty) - I(1) = \frac{\pi^2}{16} - \frac{\pi}{2}\int_0^1\frac{dt}{(t^2 + 1)\sqrt{t^2 + 2}} + \int_0^1 \frac{\tan^{-1}\left(\sqrt{t^2 + 2}\right)}{(t^2 + 1)\sqrt{t^2 + 2}}dt

The first integral is just $I(+\infty)$ and the last integral is just $I(1)$. That is, we have

I(+\infty) - I(1) = \frac{\pi^2}{16} - I(+\infty) + I(1)

and so

2 I (+\infty) - \frac{\pi^2}{16} = 2I(1)

and at last

I(1) = I(+\infty) - \frac{\pi^2}{32} = \frac{\pi^2}{12} - \frac{\pi^2}{32} = \frac{5\pi^2}{96}

Thus our original integral now becomes

\int_0^{\pi/2} \cos^{-1}\left(\frac{\cos(x)}{1 + 2\cos(x)}\right)dx = 4\int_0^1 \frac{\tan^{-1}\left(\sqrt{t^2 + 2}\right)}{(t^2 + 1)\sqrt{t^2 + 2}}dt = \frac{5\pi^2}{24}

This is very close to $\frac{\pi^2}{4}$, which two people got as answer. This is very interesting!

 

[micromass]

Thank you everybody who participated in this thread! I hope you found these integrals interesting. If you want more, get the book "Inside Interesting Integrals" by Paul J. Nahin.

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

 

[ShayanJ]

That was the longest integration I've ever seen! To Hardy:

 

[StudentOfScience]

I would've never thought of that solution. Thank you for posting the solution before I went crazy trying to figure out the integral with elementary substitution techniques. If all the integrals in that book are this involved, I'm very interested in the book.