# Thread: Higher Level Integration Marathon & Questions

1. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
$Use Inequalities to show that for all values of x in the interval -1 \leq x \leq 1 the functions:$

$P_{n}(x)= \frac{1}{\pi}\int_0^{\pi}{(x+i\sqrt{1-x^2}\cos{\theta})^n}d\theta$

$satisfies the inequality |P_{n}(x)| \leq 1$
The integrand has its modulus squared equal to (x^2 + (1-x^2)cos^2(theta))^n =< (x^2 + (1-x^2))^n=1.

The triangle inequality completes the proof.

2. ## Re: Extracurricular Integration Marathon

$\noindent \textbf{Two New Questions}$

\noindent Evalaute\\(a) \begin{align*}\int^{\frac{\pi}{2}}_0 \cot x \ln (\sec x) \, dx \,\,\,\,\,\,\,&\mbox{(b)} \,\, \int^1_0 \frac{\tan^{-1} (\sqrt{2 + x^2})}{(1 + x^2) \sqrt{2 + x^2}} \, dx\end{align*}
The Laurent series for arctan(z) looks unpromising.... We need a different approach.

3. ## Re: Extracurricular Integration Marathon

When we do Cauchy's Integrals what are we supposed to picture in our heads?

4. ## Re: Extracurricular Integration Marathon

$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$

5. ## Re: Extracurricular Integration Marathon

$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$
$I=\int_0^\infty \frac{2x^2}{x^4+2x^2+5}\, dx=\int_{\mathbb{R}}\frac{x^2}{x^4+2x^2+5}\, dx$

This integral is clearly a prime target for contour integration. (We could factorise into its real quadratic factors, use partial fractions, and integrate the simpler summands, but contour integration seems faster).

Note that since the quartic denominator is both even and real, the poles of the integrand occur at $\alpha,-\alpha,\overline{\alpha},-\overline{\alpha}$ where $\alpha$ is an arbitrarily chosen root, say the one in the first quadrant of the the complex plane.

Now if we take a semicircular contour (radius R) with diameter on the real axis centred at the origin, and semicircular arc in the upper half-plane, then the integral around this contour (with positive orientation) is just equal to I, because the integrand decays as 1/R^2, and so the contribution from the curved segment of length O(R) tends to zero.

On the other hand, for sufficiently large R this integral is just equal to

$2\pi i (\textrm{Res}(f;\alpha)+\textrm{Res}(f;-\overline{\alpha}))\\ \\ = 2\pi i\left(\frac{\alpha^2}{2\alpha(\alpha-\overline{\alpha})(\alpha+\overline{\alpha})}+\frac{\overline{\alpha}^2}{2\overline{\alpha}(\alpha-\overline{\alpha})(\alpha+\overline{\alpha})}\right)\\ \\ =\frac{i\pi}{\alpha-\overline{\alpha}}=\frac{\pi}{2\textrm{Im}(\alpha)}$

It remains to compute $\alpha$, which is just the principal square root of $-1+2i$.

Writing $\alpha=x+iy$, we get $x^2-y^2=-1,xy=1.$

The resulting biquadratic yields $x=\phi^{-1/2},y=\phi^{1/2}$, where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.

Hence

$I=\frac{\pi}{2\sqrt{\phi}}.$

6. ## Re: Extracurricular Integration Marathon

$\noindent \textbf{Two New Questions}$

\noindent Evalaute\\(a) \begin{align*}\int^{\frac{\pi}{2}}_0 \cot x \ln (\sec x) \, dx \,\,\,\,\,\,\,&\mbox{(b)} \,\, \int^1_0 \frac{\tan^{-1} (\sqrt{2 + x^2})}{(1 + x^2) \sqrt{2 + x^2}} \, dx\end{align*}
$Now, this is a wild guess, but is the second one \frac{5\pi^2}{96} ?$

$Okay, that is the answer, but proving it seems slightly difficult.$

7. ## Re: Extracurricular Integration Marathon

$\int_0^\infty \frac{(x+1)\ln^2{(x+1)}}{(4x^2 +8x +5)^{\frac{3}{2}}}\,dx$

8. ## Re: Extracurricular Integration Marathon

$Now, this is a wild guess, but is the second one \frac{5\pi^2}{96} ?$

$Okay, that is the answer, but proving it seems slightly difficult.$
$\noindent Yes, \frac{5\pi^2}{96} is the correct answer but surely you didn't just guess it?$

9. ## Re: Extracurricular Integration Marathon

$\noindent Yes, \frac{5\pi^2}{96} is the correct answer but surely you didn't just guess it?$
#ISC+MasterRace

10. ## Re: Extracurricular Integration Marathon

$Find : \int_0^1{x^x}dx$

11. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
$Find : \int_0^1{x^x}dx$
$-\sum_{n=1}^{\infty} (-n)^{-n}$

Well, I mean, Sophomore's Dream is nice and all, but there isn't a closed form...

12. ## Re: Extracurricular Integration Marathon

$-\sum_{n=1}^{\infty} (-n)^{-n}$

Well, I mean, Sophomore's Dream is nice and all, but there isn't a closed form...
Yes that's the answer how did you do it though O.o ? Did you create a taylor series and just integrate each term or?

13. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
Yes that's the answer how did you do it though O.o ? Did you create a taylor series and just integrate each term or?
Inspection

14. ## Re: Extracurricular Integration Marathon

Originally Posted by Drsoccerball
Yes that's the answer how did you do it though O.o ? Did you create a taylor series and just integrate each term or?
Yeah, it can be proved that way: https://en.wikipedia.org/wiki/Sophomore%27s_dream#Proof .

15. ## Re: Extracurricular Integration Marathon

$\int_0^\infty \frac{(x+1)\ln^2{(x+1)}}{(4x^2 +8x +5)^{\frac{3}{2}}}\,dx$
I can provide a hint if anyone is attempting this.

16. ## Re: Extracurricular Integration Marathon

$\noindent Yes, \frac{5\pi^2}{96} is the correct answer but surely you didn't just guess it?$
I have not found a complete solution to this integral, only generalised forms. Do you happen to have a solution that specifically deals with this integral?

17. ## Re: Extracurricular Integration Marathon

$Now, this is a wild guess, but is the second one \frac{5\pi^2}{96} ?$

$Okay, that is the answer, but proving it seems slightly difficult.$
Inspection
And by 'inspection' you mean 'Google' or 'Wolfram'?

18. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
And by 'inspection' you mean 'Google' or 'Wolfram'?

It was a joke -_-

I've already encountered Sophomore's Dream before, on one of my many followed blogs.

As for Ahmed's Integral, I didn't know the answer was exactly that until I put the numerical answer into ISC+.

19. ## Re: Extracurricular Integration Marathon

It was a joke -_-

I've already encountered Sophomore's Dream before, on one of my many followed blogs.
Didn't look like a joke to me. Looked like taking credit where credit was not due.

But I'll humour you. Explain the your first "wild guess" then.

I'm sure we're all fascinated to see how you guessed such an answer, whilst refusing to provide at least any sort of outline of a method.

20. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
Didn't look like a joke to me. Looked like taking credit where credit was not due.

But I'll humour you. Explain the your first "wild guess" then.

I'm sure we're all fascinated to see how you guessed such an answer, whilst refusing to provide at least any sort of outline of a method.
From above: As for Ahmed's Integral, I didn't know the answer was exactly that until I put the numerical answer into ISC+.

Also, putting the integral into Wolfram Alpha doesn't give you the closed form. Google... well you can't exactly search tex code easily. So I resorted to the above, taking the numerical value of the integral to 50 decimal places.

21. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
I'm sure we're all fascinated to see how you guessed such an answer, whilst refusing to provide at least any sort of outline of a method.
Well I would post my solution which I figured out yesterday, but I'm not sure it's worth anything at this point.

22. ## Re: Extracurricular Integration Marathon

Well I would post my solution which I figured out yesterday, but I'm not sure it's worth anything at this point.
No, I must insist.

Just a brief and general outline will do.

23. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
Just a brief and general outline will do.
$\noindent Firstly, use the fact that the domain of integration is strictly on the positive axis to invoke \tan^{-1}{x} \equiv \frac{\pi}{2} - \tan^{-1}{\frac{1}{x}} \\ \int_0^1 \frac{\tan^{-1}{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}} \,dx = \frac{\pi}{2} \int_0^1 \frac{\text{d}x}{(x^2+1) \sqrt{2+x^2}}-\int_0^1 \frac{\frac{1}{\sqrt{2+x^2}}\tan{\frac{1}{\sqrt{2+ x^2}}}}{x^2+1}\\The first integral evaluates to \frac{\pi^2}{12}, which I will leave for the reader to verify. The second integral is not as straightforward. However, observe that the numerator is extremely similar to a standard form from integration tables \frac{1}{a}\tan^{-1}{\frac{1}{a}} = \int_0^1 \frac{\text{d}y}{y^2+a^2} \\Here, we borrow the concept of iterated integrals and double integrals from the Gaussian Integral commonly seen in the harder questions of high school. \\ \int_0^1 \frac{\frac{1}{\sqrt{2+x^2}}\tan{\frac{1}{\sqrt{2+ x^2}}}}{x^2+1} = \int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+x^2+2)}$

$\noindent Observe that y^2+x^2+2 - (x^2+1) = y^2+1 The integrand can then be decomposed through the method of partial fractions to obtain \\\int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+1)} - \int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(y^2+1)(y^2+x^2+2)}\\ By the principle of bound and unbound variables for definite integrals, the integral is equal to the simpler looking integral minus itself. Thus, we have \\2\int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+x^2+2)} = \int_0^1 \int_0^1 \frac{\text{d}x\,\text{d}y}{(x^2+1)(y^2+1)}\\ This is now a double integral over the Cartesian plane with identical bounds, and so we have 2\int_0^1 \frac{\frac{1}{\sqrt{2+x^2}}\tan{\frac{1}{\sqrt{2+ x^2}}}}{x^2+1} = \left ( \int_0^1 \frac{\text{d}r}{r^2+1} \right )^2 = \frac{\pi^2}{16} \\ And so we conclude \int_0^1 \frac{\tan^{-1}{\sqrt{x^2+2}}}{(x^2+1)\sqrt{x^2+2}} \,dx = \frac{\pi^2}{12}-\frac{\pi^2}{32} = \frac{5\pi^2}{96}$

24. ## Re: Extracurricular Integration Marathon

Thank you. The spirit of this thread exists in the solution methods, not the answers.

This is still not fully consistent with the below post, but I'll drop the case as an answer has now been given, regardless of its true origins.

I have not found a complete solution to this integral, only generalised forms. Do you happen to have a solution that specifically deals with this integral?

25. ## Re: Extracurricular Integration Marathon

Originally Posted by Carrotsticks
Thank you. The spirit of this thread exists in the solution methods, not the answers.

This is still not fully consistent with the below post, but I'll drop the case as an answer has now been given, regardless of its true origins.
In saying so, I can't do anything that can only be done through contour integration, so my box of tools is much smaller than yours, or most people who are on this thread.

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