# Challenging problem: Find $a$ where $int_0^infty frac{cos(ax)ln(1+x^2)}{sqrt{1+x^2}}dx=0$.

Mathematics Asked on January 1, 2022

What is the value of $$ainmathbb{R}$$ that makes the following integral true
$$int_0^infty frac{cos(ax)ln(1+x^2)}{sqrt{1+x^2}}dx=0,?$$

This question was proposed by my friend Khalef Ruhemi and I have no idea how to approach it but all I tried is setting $$x=tantheta$$ and I don’t know how to continue after that. Also I noticed that the integrand is an even function and again I don’t know how to make use of this fact. Any help would be much appreciated.

Solution due to Khalef Ruhemi without using any kind of software:

Define

$$f(p,q)=int_0^inftyfrac{cos(qx)}{(1+x^2)^p}dx,quad p>0, qne0$$

By $$frac{1}{(1+x^2)^p}=frac{1}{Gamma(p)}int_0^infty y^{p-1}e^{-(1+x^2)y}dy$$

We have

$$f(p,q)=frac{1}{Gamma(p)}int_0^infty y^{p-1} e^{-y}underbrace{left(int_0^infty e^{-x^2y}cos(qx) dxright)}_{x^2y=t^2}dy$$

$$=frac{1}{Gamma(p)}int_0^infty y^{p-frac32} e^{-y}left(int_0^infty e^{-t^2}cosleft(frac{qt}{sqrt{y}}right)dtright)dy$$

$$=frac{sqrt{pi}}{2Gamma(p)}int_0^infty y^{p-frac32} e^{-(y+frac{q^2}{4y})}dytag1$$

$$overset{frac{q^2}{4y}=x}{=}frac{sqrt{pi}}{2Gamma(p)}left|frac{q}{2}right|^{2p-1}int_0^infty x^{-p-frac12}e^{-(x+frac{q^2}{4x})}dx$$

$$=frac{Gamma(1-q)}{Gamma(p)}left|frac{q}{2}right|^{2p-1}underbrace{left(frac{sqrt{pi}}{2Gamma(1-p)}int_0^infty x^{-p-frac12}e^{-(x+frac{q^2}{4x})}dxright)}_{=f(1-p,q) text{by} (1)}$$

Thus,

$$f(p,q)=frac{Gamma(1-q)}{Gamma(p)}left|frac{q}{2}right|^{2p-1}f(1-p,q)$$

or,

$$int_0^inftyfrac{cos(qx)}{(1+x^2)^p}dx=frac{Gamma(1-q)}{Gamma(p)}left|frac{q}{2}right|^{2p-1}int_0^inftyfrac{cos(qx)}{(1+x^2)^{1-p}}dx,quad 0

Note that $$0 follows from the fact that $$p>0$$ and $$1-p>0$$.

Next, differentiate both sides of $$(2)$$ with respect to $$p$$ then let $$pto 1/2$$ we have

$$int_0^inftyfrac{cos(qx)ln(1+x^2)}{sqrt{1+x^2}}dx=-ln|2qe^{gamma}|int_0^inftyfrac{cos(qx)}{sqrt{1+x^2}}dx$$

Finally, since the LHS integral is equal to zero, we have

$$ln|2qe^{gamma}|=0Longrightarrow q=pmfrac12e^{-gamma}.$$

Consider $$underbrace{int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^b}:dx}_{x=frac{t}{a}}=a^{2b-1}int _0^{infty }frac{cos left(tright)}{left(a^2+t^2right)^b}:dt$$ Now use the following identity that can be found here. $$K_vleft(zright)=frac{Gamma left(v+frac{1}{2}right)left(2zright)^v}{sqrt{pi }}int _0^{infty :}frac{cos left(tright)}{left(z^2+t^2right)^{v+frac{1}{2}}} dt$$ This leads to $$int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^b}:dx=a^{2b-1}K_{b-frac{1}{2}}left(aright)frac{sqrt{pi }}{Gamma left(bright)left(2aright)^{b-frac{1}{2}}}$$ This means that $$int _0^{infty }frac{cos left(axright)ln left(1+x^2right)}{sqrt{1+x^2}}:dx=-lim _{bto frac{1}{2}}frac{partial }{partial b}a^{2b-1}K_{b-frac{1}{2}}left(aright)frac{sqrt{pi }}{Gamma left(bright)left(2aright)^{b-frac{1}{2}}}$$ Using mathematica to complete the calculations we are left with $$K_0left(aright)left(-ln left(aright)+ln left(2right)+psi left(frac{1}{2}right)right)-K^{left(1,0right)}_0left(aright)$$

Now you can check here that $$K^{left(1,0right)}_0left(aright)=0$$ Proof provided below.

Meaning overall $$=K_0left(aright)left(-ln left(aright)+ln left(2right)-gamma -2ln left(2right)right)$$ $$boxed{int _0^{infty }frac{cos left(axright)ln left(1+x^2right)}{sqrt{1+x^2}}:dx=-K_0left(aright)left(ln left(aright)+gamma +ln left(2right)right)}$$ Which agrees with the results proposed above.

Now answering the main point, $$-K_0left(aright)left(ln left(aright)+gamma +ln left(2right)right)=0$$ $$ln left(2aright)+gamma =0$$ $$2a=e^{-gamma }$$

We find that $$displaystyle a=frac{e^{-gamma}}{2}$$

And so by plugging it in we can immediately see $$int _0^{infty }frac{cos left(frac{e^{-gamma }}{2}xright)ln left(1+x^2right)}{sqrt{1+x^2}}:dx=-K_0left(frac{e^{-gamma }}{2}right)left(-gamma -ln left(2right)+gamma +ln left(2right)right)=0$$

## Proof of tools used.

$$K^{left(1,0right)}_0left(aright)=0$$

$$K_vleft(aright)=int _0^{infty }e^{-acosh left(tright)}cosh left(vtright):dt$$ Differentiating with respect to $$v$$ gives us $$K_v^{left(1,0right)}left(aright)=int _0^{infty }te^{-acosh left(tright)}sinh left(vtright):dt$$ Now let $$v=0$$ $$K_0^{left(1,0right)}left(aright)=int _0^{infty }te^{-acosh left(tright)}sinh left(0right):dt=0$$

$$displaystyle K_vleft(aright)=frac{Gamma left(v+frac{1}{2}right)left(2aright)^v}{sqrt{pi }}int _0^{infty }frac{cos left(tright)}{left(a^2+t^2right)^{v+frac{1}{2}}}:dt=int _0^{infty }e^{-acosh left(tright)}cosh left(vtright):dt$$

First consider $$Ileft(aright)=int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^v}:dx$$

Well make use of the following gamma function representation $$Gamma(v)={left(1+x^{2}right)}^{v}int_{0}^{infty}e^{-left(1+x^{2}right)u} u^{v-1}du$$ Multiply $$Ileft(aright)$$ by $$Gamma(v)$$ $$Gamma(v)I(a)=int_{0}^{infty}cos(ax)int_{0}^{infty}e^{-left(1+x^{2}right)u} u^{v-1}dudx$$ $$=int_{0}^{infty}u^{v-1}e^{-u}int_{0}^{infty}e^{-x^{2}u}cos(ax)dxdu=frac{1}{2}sqrt{{pi}}underbrace{int_{0}^{infty}u^{v-frac{2}{2}}e^{-u-frac{a^{2}}{4u}}du}_{u=left(frac{a}{2}right)e^t}$$ $$=frac{sqrt{pi}}{2}frac{1}{Gamma(v)}{left(frac{a}{2}right)}^{v-frac{1}{2}}int_{-infty}^{infty}e^{-acosh(t)}e^{left(v-frac{1}{2}right)t} dt$$ $$=frac{sqrt{pi}}{Gamma(v)}{left(frac{a}{2}right)}^{v-frac{1}{2}}int_{0}^{infty}e^{-acosh(t)}cosh{left(left(v-frac{1}{2}right)tright)} dt$$ $$frac{Gamma left(vright)}{sqrt{pi }}:left(frac{2}{a}right)^{v-frac{1}{2}}int _0^{infty }frac{cos left(axright)}{left(1+x^2right)^v}:dx=int_{0}^{infty}e^{-acosh(t)}cosh{left(left(v-frac{1}{2}right)tright)} dt$$ $$frac{Gamma left(v+frac{1}{2}right)}{sqrt{pi }}:left(frac{2}{a}right)^vint _0^{infty }frac{cos left(axright)}{left(1+x^2right)^{v+frac{1}{2}}}:dx=int _0^{infty }e^{-acosh left(tright)}cosh left(vtright):dt$$

Answered by Dennis Orton on January 1, 2022

The integral equals $$K_0(a) (gamma+log(2)+log(a)) tag{*}$$ (where $$K_0(a)$$ is a modified Bessel function, assume $$a>0$$, $$a<0$$ follows by symmetry), which can be shown by integration under the integral sign together with $$K_0'(a)=-K_1(a)$$.

Since Bessel K's have no zeros, we can equate the bracket in (*) to zero and get

$$a=pmfrac{e^{-gamma}}{2}approxpm 0.28073,,$$

which is the same as numercis suggests (see comments to question).

Answered by Besselsslave on January 1, 2022

## Related Questions

### $limlimits_{Rto0^+}intlimits_{x^2+y^2le R^2}e^{-x^2}cos(y)dxdy=?$

1  Asked on December 8, 2021

### What is the difference between ${3 choose 2}$ and ${3 choose 1}{2 choose 1}$?

3  Asked on December 8, 2021

### Computing sparse eigenvectors of psd matrices

1  Asked on December 8, 2021

### Express the accumulated amount in the bank as an explicit formula

0  Asked on December 8, 2021 by tay-ivan

### Which of the following sets of boolean functions is functionally complete?

1  Asked on December 8, 2021 by play4u

### Find the probability using the formula for the number of permutations with repetitions

2  Asked on December 8, 2021

### How to decompose a number sequence into convergent subsequences?

1  Asked on December 8, 2021 by user779130

### How to approximate the peak of an epidemic (depending on epidemiological parameters)?

1  Asked on December 8, 2021

### Two conditional expectations equal almost everywhere

1  Asked on December 8, 2021 by yy66yy

### Evaluate $cos a cos 2 a cos 3 a cdots cos 999 a$ where $a=frac{2 pi}{1999}$

1  Asked on December 8, 2021

### Uniformly continuous and differentiable

0  Asked on December 8, 2021

### Is there a thing as a “Negative Dimensional Space”?

0  Asked on December 8, 2021

### Are the rationals minus a point homeomorphic to the rationals?

3  Asked on December 8, 2021 by cheerful-parsnip

### Edited: Let $I=int_{0}^{pi/2}(sin 2x)^{1/3}sin x dx$ $J=int_{0}^{pi/2}(cos 2x)^{1/3}cos x dx$. Find $I/J$

3  Asked on December 8, 2021 by aspirant

### Abelian finite groups and their subgroups

2  Asked on December 8, 2021 by markvs

### Painting the edges of an $n$-gonal prism with 3 colors all the edges of each vertex have different colors if $n= 2018$ and $n= 2019$?

2  Asked on December 8, 2021

### A question for a measurable function $g$ on a finite measure space such that $fgin L^p$ for all $fin L^p$

1  Asked on December 8, 2021 by blancket

### Integral of a logarithmic derivative of a complex polynomial over the real line

1  Asked on December 8, 2021 by user302934

### Relation between spectral radius when the norms are equivalent

1  Asked on December 8, 2021 by math-lover

### Financial Mathematics : Annuity loan with a different first period?

2  Asked on December 8, 2021