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bw0young0math
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- Jun 14, 2013
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Let's consider the second integral. A primitive of the function is...Hello. I'm studying improper integrals in real analysis. However, two problems are very difficult to me. If you are OK, please help me.(heart)
Regarding to the first integral 'Monster Wolfram' supplies...Hello. I'm studying improper integrals in real analysis. However, two problems are very difficult to me. If you are OK, please help me.(heart)
May be that the following 'forgotten formula' ...We have to consider \(\displaystyle \alpha>-1 \) to solve the first integral by Laplace transform .
I did not see the answer but I think it should beRegarding to the first integral 'Monster Wolfram' supplies...
int e^(- x) sin x x^a from 0 to infinity - Wolfram|Alpha
How to obtain that result however is a difficult task...
Kind regards
$\chi$ $\sigma$
An Italian adage says 'The Devil is not so ugly as You paint him' so that we try to compute the integral...May be that the following 'forgotten formula' ...
http://mathhelpboards.com/analysis-50/laplace-transform-7057.html#post32097
$\displaystyle \mathcal {L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F_{1} (z)\ F_{2}(s - z)\ dz\ (1)$
... is useful in this case. Here is $\displaystyle f_{1} (t) = t^{\alpha} \implies F_{1} (s) = \frac{\Gamma (1 + \alpha)}{s^{1 + \alpha}}$ and $\displaystyle f_{2} (t) = \sin t \implies F_{2}(s)= \frac{1}{1 + s^{2}}$. In any case that is not a trivial task...
\(\displaystyle (1-i) = \sqrt{2} e^{-i \frac{\pi }{4}}\)I did not see the answer but I think it should be
\(\displaystyle \Im \left (\frac{\Gamma (\alpha +1)}{(1-i)^{\alpha +1}}\right)\)