sol := dsolve({diff(f(x), x) = f(1/x), f(0) = 0}, numeric, delaymax = 1.0, delaypts = 2000);
plots:-odeplot(sol, [x, f(x)], x = -1 .. 1);

Maybe you can try with another initial condition f(0)=1 ?

evalf[30](subs(z = 2 + 3*I, Zeta(-z) + 2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1)));

rtable([seq(evalf[2^m](subs(z = 2 + 3*I, Zeta(-z) + 2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1))), m = 1 .. 10)], subtype = Vector[column]);#Depending of the precise

I have another sum with Laguerre polynomials to  give exp(z).

Sum((m + 1)!*z^k*LaguerreL(k, k + m, z)/(k + m + 1)!, k = 0 .. infinity) = exp(z)

We can check:

restart;
f := (z, m) -> sum((m + 1)!*z^k*LaguerreL(k, k + m, z)/(k + m + 1)!, k = 0 .. infinity)
interface(rtablesize = 100);
rtable([seq([m, evalf(f(1, m))], m = 0 .. 20)], subtype = Vector[column]);

And for your example:

Digits := 30;
f := (x, m) -> sum(m!*x^(m - n)*LaguerreL(m, n - m, x)^2/n!, n = 0 .. infinity);
interface(rtablesize = 100);
rtable([seq([m, evalf(f(1, m))], m = 0 .. 20)], subtype = Vector[column]);

From this book on page 152 I borrowed the code.

See attached files.

fracdiff.mw

fracdiff_for_alpha=1_test.mw

For first question:

Int(exp(-(abs(x - mu)/sigma)^beta), x = -infinity .. s) = piecewise(Or(s = mu, s <= 0), sigma*GAMMA(1 + 1/beta), And(mu < s, 0 < s), ((-s + mu)*Ei(-(1 + beta)/beta, (s - mu)^beta*sigma^(-beta)) + 2*sigma*GAMMA(1/beta))/beta, sigma*GAMMA(1/beta, (-s + mu)^beta*sigma^(-beta))/beta)

Try this:

eq := 2*exp(-2*t) + 4*t = 127;
solve([eq, 0 < t], {t});
limit(LambertW(-exp(-x))/2 + 127/4, x = infinity);
interface(rtablesize = 100);
rtable([evalf[100](seq(limit(LambertW(-exp(-x))/2 + 127/4, x = m), m = 2 .. 200, 5))], subtype = Vector[column]);

This integral, in most cases, cannot be expressed in terms of elementary functions,but we can expressed in terms of GAMMA function.

data_v1.mw
 

Download data_v1.mw

Download invLaplace.mw

Do you have any reason to think there is a closed form?

Most integrals don't have one.

Maybe the best you can do is numerical methods.

See atthached file.

ode.mw

See teory of First Order Differential Equations. Only one initial value problem can be not two.

See attached file:

integral.mw

See attached file:

PDE_by_Elziki_Transform.mw

For first question: 

f := x -> 36*x^6 + 2665*x^4 + 240*x - 675 + 4534*x^2 - 5836*x^3 - 516*x^5;

minimize(f(x), x = 0 .. 4, location);
#-675, {[{x = 0}, -675]}

evalf(maximize(f(x), x = 0 .. 4, location));
#703.9550742, {[{x = 3.800387934}, 703.9550742]}

Try:

ode := diff(U(z), z $ 4) + c^2*diff(U(z), z $ 2) + k*c*diff(U(z), z $ 2) - (3*U(z)^2 + a)*diff(U(z), z $ 2) = 0;
Order := 5;dsolve(ode, U(z), type = 'series');


#U(z) = U(0) + D(U)(0)*z + 1/2*(D@@2)(U)(0)*z^2 + 1/6*(D@@3)(U)(0)*z^3 + (U(0)^2*(D@@2)(U)(0)/8 - c^2*(D@@2)(U)(0)/24 - k*c*(D@@2)(U)(0)/24 + (D@@2)(U)(0)*a/24)*z^4 + O(z^5)

With initial conditions 

Order := 5;dsolve([ode, U(A) = A1, D(U)(A) = B1, (D@@2)(U)(A) = C1], U(z), type = 'series');

#U(z) = A1 + B1*(z - A) + 1/2*C1*(z - A)^2 + 1/6*(D@@3)(U)(A)*(z - A)^3 + (1/8*A1^2*C1 - 1/24*c^2*C1 - 1/24*k*c*C1 + 1/24*C1*a)*(z - A)^4 + O((z - A)^5)

As a workround using fourier transform:

(inttrans:-invfourier(int((inttrans:-fourier(sin(p*r), p, s) assuming (0 <= r))*sin(q*r)/(p*q), r = 0 .. infinity), s, p) assuming (q < p));

#-Pi*Dirac(p + q)/(2*p*q)

Try:

simplify(pdetest(sol, sys));

gives:

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]