Integral Proofing: Proving Existence and Continuity of I(x)

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Homework Statement



let f(x,t)=xe^(-xt).show that the integral I(x)=∫f(x,t)dt (integration from 0 to infinite)exists for all x>=0 . is x->I(x) continuous on [0,infinite)



The Attempt at a Solution


what should i use here to prove the integral exist ?once i prove that exist, can i use the specific integration to see its continuity?
 
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For a fixed value of x, are you able to show that the function g:\mathbb{R}\rightarrow\mathbb{R} defined by g(t) = xe^{-xt} is continuous for all t? If so, then it can be integrated on any finite interval, so \int_{0}^{T} g(t) dt exists and is finite for all T > 0. You should be able to evaluate this integral explicitly to get some function G(x,T). Then check whether
\lim_{T \rightarrow 0}G(x,T)
exists.
 
jbunniii said:
For a fixed value of x, are you able to show that the function g:\mathbb{R}\rightarrow\mathbb{R} defined by g(t) = xe^{-xt} is continuous for all t? If so, then it can be integrated on any finite interval, so \int_{0}^{T} g(t) dt exists and is finite for all T > 0. You should be able to evaluate this integral explicitly to get some function G(x,T). Then check whether
\lim_{T \rightarrow 0}G(x,T)
exists.

yes, i have solved it out the limit is 1 , then means I(x)>0 but i feel confused about the continuous part,what should i do in the secound part?
 
cummings12332 said:
yes, i have solved it out the limit is 1 , then means I(x)>0 but i feel confused about the continuous part,what should i do in the secound part?

Ok, so I(x)=1 if x>0. What's I(0)?
 
Dick said:
Ok, so I(x)=1 if x>0. What's I(0)?

ok, i get it , thanks so much
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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