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:
Solution 1 (by Skipjack) Quote:
Problem 2 $$ \int_0^\infty x e^{1x}  \left\lfloor{x}\right\rfloor e^{1  \left\lfloor{x}\right\rfloor} \, dx$$ 
\begin{align}\int_0^\infty x e^{1x}  \left\lfloor{x}\right\rfloor e^{1  \left\lfloor{x}\right\rfloor} \, \mathrm dx &= \int_0^\infty x e^{1x} \,\mathrm dx  \left( 1+2e^{1} + 3e^{2} + \ldots \right) \\ &= xe^{1x}\bigg_0^\infty + \int_0^\infty e^{1x} \,\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^{1x}\bigg_0^\infty  \left(\frac{\mathrm d}{\mathrm dx} (1x)^{1}\right) _{x=e^{1}} \\ &= e\left((1x)^{2}\right) _{x=e^{1}} \\ &= e + \left( \frac1{1e^{1}}\right)^2 \\ &= e +\frac{e^2}{e^22e+1} \\ &= \frac{{e^32e^2+e}+e^2}{e^22e+1} \\ &= \frac{e^3e^2+e}{e^22e+1} \end{align} 
Rewrite it as a sum \[ \int_0^\infty x^{1x}  \lfloor x \rfloor e^{1\lfloor x \rfloor} = \sum_{n=0}^\infty \int_n^{n+1} x^{1x} \ dx  ne^{1n} \] Now compute the integral via $\frac{d}{dx}(x+1)e^{1x} = xe^{1x}$ and evaluate on $[n,n+1]$. This simplifies the sum to \[ \sum_{n=0}^\infty (2n)e^{1n} \] but this is just the series expansion for the derivative of $\frac{e^2}{1x}$ evaluated at $x = e^{1}$. Since this series converges uniformly for $x < 1$, it follows that \[ \sum_{n=0}^\infty (2n)e^{1n} = \frac{e^2}{(1e^{1})^2}\] 
Alternate solutions are good but please provide a new problem afterwards so we can keep it going :). 
$$\int \frac{\sqrt{x+1}  \sqrt{x1}}{\sqrt{x+1}+\sqrt{x1}} \,\mathrm dx$$ Answers without trigonometric or hyperbolic functions. 
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. 
I don't mean that you can't use them to get to your answer, but the final answer shouldn't contain them. 
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Start by noticing the factorization $\displaystyle \frac{2}{\sqrt{x+1}+\sqrt{x1}} = {\sqrt{x+1}\sqrt{x1}}$. 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 $f = \sqrt{x^21}$ we get $\displaystyle F = x  \int \sqrt{x^2  1} \ dx = x  x\sqrt{x^21} + \int \frac{x^2}{\sqrt{x^21}} \ dx $ Write $\displaystyle \frac{x^2}{\sqrt{x^21}} = \sqrt{x^21} + \frac{1}{\sqrt{x^21}}$ so we can solve for the integral of $f$: $\displaystyle 2 \int \sqrt{x^21}\,dx = x \sqrt{x^21}  \int \frac{1}{\sqrt{x^21}}\,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^21}  \log{\large(}\sqrt{x^21} + 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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