mjhjmtu

So my friend sent me this really interesting problem. It goes:

Evaluate the following expression:

$$ sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx .$$

Here is my approach:

First evaluate the integral:

$$ frac1b! int_0^infty fracx^be^ax dx.$$

This can be done using integration by parts and we get:

$$ frac1b! fracba int_0^infty fracx^b-1e^ax dx.$$

We can do this $ b $ times until we get:

$$ frac1b! frac(b)(b-a).....(b-b+1)a^b int_0^infty fracx^b-be^ax dx.$$

and hence we end up with:

$$ frac1b! fracb!a^bqquadleft(frac-1 e^-axaBig|_0^inftyright) = frac1a^b+1.$$

Now we can apply the sum of GP to infinity formula and we get:

$$ sum_a=2^infty sum_b=1^infty frac1a^b+1 = sum_a=2^infty fracfrac1a^21-frac1a.$$

This is a telescoping series and we end up with $$ frac1a-1 = frac12-1 = 1.$$

Do you guys have any other ways of solving this problem? Please do share it here.

since $fracx^be^ax b!$ is non-negative, Tonelli's theorem for iterated integrals/sums allows us to interchange integrals and sums without worry. Then:
beginalign
&sum_a=2^infty sum_b=1^infty int_0^infty fracx^be^ax b! dx \
&=int_0^infty sum_a=2^infty e^-ax sum_b=1^infty fracx^b b! dx \
&= int_0^infty underbraceleft(sum_a=2^infty (e^-x)^aright)_textgeometric series overbraceleft(sum_b=0^infty fracx^b b!-1right)^textseries definition of $e^x$ dx \
&= int_0^infty frac1e^x(e^x-1)(e^x-1)dx \
&= int_0^infty e^-x dx \&= 1.endalign

The integral is of the Gamma type,

$$int_0^infty fracx^be^ax dx=frac1a^b+1int_0^infty t^be^-t dx =fracb!a^b+1.$$

Then

$$sum_a=2^infty sum_b=1^inftyfrac1a^b+1=sum_a=2^infty frac1a^2left(1-dfrac1aright)=sum_a=2^infty frac1a(a-1)$$ is indeed a telescoping sum, giving

$$frac12-1.$$

For the first part I often use the Laplace transform:
$$frac1b! int_0^infty fracx^be^ax dx = frac1b! int_0^infty x^be^-ax dx = frac1b! cal L(x^b)Big|_s=a = frac1b! fracb!a^b+1 = frac1a^b+1$$
this make it easier.

Also note:
$$S=sum_a=2^inftysum_b=1^inftyint_0^inftyfracx^be^axb!dx=sum_a=2^inftysum_b=1^inftyfrac1b!int_0^infty x^be^-axdx$$
now let:
$$u=ax$$
$$dx=fracdua$$
so:
$$S=sum_a=2^inftysum_b=1^inftyfrac1a*b!int_0^infty left(fracuaright)^be^-udu=sum_a=2^inftysum_b=1^inftyfraca^-(b+1)b!int_0^infty u^be^-udu$$
and we know that:
$$(n-1)!=Gamma(n)=int_0^infty e^-tt^n-1dt$$
so our summation now simplifies to:
$$S=sum_a=2^inftysum_b=1^inftyfraca^-(b+1)b!b!=sum_a=2^inftysum_b=1^inftyfrac1a^b+1=sum_a=2^inftysum_c=2^inftyfrac1a^c=sum_a=2^inftyleft(sum_c=1^inftyfrac1a^c-frac1aright)=sum_a=2^inftysum_c=1^inftyfrac1a^c-sum_a=2^inftyfrac1a$$
I know this is correct up to the second summation on the final line but after this I am not sure.