definite integration

outermeasure · **Posted:** Sun, 19 Aug 2012 11:13:07 UTC

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

, and take appropriate limit as $z\uparrow 0$ .

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1271

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:

outermeasure · **Posted:** Sun, 19 Aug 2012 16:42:02 UTC

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 $(-\infty)-(-\infty)$ .

outermeasure · **Posted:** Mon, 20 Aug 2012 14:50:22 UTC

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

, 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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definite integration

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