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-   -   Challenging integral marathon! (http://mymathforum.com/calculus/345077-challenging-integral-marathon.html)

Joppy October 4th, 2018 06:42 PM

Challenging integral marathon!
 
The idea is poached from the AOPS forum.

Rules: Solve the integral in the previous post (show your working) and provide the next challenging integral for the next person. This is open to suggestion/refinement, I'm just getting us started..

Problem 1 (by greg1313)

Quote:

Originally Posted by greg1313 (Post 600352)

$$\int\frac{x-1}{x+x^2\log(x)}\,dx$$


Solution 1 (by Skipjack)

Quote:

Originally Posted by skipjack (Post 600367)
For $x$ > 0, $\displaystyle \!\int\! \frac{x - 1}{x + x^2\ln(x)}\,dx = \!\int\!\left(\frac{1 + \ln(x)}{1 + x\ln(x)} - \frac{1}{x}\right)dx = \ln(1 + x\ln(x)) - \ln(x) + \mbox{C}$, where $\mbox{C}$ is a constant.


Problem 2

$$ \int_0^\infty x e^{1-x} - \left\lfloor{x}\right\rfloor e^{1 - \left\lfloor{x}\right\rfloor} \, dx$$

v8archie October 4th, 2018 07:39 PM

\begin{align}\int_0^\infty x e^{1-x} - \left\lfloor{x}\right\rfloor e^{1 - \left\lfloor{x}\right\rfloor} \, \mathrm dx &= \int_0^\infty x e^{1-x} \,\mathrm dx - \left( 1+2e^{-1} + 3e^{-2} + \ldots \right) \\ &= -xe^{1-x}\bigg|_0^\infty + \int_0^\infty e^{1-x} \,\mathrm dx - \left( 1\left(e^{-1}\right)^0+2\left(e^{-1}\right)^1 + 3\left(e^{-1}\right)^2 + \ldots \right) \\ &= 0 - e^{1-x}\bigg|_0^\infty - \left(\frac{\mathrm d}{\mathrm dx} (1-x)^{-1}\right) _{x=e^{-1}} \\ &= e-\left(-(1-x)^{-2}\right) _{x=e^{-1}} \\ &= e + \left( \frac1{1-e^{-1}}\right)^2 \\ &= e +\frac{e^2}{e^2-2e+1} \\ &= \frac{{e^3-2e^2+e}+e^2}{e^2-2e+1} \\ &= \frac{e^3-e^2+e}{e^2-2e+1}
\end{align}

SDK October 4th, 2018 08:38 PM

Rewrite it as a sum
\[ \int_0^\infty x^{1-x} - \lfloor x \rfloor e^{1-\lfloor x \rfloor} = \sum_{n=0}^\infty \int_n^{n+1} x^{1-x} \ dx - ne^{1-n} \]
Now compute the integral via $\frac{d}{dx}(x+1)e^{1-x} = -xe^{1-x}$ and evaluate on $[n,n+1]$. This simplifies the sum to
\[ \sum_{n=0}^\infty (2-n)e^{1-n} \]
but this is just the series expansion for the derivative of $\frac{e^2}{1-x}$ evaluated at $x = e^{-1}$. Since this series converges uniformly for $|x| < 1$, it follows that \[ \sum_{n=0}^\infty (2-n)e^{1-n} = -\frac{e^2}{(1-e^{-1})^2}\]

Joppy October 4th, 2018 08:54 PM

Alternate solutions are good but please provide a new problem afterwards so we can keep it going :).

v8archie October 4th, 2018 09:48 PM

$$\int \frac{\sqrt{x+1} - \sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} \,\mathrm dx$$

Answers without trigonometric or hyperbolic functions.

SDK October 6th, 2018 07:49 AM

This one is quite good. I still haven't found a solution without trig/complex exponentials. Looking forward to reading the solution when somebody smarter than me figures it out.

v8archie October 6th, 2018 08:53 AM

I don't mean that you can't use them to get to your answer, but the final answer shouldn't contain them.

skipjack October 6th, 2018 09:14 AM

Quote:

Originally Posted by v8archie (Post 600470)
Answers without trigonometric or hyperbolic functions.

Any final answer that happens to involve trigonometric or hyperbolic functions explicitly can be rewritten using exponential or logarithmic functions instead, so that's not a significant restriction.

SDK October 6th, 2018 12:53 PM

Quote:

Originally Posted by v8archie (Post 600566)
I don't mean that you can't use them to get to your answer, but the final answer shouldn't contain them.

I finally have a solution without using any trig/complex exponentials but I don't have a good integral to suggest to replace it so I'll defer answering until I think of something good.

SDK October 6th, 2018 03:41 PM

Quote:

Originally Posted by v8archie (Post 600470)
$$\int \frac{\sqrt{x+1} - \sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} \ dx$$

Answers without trigonometric or hyperbolic functions.

Let $F$ denote the function given by this integral defined on a domain with $x > 1$.
Start by noticing the factorization $\displaystyle \frac{2}{\sqrt{x+1}+\sqrt{x-1}} = {\sqrt{x+1}-\sqrt{x-1}}$. Substituting for the denominator of the integrand and simplifying, we have $\displaystyle F = \int 1 - \sqrt{x^2 - 1} \ dx = x - \int \sqrt{x^2 - 1} \ dx$
Next, we recall the integral identity $\displaystyle \int f(x) \ dx = xf(x) - \int xf'(x) dx $ which holds for $f \in C^1\!$. Applying this to $f = \sqrt{x^2-1}$ we get $\displaystyle F = x - \int \sqrt{x^2 - 1} \ dx = x - x\sqrt{x^2-1} + \int \frac{x^2}{\sqrt{x^2-1}} \ dx $

Write $\displaystyle \frac{x^2}{\sqrt{x^2-1}} = \sqrt{x^2-1} + \frac{1}{\sqrt{x^2-1}}$ so we can solve for the integral of $f$:

$\displaystyle 2 \int \sqrt{x^2-1}\,dx = x \sqrt{x^2-1} - \int \frac{1}{\sqrt{x^2-1}}\,dx $

Finally, we can solve this last integral by trig substitution with $x = \sec (s)$ and noting that $\int \sec s \ ds = \log( \tan s + \sec s)$.
Adding up this mess, we get $\displaystyle F = x - \frac{1}{2} \left(x \sqrt{x^2-1} - \log{\large(}\sqrt{x^2-1} + x{\large)} \right) $.

Sorry about the formatting. For some reason, only inline math mode will compile properly.

Anyways, I remembered the following integral which kept me busy for a few hours (or more) in grad school. Enjoy.

Compute
$\int_0^{2\pi} \log \Gamma (\frac{x}{2 \pi}) \exp(\cos x) \sin(x + \sin x) \ dx $ where $\Gamma$ is the usual gamma function.


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