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Need help with integration $int_1^inftyx^-azcdot e^-bxcdot dx$
Bounding the integral $ int_1^infty frac (log(y))^n y^2 dy $Help needed with an integration problem.Evaluating $int_1^inftyx: texterfc(a+b log (x)) , dx$I need help to compute $int_1^+infty frac1tsqrtt^2 -1 dt$Need help with this ODE $y' + fracyx log y = fracyx^2 cdot (log y)^2$Need help with integration…Prove the convergence of $int_1^infty x^-x,dx$$int_1^inftysqrtfraclog xx^4+1dx=$??Does $int_1^infty f(x)ln(x)dx$ converge if $int_1^infty f(x)dx $ converges?Compute $int_1^infty frac1x^alpha + x^betadx$
$begingroup$
Need help with the following problem:
$$int_1^inftyx^-azcdot e^-bxcdot dx$$
Where $ z in mathbbC$ and $a, b in mathbbR$ and $ a>0, b>0 $
I thought of turning this into:
$$int_1^inftye^-azlog(x)cdot e^-bxcdot dx$$
$$=int_1^inftye^-azlog(x)-bxcdot dx$$
$$=int_1^inftye^f(x)cdot dx$$
$$=frace^f(x)f’(x) |_1^infty$$
But I am not sure if this makes sense for non-linear $f(x)$
integration
asked 2 days ago
ShreeShree
1067
$endgroup$
add a comment |
$begingroup$
Need help with the following problem:
$$int_1^inftyx^-azcdot e^-bxcdot dx$$
Where $ z in mathbbC$ and $a, b in mathbbR$ and $ a>0, b>0 $
I thought of turning this into:
$$int_1^inftye^-azlog(x)cdot e^-bxcdot dx$$
$$=int_1^inftye^-azlog(x)-bxcdot dx$$
$$=int_1^inftye^f(x)cdot dx$$
$$=frace^f(x)f’(x) |_1^infty$$
But I am not sure if this makes sense for non-linear $f(x)$
integration
asked 2 days ago
ShreeShree
1067
$endgroup$
$begingroup$
This can be written in terms of the incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function
$endgroup$
– Andrei
2 days ago
1
$begingroup$
Is there any sense to use $az $ instead of $z $?
$endgroup$
– user
2 days ago
add a comment |
$begingroup$
Need help with the following problem:
$$int_1^inftyx^-azcdot e^-bxcdot dx$$
Where $ z in mathbbC$ and $a, b in mathbbR$ and $ a>0, b>0 $
I thought of turning this into:
$$int_1^inftye^-azlog(x)cdot e^-bxcdot dx$$
$$=int_1^inftye^-azlog(x)-bxcdot dx$$
$$=int_1^inftye^f(x)cdot dx$$
$$=frace^f(x)f’(x) |_1^infty$$
But I am not sure if this makes sense for non-linear $f(x)$
integration
asked 2 days ago
ShreeShree
1067
$endgroup$
Need help with the following problem:
$$int_1^inftyx^-azcdot e^-bxcdot dx$$
Where $ z in mathbbC$ and $a, b in mathbbR$ and $ a>0, b>0 $
I thought of turning this into:
$$int_1^inftye^-azlog(x)cdot e^-bxcdot dx$$
$$=int_1^inftye^-azlog(x)-bxcdot dx$$
$$=int_1^inftye^f(x)cdot dx$$
$$=frace^f(x)f’(x) |_1^infty$$
But I am not sure if this makes sense for non-linear $f(x)$
integration
integration
asked 2 days ago
ShreeShree
1067
asked 2 days ago
ShreeShree
1067
edited 2 days ago
Shree
asked 2 days ago
ShreeShree
1067
asked 2 days ago
ShreeShree
1067
asked 2 days ago
ShreeShree
1067
1067
$begingroup$
This can be written in terms of the incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function
$endgroup$
– Andrei
2 days ago
1
$begingroup$
Is there any sense to use $az $ instead of $z $?
$endgroup$
– user
2 days ago
add a comment |
$begingroup$
This can be written in terms of the incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function
$endgroup$
– Andrei
2 days ago
1
$begingroup$
Is there any sense to use $az $ instead of $z $?
$endgroup$
– user
2 days ago
$begingroup$
This can be written in terms of the incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function
$endgroup$
– Andrei
2 days ago
$begingroup$
This can be written in terms of the incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function
$endgroup$
– Andrei
2 days ago
1
1
$begingroup$
Is there any sense to use $az $ instead of $z $?
$endgroup$
– user
2 days ago
$begingroup$
Is there any sense to use $az $ instead of $z $?
$endgroup$
– user
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using the substitution $u=bx$ transforms the integral into
$$b^az-1int_b^infty u^-aze^-udu=b^az-1Gamma(1-az, b)$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
votes
$begingroup$
Using the substitution $u=bx$ transforms the integral into
$$b^az-1int_b^infty u^-aze^-udu=b^az-1Gamma(1-az, b)$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
$endgroup$
add a comment |
$begingroup$
Using the substitution $u=bx$ transforms the integral into
$$b^az-1int_b^infty u^-aze^-udu=b^az-1Gamma(1-az, b)$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
$endgroup$
add a comment |
$begingroup$
Using the substitution $u=bx$ transforms the integral into
$$b^az-1int_b^infty u^-aze^-udu=b^az-1Gamma(1-az, b)$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
$endgroup$
Using the substitution $u=bx$ transforms the integral into
$$b^az-1int_b^infty u^-aze^-udu=b^az-1Gamma(1-az, b)$$
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
answered 2 days ago
Peter ForemanPeter Foreman
3,6751216
3,6751216
add a comment |
add a comment |
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$begingroup$
This can be written in terms of the incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function
$endgroup$
– Andrei
2 days ago
1
$begingroup$
Is there any sense to use $az $ instead of $z $?
$endgroup$
– user
2 days ago