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I successfully solved the following integral of $\int _0^\infty e^{-y}e^{-xy}y^2dy$ by using partial integration two times, which was pretty hell long, and since solving this integral was not the main issue in the exercise I was doing, I wondered if there's an easier way to integrate it. Is there? Thanks! (btw my result was $\frac{2}{(x+1)^3}$)

P.S I am not familiar with gamma functions or anything beyond "common B.SC calculus".

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  • $\begingroup$ This integral looks very definite... $\endgroup$ – user Jan 23 '19 at 21:26
  • $\begingroup$ @user I edited it to include limits to match the stated result. I'll fix the title. $\endgroup$ – J.G. Jan 23 '19 at 21:31
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To prove by induction that $\int_0^\infty y^n e^{-zy}dy=\frac{n!}{y^{n+1}}$ (for $\Re z>0$), note that $$\int_0^\infty e^{-zy}dy=\frac{1}{z},\,\int_0^\infty y^{n+1} e^{-zy}dy=-\partial_z\int_0^\infty y^{n+1} e^{-zy}dy.$$For your stated problem, you only need to differentiate the base case twice; the induction is in case you want to generalise it.

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\begin{align} \int_0^\infty e^{-(1+x)y}y^2dy &= \frac{d^2}{dx^2}\int_0^\infty e^{-(1+x)y}dy \\ &= \frac{d^2}{dx^2}\frac{1}{(1+x)}e^{-(1+x)y}\bigg|_0^\infty \\ &= \frac{d^2}{dx^2}\frac{1}{(1+x)} \\ &= \frac{2}{(1+x)^3} \end{align}

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  • $\begingroup$ Thanks! can you please explain the first step? $\int_0^\infty e^{-(1+x)y}y^2dy =\frac{d^2}{dx^2}\int_0^\infty e^{-(1+x)y}dy $ $\endgroup$ – superuser123 Jan 24 '19 at 7:40
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    $\begingroup$ First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$\frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $\int_0^\infty \frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$\frac{d^2}{dx^2} \int_0^\infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick. $\endgroup$ – suneater Jan 24 '19 at 8:55
  • $\begingroup$ Why does it bring down the factors of y? I just don't get it :( $\endgroup$ – superuser123 Jan 24 '19 at 13:01
  • $\begingroup$ Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$\frac{d}{dx} e^{ax} = a e^{ax}$$ $$\frac{d^2}{dx^2} e^{ax} = a \frac{d}{dx} e^{ax} = a^2 e^{ax}$$ $\endgroup$ – suneater Jan 24 '19 at 20:43
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Since your questions are related to random variables, this answer uses an exponential RV to compute the integral. Note that for an exponential RV, $Y$, with pdf $$f(y) = \lambda e^{-\lambda y}, \quad \lambda,y>0,$$ we have $$E[Y] = \frac{1}{\lambda},$$ $$var(Y) = \frac{1}{\lambda^2} \implies E[Y^2] = \frac{2}{\lambda^2}.$$ That is, $$\lambda\int_0^{\infty} y^2 e^{-\lambda y} dy = \frac{2}{\lambda^2} \implies \int_0^{\infty} y^2 e^{-\lambda y} dy = \frac{2}{\lambda^3}.$$ Can you guess $\lambda$ in your problem?

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  • $\begingroup$ How would you prove $E[Y^2]=2\lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.) $\endgroup$ – J.G. Jan 23 '19 at 21:21
  • $\begingroup$ I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = \frac{n!}{\lambda^n}$ which is quite easy to evaluate. $\endgroup$ – Math Lover Jan 23 '19 at 21:26
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$I(x) = \int_0^{\infty} e^{-y}e^{-xy}y^2 dy$

Integration under the integral

$I(x) = \frac {d}{dx} \int_0^{\infty} e^{-y}y^2(\int e^{-xy}\ dx) dy = \frac {d}{dx}\int_0^{\infty} e^{-y(1+x)}y\ dy$

And one more time.

$I(x) = \frac {d^2}{dx^2}\int_0^{\infty} e^{-y(1+x)} dy = \frac {d^2}{dx^2}\frac {1}{1+x} = \frac {1}{2(1+x)^3}$

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Try separating the integral like this $$\int _0^\infty e^{-y(x+1)}y^2dy=\int _0^1 e^{-y(x+1)}y^2dy+\int _1^\infty e^{-y(x+1)}y^2dy$$

you can see that the first integral is a proper one hence the integral converges. For the second one try taking the cases when $x+1 \lt0 $, $x+1 =0$ and $x+1 \gt0$

for $x+1 \gt0$ from $$\lim_{y\to \infty}\frac{e^{-y(x+1)}y^2}{\frac{1}{y2}}=0$$ this implies that $$\int _1^\infty e^{-y(x+1)}y^2dy\le\int _1^\infty \frac{1}{y^2}dy$$

the last integral converges, from the comparison test so will the integral $\int _1^\infty e^{-y(x+1)}y^2dy$

Can you try the other cases?

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