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Jan
2
comment Finding in terms of via integration
Hint : use that $\sin^{98}(x)\sin^2(x) = \sin^{98}(x) (1 - \cos^2(x) )$ and then use an integration by parts knowing that $\frac{(\sin^{99}(x))}{99} '= \sin^{98}\cos(x)$
Nov
2
revised Evaluation of $I_{a,b} = \int_{1}^{+\infty} \frac{\ \exp{(-at)}}{ 1-bt} \ \mathrm{d}t $
deleted 13 characters in body
Oct
18
accepted Evaluate or Simplify $\int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$
Jun
4
revised Double integral $\int_{z=u}^{+\infty}\int_{t=u}^{+\infty}\frac{e^{-Az}}{z+B}\frac{te^{-tD}}{t-zC}\,dtdz$
deleted 8 characters in body
Jun
4
answered Double integral $\int_{z=u}^{+\infty}\int_{t=u}^{+\infty}\frac{e^{-Az}}{z+B}\frac{te^{-tD}}{t-zC}\,dtdz$
Jun
2
revised Evaluate or Simplify $\int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$
edited body
Jun
1
comment Evaluate or Simplify $\int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$
@GEdgar in that case the integral does not converge.
Jun
1
asked Evaluate or Simplify $\int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$
Jun
1
revised Evaluation of $I_{a,b} = \int_{1}^{+\infty} \frac{\ \exp{(-at)}}{ 1-bt} \ \mathrm{d}t $
Corrected title
May
31
revised Exact value of an integral
remove an impolite expression
May
31
suggested suggested edit on Exact value of an integral
May
31
answered Solutions of an integral equation
May
28
awarded  Suffrage
May
28
accepted Evaluation of $I_{a,b} = \int_{1}^{+\infty} \frac{\ \exp{(-at)}}{ 1-bt} \ \mathrm{d}t $
May
28
accepted How demonstrate the Craig representation for the Gaussian probability function?
May
28
comment How demonstrate the Craig representation for the Gaussian probability function?
Alright then, thanks anyway :)
May
27
comment How demonstrate the Craig representation for the Gaussian probability function?
Well Done @Didier, About the Craig's proof I refered to this [pdf document][1]. When I searching I found a book called : "Probability Distributions Involving Gaussian Random Variables" state that : <<The form in (l) is not readily obtainable by a change of variables directly in (2). However, by first extending (1) to two dimensions (x and y) where one of the dimensions (y) is integrated over the half plane, a change of variables from rectangular to polar coordinates readily produces (2).>> pg.123. If this is clear for you, how can be done? [1]:wsl.stanford.edu/~ee359/craig.pdf
May
27
revised How demonstrate the Craig representation for the Gaussian probability function?
improved formatting
May
27
comment How demonstrate the Craig representation for the Gaussian probability function?
Thanks @DavideGiraudo for fixing it.
May
27
revised How demonstrate the Craig representation for the Gaussian probability function?
edited body