Challenge Micromass' big integral challenge

[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.