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Articles by Yagna Patel
One Step Forward from Gaussian Integral
Note: Not an Answer but a Generalization Consider the integral, $\displaystyle I = \int_0^\infty e^{-x^A} \text{ d}x $. Substitute $\displaystyle x = u^{\frac{1}{A}} \Rightarrow \text{ d}x = \frac{1}{A} u^{\frac{1-A}{A}} \text{ d}u $.
Limit of an Integral, Then taking Sum
Yes, you are right about the first step. First of all, let us find the recurrent relation for the $I_n$. Integrating by parts we get: $$I_n=\int^1_0x^ne^x\,dx=e-\int^1_0nx^{n-1}e^x\,dx=e-nI_{n-1}$$ This is equivalent to $I_n+nI_{n-1}=e$. The last equality implies $I_{n+1}+(n+1)I_n=I_n+nI_{n-1}$ or $I_{n+1}=n(I_{n-1}-I_n)$. Now lets analyze our sum. We obtain: $$\sum_{k=1}^n\frac{I_{k+1}}{k}=\sum_{k=2}^n(I_{k-1}-I_k)+I_2=I_1-I_n+I_2$$ Now we must find to which number does $I_n$ tend.
Trigonometry to the 24th power
Considering that the answer is apparently $$ \frac{7 \left(489857+280140 \sqrt{3}\right)}{8388608}, $$ I doubt there's an easy way. You know $\sin{\frac{\pi}{6}}=\frac{1}{2}$, $\cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}$, so using the half-angle formulae $$ 2\sin^2{\frac{x}{2}} = 1-\cos{x},\\ 2\cos^2{\frac{x}{2}} = 1+\cos{x} $$ will get us to $\sin{\frac{\pi}{24}}$ and $\cos{\frac{\pi}{24}}$, when applied twice.
Tricky Rectangle Problem
Here is a more combinatorial/algebraic approach: Let the Horizontal lines from bottom to top be $h_{0}, h_{1}, ..., h_{6}$ and Vertical lines from left to right be $v_{0}, v_{1}, ..., v_{6}$ Rectangles that contain bottom left yellow square is all rectangles that can be formed by choosing one horizontal line from $\{h_{0}, h_{1}\}$, another horizontal line from $\{h_{2}, ..., h_{6}\}$ and one vertical line from $\{v_{0}, v_{1}\}$, another vertical line from $\{v_{2}, ..., v_{6}\}$.
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