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 Post subject: definite integration
PostPosted: Sat, 18 Aug 2012 19:19:32 UTC 
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$\int_0^1\frac{x^3-1}{logx}$

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 Post subject: Re: definite integration
PostPosted: Sun, 19 Aug 2012 11:13:07 UTC 
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mun wrote:
$\int_0^1\frac{x^3-1}{logx}$


Obviously letting t=\log x gives \displaystyle\int_{-\infty}^0\frac{e^{3t}-1}{t} e^t\,\mathrm{d}t. Now recall the exponential integral \displaystyle\mathrm{Ei}(z)=\int_{-\infty}^z \frac{e^t}{t}\,\mathrm{d}t has the convergent series \displaystyle \mathrm{Ei}(z)=\gamma+\log\lvert z\rvert+\sum_{k=1}^\infty\frac{z^k}{k\cdot k!} for real z, and take appropriate limit as z\uparrow 0.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: definite integration
PostPosted: Sun, 19 Aug 2012 15:07:10 UTC 
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sir at this solution where is the error
\int_0^1(\frac{x^3}{logx}-\frac{1}{logx})
now when i put x^4=t in the first integral ,we get \int_0^1(\frac{1}{logt}).

the whole answer reduces to 0.where is the error. :shock:

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 Post subject: Re: definite integration
PostPosted: Sun, 19 Aug 2012 16:42:02 UTC 
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mun wrote:
sir at this solution where is the error
\int_0^1(\frac{x^3}{logx}-\frac{1}{logx})
now when i put x^4=t in the first integral ,we get \int_0^1(\frac{1}{logt}).

the whole answer reduces to 0.where is the error. :shock:


\displaystyle\int_0^1\frac{1}{\log t}\,\mathrm{d}t=-\infty, and your naïve method would yield (-\infty)-(-\infty).

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: definite integration
PostPosted: Mon, 20 Aug 2012 14:50:22 UTC 
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outermeasure wrote:
mun wrote:
$\int_0^1\frac{x^3-1}{logx}$


Obviously letting t=\log x gives \displaystyle\int_{-\infty}^0\frac{e^{3t}-1}{t} e^t\,\mathrm{d}t. Now recall the exponential integral \displaystyle\mathrm{Ei}(z)=\int_{-\infty}^z \frac{e^t}{t}\,\mathrm{d}t has the convergent series \displaystyle \mathrm{Ei}(z)=\gamma+\log\lvert z\rvert+\sum_{k=1}^\infty\frac{z^k}{k\cdot k!} for real z, and take appropriate limit as z\uparrow 0.


Actually one doesn't need the full series, just the divergent part \displaystyle\int_{-\infty}^z\frac{e^t}{t}\,\mathrm{d}t=\log\lvert z\rvert+\text{(finite) continuous at 0} near 0.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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