The overpowered Laplace technique for summing series. 

even if a and b are both positive? r/b-a will blow up but Xᵃ - Xᵇ will go to zero. hmmmm its not apparent that a=b is invalid from the series alone

you can take a limit as b approaches a and notice it is the derivative of x^(a) with respect to a

@@aliinci1874 can confirm. Evaluating for b approaches a at a=0 r=1 you find the solution to the basel problem

@@GeoffryGifari of a=b u can't use partial fraction decomposition, not in the way Michael did

a=b is a special case of decomposition.

Computer scientist spotted!!!!!!!!! Also just take the original problem and consider the case a=b and you kind of just solve it from there

Say a == b, not a = b.

Well... He assumed a,b to be nonnegative integers right at the start, didn't he?

Both x and u are positive on the interval (0, 1).

The integration is within the interval which is more or equal to zero ,i.e. from 0 to 1

@@michaelguenther7105 Of course! Thanks.

@@IoT_ Thank you!

Actually if u=-sqrtx, x=1,u=-1,u can be negative However, u=sqrtx and u=-sqrtx are the same integral Let x=u^2 and -sqrtx=u dx=2udu x=0,u=0 x=1,u=-1 ∫_(0,1) sqrtx/(x-1)dx =∫_(0,-1) -u/(u^2-1)*2udu =∫_(-1,0) 2u^2/(u^2-1)du =∫_(0,1) 2u^2/(u^2-1)du Because it is an even function

By the same method on the video I reduced -sum_{n=1}^{infty} (-1)^n/(n+x) to int_0^infty e^-(1+x)t/(1+e^(-t)) dt. I'm not sure if I did it right and if it is how it is related to gamma(1-x)gamma(1+x).

When plotted on Desmos the integral is a really got approximation for gamma(1-x)gamma(1+x) in the interval (-1,-1/2]

I think you forgot to multiply the dx terms.

No, it is right because of the du term.

​@@megauser8512 Sorry, you are right.

There is a theorem about this which has to do with different types of convergence, I think when you have "absolute" convergence then it is okay to make the switch. Maybe someone can expand on my answer.

The "Dominated Convergence Theorem". Basically says as long as your series is 'nicely behaved', you can swap the order. (*very* oversimplified explanation of 'nice' in this case is absolutely bounded and converges at every value of the summation index. don't use this explanation in your homework.)

Iirc the condition for dominanted convergence is that (or at least in one of the versions) there is a nonnegative function such that at each point within the integration bounds, the expression doesn’t exceed the function regardless of n, and that the function is integrable

Since the original series is absolutely convergent, then any change of variables or order is valid.

Well, you can tear the former integral into two, do the change of variable for each of them as Michael has done. The variables used in the change should probably differ as they result from different changes, but we can rename the variable we integrate over as we please. Then we would be left with the addition of the two integrals over the same integral, which can be joined again. Schematically: int_0^1 [ f(x) + g(x) ] dx = int_0^1 f(x) dx + int_0^1 g(x) dx = int_0^1 h(u) du + int_0^1 k(v) dv = int_0^1 h(u) du + int_0^1 k(u) du = int_0^1 [ h(u) + k(u) ] du The thing is we are able to do this because both of the changes of variables give the same interval (0, 1) "by coincidence".