My Math Forum Challenging integral marathon!

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October 4th, 2018, 06:42 PM   #1
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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 $$\int\frac{x-1}{x+x^2\log(x)}\,dx$$

Solution 1 (by Skipjack)

Quote:
 Originally Posted by skipjack 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

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

 October 4th, 2018, 07:39 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra \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}
 October 4th, 2018, 08:38 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 473 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics 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 $\sum_{n=0}^\infty (2-n)e^{1-n} = -\frac{e^2}{(1-e^{-1})^2}$
 October 4th, 2018, 08:54 PM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,717 Thanks: 597 Math Focus: Yet to find out. Alternate solutions are good but please provide a new problem afterwards so we can keep it going . Thanks from SenatorArmstrong and ProofOfALifetime
 October 4th, 2018, 09:48 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra $$\int \frac{\sqrt{x+1} - \sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} \,\mathrm dx$$ Answers without trigonometric or hyperbolic functions.
 October 6th, 2018, 07:49 AM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 473 Thanks: 262 Math Focus: Dynamical systems, analytic function theory, numerics 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.
 October 6th, 2018, 08:53 AM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra I don't mean that you can't use them to get to your answer, but the final answer shouldn't contain them.
October 6th, 2018, 09:14 AM   #8
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Quote:
 Originally Posted by v8archie 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.

October 6th, 2018, 12:53 PM   #9
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Quote:
 Originally Posted by v8archie 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.

October 6th, 2018, 03:41 PM   #10
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Quote:
 Originally Posted by v8archie $$\int \frac{\sqrt{x+1} - \sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}} \ dx$$ 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.

Last edited by skipjack; October 6th, 2018 at 05:20 PM.

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