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Suppose $\displaystyle f(x,\alpha )=\begin{cases}
\dfrac{e^{-\alpha x}-e^{-2\alpha x}}{x} & \text{ if } x\neq 0 \\\\
\alpha & \text{ if } x=0
\end{cases}$.
Is $\displaystyle \int_{0}^{\infty }f(x,\alpha )dx$ uniformly convergent for all $\alpha \geq 0$ ? If yes, then how? Please help.

Thank you!!

\[ \int_{0}^{\infty }\frac{\ln^{2}x}{x^2+2x\cos\theta+1 }dx=?\ where\ \theta\in [0,\pi ] \]

\[ \int_{0}^{\infty }\left \lfloor{2016\,e^{-x}}\right \rfloor dx=?
\]

By using Discrete Fourier Transform (used in the http://www.mathimatikoi.org/forum/viewt ... f=27&t=686 )

\(\mathcal{S}(x)=\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}\)

\(= \frac{1}{2}\left ( \cosh x +\cos x \right )\)

Simply Beautiful Solution!!!
Thank you soooo much!

\[\sum_{{n=0}}^{\infty} \frac{1}{\left ( 3n \right )!}= ?\]

$$\sum_{n=2}^{\infty}(-1)^n \frac{\ln n}{n} =?$$

A Problem by a Great Teacher and Mathematician!
Solve using Elliptic Integrals:
$$ \int_{0}^{a}\frac{1}{\sqrt{(\sin x+\cos x)(1+2\sin 2x-4\sin^22x)}} dx=? $$

We have that: $$\int \left [ \left ( \frac{3}{x^2}-1 \right )\sin x - \frac{3}{x}\cos x \right ]^2 \, dx = \int \left ( \frac{9}{x^4}\sin^2 x +\sin^2 x - \frac{6}{x^2}\sin^2 x + \frac{9}{x^2}\cos^2 x -\frac{18}{x^3}\sin x \cos x + \frac{3}{x}\sin 2x \right )\, dx$$ Using Tabular Integration, we have...