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3

Start from the first Binet's formula: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\int_0^\infty\left(\frac12-\frac1t+\frac1{e^t-1}\right)\frac{e^{-t\,z}}t dt.\tag1$$ Change variable $t=-2\ln x$: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\frac12\int_0^1x^{2z-1}\frac{1-x^2+(1+x^2)\ln x}{(x^2-1)\ln^2x}dx.\tag2$$ ...

3

See http://en.wikipedia.org/wiki/Lists_of_integrals#Definite_integrals_lacking_closed-form_antiderivatives. This link tells you a list of integrals not expressible in a closed form expression.

2

Substitute $x=\tan(\theta)$: \begin{align} \int_0^1\frac{\log(1+x^2)}{1+x^2}\,\mathrm{d}x &=2\int_0^{\pi/4}\log(\sec(\theta))\,\mathrm{d}\theta\\ &=-2\int_0^{\pi/4}\log(\cos(\theta))\,\mathrm{d}\theta\\ &=-\int_0^{\pi/4}\left[\log(1+e^{i2\theta})+\log(1+e^{-i2\theta})-2\log(2)\right]\,\mathrm{d}\theta\\ ... 2 If you're talking about power series, any function expressible by a power series that converges in some open interval (a,b) is analytic in a (complex) neighbourhood of that interval and so are its derivative and antiderivative. So a counterexample would be a any function that is integrable but not analytic. For an example of a function that is smooth ... 2 Sub x=a u; then\int_0^a dx \, \log{x} \, \log{(a-x)} = a \int_0^1 du \left [\log^2{a} + \log{a} \left (\log{u} + \log{(1-u)}\right ) + \log{u} \, \log{(1-u)}\right ] $$The first three integrals are straightforward; the middle two may be evaluated using the antiderivative$$\int dx \, \log{x} = x \log{x} - x +C$$For the final integral, you can ... 2 I rather dislike the way a lot of textbooks present the material on surfaces and solids of revolution because they will at some point present a table of formulas without having clarified the underlying reasoning. Students are confronted with several rather similar-looking equations and wonder how they are going to memorize them. What the surface area ... 1 A classical method, going back to Newton himself, is to expand the integrand into a power series and integrate term by term. If the resulting series converges, it converges to the value of the integral (this is easy to show if the interval of integration is within the open interval of convergence; in your case, since 1 is right at the edge of convergence, ... 1 Here is another proof, the main part of which was communicated to me by Dr. Peter Otte of Bochum University: $$I_n := \int_{[0,1]^n}\mathrm{d}u\,\delta(1-\lvert u\rvert_1) \frac{1}{\prod_{j=1}^n (u_j + u_{j+1})} = (2\pi)^{n-2} \frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}.$$ First, define$$J_n(t) := ...

1

In the proof that you cite there is an integral from $0$ to $\frac{\pi}{2}$, which for a circle means that it is in in the first quadrant (which makes sense, considering they were looking for the area of the quarter circle formed in the first quadrant), and because the $y$ values are positive, we use a positive square root. For a more graphic ...

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The parametric integrand you want is as follows. Put $$f(p)=\int_0^1\frac{x^p-1}{\ln x}dx$$ Then $$f'(p) =\int_0^1\frac{\partial}{\partial p}\left(\frac{x^p-1}{\ln x}\right)dx=\int_0^1 x^p\;dx=\left.\frac{x^{p+1}}{p+1}\right|_0^1=\dfrac{1}{p+1}$$ so $f(p)= \ln(p+1)+ C$, where $C$ is to be determined. To find $C$, just note that $f(0)=0$ since the ...

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With $r$ constant, the differential element of area is $r^{2}\sin\theta~ d\theta~ d\phi$. The cone you have given is with $\theta=45^{o}$. Integrating as below $$\int^{\pi/4}_{\theta=0}\int^{2\pi}_{\phi=0} r^{2}\sin\theta~ d\theta~ d\phi=2\pi r^{2}(1-\cos{\pi/4})$$

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