divergent series – Regularization of “harmonic integral” \$int_0^infty frac1xdx\$ and several others

Can we ascribe a regularized value to the integral $$int_0^infty frac1xdx$$? Well, I tried, and here is how.

Let’s consider the transform $$mathcal{L}_t(t f(t))(x)$$. It is notable by the fact that it preserves the area under the curve:
$$int_0^infty f(x)dx=int_0^infty mathcal{L}_t(t f(t))(x) dx$$

But more interesting, it also works well with divergent integrals, allowing us to define the equivalence classes of divergent integrals. Particularly, by successfully applying this transform to $$int_1^infty frac1xdx=int_0^inftyfrac{theta (x-1)}{x}dx$$, one can obtain the following equivalence class of divergent integrals:

$$int_0^inftyfrac{theta (x-1)}{x}dx=int_0^inftyfrac{e^{-x}}{x}dx=int_0^inftyfrac{dx}{x+1}=int_0^inftyfrac{e^x x text{Ei}(-x)+1}{x}dx=int_0^inftyfrac{x-log (x)-1}{(x-1)^2}dx$$

On the other hand, applying the transform to $$int_0^1frac1x dx=int_0^infty frac{theta (1-x)}{x}dx$$ one can obtain another set of equal integrals:

$$int_0^inftyfrac{theta (1-x)}{x}dx=int_0^inftyfrac{1-e^{-x}}{x}dx=int_0^inftyfrac{1}{x^2+x}dx=int_0^infty-e^x text{Ei}(-x)dx=int_0^infty-frac{x+x (-log (x))-1}{(x-1)^2 x}dx$$

Now, having postulated equivalence of divergent integrals in each class, we can pick two integrals, one from each class and compare them. Well, it seems, the integrals in the second class are bigger by an Euler’s constant:

$$int_0^{infty } left(frac{1-e^{-x}}{x}-frac{1}{x+1}right) , dx=gamma$$

And thus, we can conclude that $$int_0^1frac1xdx=gamma+int_1^infty frac1xdx$$. Surprising, is not it, given that one would naively expect $$ln 0=-lninfty$$?

But we also have an identity $$gamma = lim_{ntoinfty}left(sum_{k=1}^n frac1{k}-int_1^nfrac1t dtright)$$ and know the regularized value of harmonic series $$operatorname{reg}sum_{k=1}^n frac1{k}=gamma$$. Thus, $$int_0^1frac1xdx=gamma+int_1^inftyfrac1xdx=sum_{k=1}^inftyfrac1k$$.

Adding up, we obtain $$int_0^infty frac1xdx=gamma+2int_1^inftyfrac1xdx=-gamma+2sum_{k=1}^inftyfrac1k$$. Its regularized value is $$operatorname{reg}int_0^inftyfrac1xdx=gamma$$.

That said, I wonder whether anyone before proposed this regularization of “harmonic integral” to the Euler-Mascheroni constant? Does it contradict any known facts?

multivariable calculus – Interchange of order of integration for \$int_0^infty y^{s-1} e^{-ay} int_0^infty frac{sin(2yx)cos(pi x^2)}{sinh(pi x)}dx dy \$

I have been trying to interchange the order of integration for the integral: $$int_0^infty y^{s-1} e^{-ay} int_0^infty frac{sin(2yx)cos(pi x^2)}{sinh(pi x)}dx dy$$

I am unable to find a proof for the same. Using Mathematica I was able to see that the values are the same even when we interchange the order of integration upto some $$s,a in mathbb{R}$$ with $$a>0$$.

But for larger values of $$s$$, mathematica says it is divergent, so I am interested in the condition on $$s,a$$ such that we can interchange the order of integration.

Any help is highly appreciated.

\$ int_{0}^{infty} x^{(m+n)/2- 1} e^{-x(frac{mz+n}{2})} dx \$ (density of \$F\$ distribution)

I’m deriving the density of the $$F_{n, m}$$ distribution and I ended up with the following integral

$$int_{0}^{infty} x^{(m+n)/2- 1} e^{-x(frac{mz+n}{2})} dx$$

This integral looks like it could be molded into a Gamma function but I don’t see how. I can add the entire context to this question but I don’t think it would help.

numerical integration – Finding closed or numeric value for \$int_0^{infty } frac{x csc (a x)}{x^2+b^2} , dx\$

According to Gradshteyn 3.747-3 $$int_0^{infty } frac{x csc (a x)}{x^2+b^2} , dx=frac{pi}{2sinh(ab)}$$ $$b>0$$

so I’m trying to get some numeric results from Mathematica 12.

``````a = 6; b = 12;

NIntegrate((x/((x^2 + b^2) Sin(a x))), {x, 0, Infinity}, MaxRecursion -> 12)
``````

However this does not match the values given by

``````Pi/(2*Sinh(a*b))
``````

So it is not clear where I’m going wrong.

Compute \$ int_{0}^{+infty} bigg( frac{ln(x)}{e^x}bigg)^2 dx \$

How can I compute this integral?
$$int_{0}^{+infty} bigg( frac{ln(x)}{e^x}bigg)^2 dx$$

prove / disprove the convergence of \$int_0^infty x*sin(x^3), dx\$

I need to check if this integral converges / diverges conditionally. I have tried integrating by parts and it didn’t work.
Will appreciate any help :].

definite integrals – Value of \$int_{0}^{infty} sin^2xcdot f(x) dx\$ and \$int_{0}^{infty} cos^2xcdot f(x) dx\$

I got these value by WolframAlpha:

$$int_{0}^{infty} sin^2xcdot e^{-x} dx=frac{2}{5}$$
,$$int_{0}^{infty} cos^2xcdot e^{-x} dx=frac{3}{5}$$
,$$int_{0}^{infty} sin^2xcdot e^{-x}x dx=frac{14}{25}$$
,$$int_{0}^{infty} cos^2xcdot e^{-x}x dx=frac{11}{25}$$

I tried to generalize and explain these, but I stuck on generalization of their partial integrations.

calculus – Why \$lim_{epsilon to 0} epsilon int_0^infty frac{zdz}{(z-1)^2 + epsilon^2z^3} = pi\$?

I need help evaluating the following limit:
$$lim_{epsilon to 0}2epsilon int_0^infty frac{x^3dx}{(x^2-epsilon^2)^2 + epsilon^2x^6}$$
Making the substitutions $$y = x^2$$ and $$z = y/epsilon$$ I was able to put the integral in a form that seems more tractable:
$$lim_{epsilon to 0}epsilon int_0^infty frac{zdz}{(z-1)^2 + epsilon^2z^3}$$
I have two reasons to think that the limit evaluates to $$pi$$. The first one is that it’s necessery for the result I’m trying to reach in a physics problem, which obviously isn’t a very good justification. The second one ist that I made a numerical calculation that gave me a result very close to $$pi$$: I restricted the integral to the interval $$(0,10)$$, because the integrand has a very sharp peak at $$z=1$$. Then, I divided that interval in $$2 cdot 10^6$$ subintervals and used Simpson’s method to calculate the integral. With $$epsilon = 10^{-4}$$ I got the result $$3.1417$$.

I tryed to calculate it using the residues method, but I couldn’t find the roots of the third degree polynomial in the denominator.

Does anyone have an idea on how to evaluate the limit analytically? Any help is appreciated.

real analysis – Use of Tonelli’s Theorem to calculate \$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)\$

Define $$f: [0,infty[times [0,infty[ to bar{mathbb R}$$, $$f(x,y):= xe^{-x^{2}(1+y^{2})}$$ it is clear that $$f(x,y)geq 0$$ $$lambda-$$a.e. Therefore, Tonelli’s theorem holds. This means:

$$int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (x)) dlambda (y)=int_{[0,infty[} (int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)) dlambda (x)$$

Looking exclusively at $$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)$$, the only thing that comes to mind is “Riemann Integration”, namely:

$$int_{[0,infty[}xe^{-x^{2}(1+y^{2})}dlambda (y)=lim_{nto infty}int_{[0,n]}xe^{-x^{2}(1+y^{2})}dy=lim_{nto infty}frac{1}{-x^2times2y}e^{-x^2y^2}vert_{0}^{n}$$

Which gets me stuck as I have a $$y$$ in the denominator, however, I see no other way of doing it. Any other ideas?