# Check proof of $\int_{0}^{\infty}\sin(x\sinh\alpha)\sin(x\cosh\alpha)\ln(x)\frac{dx}{x}=-\alpha\gamma$

The question of the proof of this identity was raised today by user567627, but unfortunately the question was deleted by "community" without providing a reason. My answer (which had already received an upvote 1) and my comment were thus also hidden from the participants of this forum.

Hence I rephrase the question here:

Validate the proof of the following relation

$$\int_{0}^{\infty}\sin(x\sinh\alpha)\sin(x\cosh\alpha)\ln(x)\frac{dx}{x}=-\alpha\gamma\tag{1}$$

and give hints to improve the proof.

Here's my proof:

Using the trigonometric identity $\sin (u) \sin (v)=-\frac{1}{2} (\cos (u+v)-\cos (u-v))$ and expanding the $\cos$ in a power series the integrand becomes

$$-\frac{1}{2} \sum _{k=1}^{\infty } \frac{(-1)^k \left(e^{2 \alpha k}-e^{-2 \alpha k}\right) x^{2 k} \log (x)}{x (2 k)!}$$

Now we do the integration over x from 0 to some $t\gt 0$

$$\int_0^t \frac{x^{2 k} \log (x)}{x} \, dx = \frac{t^{2 k} \log (t)}{2 k}-\frac{t^{2 k}}{4 k^2}$$

The sum is then found by mathematica to be this (horrible) expression

$$-\frac{1}{32} e^{-2 a} \left(e^{4 a} t^2 \, _3F_4\left(1,1,1;\frac{3}{2},2,2,2;-\frac{1}{4} e^{2 a} t^2\right)-t^2 \, _3F_4\left(1,1,1;\frac{3}{2},2,2,2;-\frac{1}{4} e^{-2 a} t^2\right)-8 e^{2 a} \log (t) \text{Ci}\left(e^{-a} t\right)+8 e^{2 a} \log (t) \text{Ci}\left(e^a t\right)+8 e^{2 a} \log (t) \log \left(e^{-a} t\right)-8 e^{2 a} \log (t) \log \left(e^a t\right)\right)$$

In the limit $t\to\infty$ this is equivalent to

$$\frac{e^{2 a} \log \left(\frac{1}{t}\right) \cos \left(e^{-a} t\right)}{4 t^2}-\frac{e^{-2 a} \log \left(\frac{1}{t}\right) \cos \left(e^a t\right)}{4 t^2}+\frac{e^{-a} \log \left(\frac{1}{t}\right) \sin \left(e^a t\right)}{4 t}-\frac{e^a \log \left(\frac{1}{t}\right) \sin \left(e^{-a} t\right)}{4 t}-\frac{\gamma a}{4}+\frac{1}{8} a \log (16)+\frac{1}{4} a \psi ^{(0)}\left(\frac{1}{2}\right)$$

Letting $t\to\infty$ this gives

$$-\frac{\gamma a}{4}+\frac{1}{8} a \log (16)+\frac{1}{4} a \psi ^{(0)}\left(\frac{1}{2}\right)$$

which can be simplified to

$$-\frac{1}{2} (\gamma a)$$

thus proving the relation (1). QED.

Observation (my deleted comment):

It is interesting that the similar integral using $\sin$ and $\cos$ and the cmplementary hperbolic functions in the arguments is independent of $\alpha$: $\int_0^{\infty } \frac{\log (x) \cos (x \sinh (\alpha )) \sin (x \cosh (\alpha ))}{x} \, dx=-\frac{1}{2} (\gamma \pi )$

• Wait so is your question, "Is my proof of this problem valid?" – Frank W. Jun 24 '18 at 20:08
• For those with enough rep, the deleted question is this: math.stackexchange.com/questions/2830121/… – Hans Lundmark Jun 24 '18 at 20:15
• @ Frank W IMHO the question was clearly stated, quote "Validate the proof of the following relation ... and give hints to improve the proof." Lokking forward to seeing your ideas for an answer. – Dr. Wolfgang Hintze Jun 25 '18 at 9:18
• Hello to the downvoter (in the background): you might wish to have a look at the many related questions of almost exactly the same type on the margin which have received plenty of upvotes. So please tell us the special reason for your different vote. – Dr. Wolfgang Hintze Jun 25 '18 at 9:23

## 1 Answer

A nice thing about your proof is of course the fact that it provides not only the value of the definite integral but also the entire antiderivative (even though it is rather complicated). You seem to have an extra factor of $\frac{1}{2}$ in your calculations and in the result, but apart from that plots of the antiderivative and your expression agree, as far as I can tell.

I am not exactly well versed in the theory of hypergeometric functions, however, so I am unable to confirm the more complicated steps of your computation. The best I can do to validate your proof is give an alternative proof to confirm the final result. I had posted this as an answer to the original question as well.

Step 1: A generalised sine integral

First note that for $t \in (0,1)$ we can use the substitution $u = x^{1-t}$ to find $$f(t) \equiv \int \limits_0^\infty \frac{\sin(x)}{x^t} \, \mathrm{d} x = \frac{1}{1-t} \int \limits_0^\infty \sin \left(u^{\frac{1}{1-t}}\right) \, \mathrm{d} u = \frac{\pi}{2 \Gamma(t) \sin \left(\frac{\pi}{2} t\right)} \, ,$$ where the last step follows from the result for the generalised Fresnel integral from this question. Both the left-hand side and the right-hand side are holomorphic functions of $t$ on the strip $$S = \{z \in \mathbb{C}: 0 < \operatorname{Re} (z) < 2\} \, ,$$ so by analytic continuation the holomorphic function $f:S \rightarrow \mathbb{C}$ is given by the above expression on the whole domain $S$ .

Step 2: An easier integral

For $a,b > 0$ we define $g_{a,b}: S \rightarrow \mathbb{C}$ by $$g_{a,b} (t) = \int \limits_0^\infty \frac{\cos(a x) \sin(b x)}{x^t} \, \mathrm{d} x \, .$$ Using a trigonometric identity we obtain \begin{align} g_{a,b} (t) &= \frac{1}{2} \int \limits_0^\infty \frac{\sin[(a+b) x] - \operatorname{sgn}(a - b) \sin[|a-b|x]}{x^t} \, \mathrm{d} x \\ &= \frac{1}{2} \left[(a+b)^{t-1} - \operatorname{sgn}(a - b) |a-b|^{t-1}\right] f(t) \\ &= \frac{\pi}{4 \Gamma(t) \sin \left(\frac{\pi}{2} t\right)} \left[(a+b)^{t-1} - \operatorname{sgn}(a - b) |a-b|^{t-1}\right] \, . \end{align} We define $\operatorname{sgn}(0) |0|^{t-1} = 0$ for any $t \in S$ to correctly treat the case $a = b$. Assuming that we can differentiate under the integral sign, we can use $\Gamma'(1) = - \gamma$ to compute \begin{align} I (a,b) &\equiv \int \limits_0^\infty \frac{\cos(a x) \sin(b x) \ln (x)}{x} \, \mathrm{d} x = - g_{a,b}' (1) \\ &= \frac{\pi}{4} \left[ \ln \left( \frac{a+b}{|a-b|^{\operatorname{sgn}(a-b)}} \right) - (1 - \operatorname{sgn}(a-b)) \gamma \right] \, . \end{align} The convention $0^0 = 1$ yields the correct result for $a=b$. In particular, we recover the result $$I (\sinh(\alpha), \cosh(\alpha)) = \frac{\pi}{4} \left[ \ln \left( \cosh^2 (\alpha) - \sinh^2 (\alpha) \right) - 2 \gamma \right] = - \frac{\pi \gamma}{2}$$ from your observation.

Step 3: The actual integral

Now let us define $h_{a,b}: S \rightarrow \mathbb{C}$ by $$h_{a,b} (t) = \int \limits_0^\infty \frac{\sin(a x) \sin(b x)}{x^t} \, \mathrm{d} x \, .$$ We want $a, b >0$ but also $a \neq b$, since the integral diverges for $a = b$ and $\operatorname{Re} (t) \leq 1$ . For $0 < \operatorname{Re} (t) < 1$ we can use integration by parts to find \begin{align} h_{a,b} (t) &= \frac{t}{a^2 - b^2} \int \limits_0^\infty \frac{b \sin(a x) \cos(b x) - a \cos(a x) \sin(b x)}{x^{t+1}} \, \mathrm{d} x \\ &= \frac{t}{a^2 - b^2} [b g_{b,a} (t+1) - a g_{a,b}(t+1)] \\ &= \frac{\pi t}{4(a^2 - b^2)\Gamma(t+1) \cos\left(\frac{\pi}{2}t \right)} [(a+b) \operatorname{sgn}(a-b) |a-b|^t - (a-b) (a+b)^t] \\ &= \frac{\pi }{4\Gamma(t)} \frac{|a-b|^{t-1} - (a+b)^{t-1}}{\cos\left(\frac{\pi}{2}t \right)} \, . \end{align} We are interested in $h_{a,b}'(1)$. Luckily, the singularity of the right-hand side at $t=1$ is removable and we obtain the power series expansion \begin{align} h_{a,b} (t) &= \frac{\pi }{4\Gamma(t)} \frac{ \ln \left(\frac{|a-b|}{a+b}\right) (t-1) + \frac{1}{2} \ln \left(\frac{|a-b|}{a+b}\right) \ln \left(|a^2-b^2|\right) (t-1)^2 + \mathcal{O} ((t-1)^3)}{- \frac{\pi}{2} (t-1) + \mathcal{O} ((t-1)^3)} \\ &= \frac{1 }{2\Gamma(t)} \left[\ln \left(\frac{a+b}{|a-b|}\right) + \frac{1}{2} \ln \left(\frac{a+b}{|a-b|}\right) \ln \left(|a^2-b^2|\right) (t-1) + \mathcal{O} ((t-1)^2) \right] \, , \end{align} which is valid for any $t \in S$ by analytic continuation. Differentiation under the integral sign now yields \begin{align} J (a,b) &\equiv \int \limits_0^\infty \frac{\sin(a x) \sin(b x) \ln (x)}{x} \, \mathrm{d} x = - h_{a,b}' (1) \\ &= \frac{1}{2} \ln \left(\frac{|a-b|}{a+b}\right) \left[\frac{1}{2} \ln \left(|a^2-b^2|\right) + \gamma\right] \, . \end{align} In particular, $$J(\sinh(\alpha),\cosh(\alpha)) = \frac{1}{2} \ln \left(\frac{\mathrm{e}^{-\alpha}}{\mathrm{e}^\alpha}\right) \left[\frac{1}{2} \ln \left(\cosh^2 (\alpha) - \sinh^2 (\alpha)\right) + \gamma\right] = - \gamma \alpha$$ holds for $\alpha > 0$ (and in fact for any $\alpha \in \mathbb{R}$ by symmetry).

Step 4: Justification for differentiating under the integral sign

For $\operatorname{Re}(t) > 1$ the integrals defining $f(t)$, $g_{a,b}(t)$ and $h_{a,b}(t)$ converge absolutely and the differentiation can be justified using the dominated convergence theorem. For $\operatorname{Re}(t) \leq 1$ the situation is more complicated, as we are dealing with a conditionally convergent (Riemann) integral on an unbounded domain.

However, there is a simple trick, illustrated for the function $f$ here: For any $t \in S$ we can split the defining integral and integrate by parts to obtain $$f(t) = \int \limits_0^1 \frac{\sin(x)}{x^t} \, \mathrm{d} x + \cos(1) - t \int \limits_1^\infty \frac{\cos(x)}{x^{t+1}} \, \mathrm{d} x \, .$$ Note that both remaining integrals are absolutely convergent and this does not change if the integrands are differentiated with respect to $t$. We can now easily find a dominating non-negative function (at least locally in $t$).

Then the dominated convergence theorem implies that $f$ is holomorphic and that we may interchange integration and differentiation. But now, again integrating by parts, we find \begin{align} f'(t) &= \int \limits_0^1 \frac{- \ln(x) \sin(x)}{x^t} \, \mathrm{d} x - \int \limits_1^\infty \frac{\cos(x)}{x^{t+1}} \, \mathrm{d} x + t \int \limits_1^\infty \frac{\ln(x) \cos(x)}{x^{t+1}} \, \mathrm{d} x \\ &= \int \limits_0^1 \frac{- \ln(x) \sin(x)}{x^t} \, \mathrm{d} x - \int \limits_1^\infty \frac{\cos(x)}{x^{t+1}} \, \mathrm{d} x + \int \limits_1^\infty \frac{\frac{\cos(x)}{x} - \ln(x) \sin(x)}{x^{t}} \, \mathrm{d} x \\ &= \int \limits_0^\infty \frac{- \ln(x) \sin(x)}{x^t} \, \mathrm{d} x \, , \end{align} so differentiating under the integral sign is also valid for the original integral.

Similar considerations apply to $g_{a,b}$ and $h_{a,b}$.

• @ ComplexYetTrivial Thank you for the comprehensive derivations. BTW I have noticed that "they" are about to close this question as well - again with no sign of a reason. Your work will then vanish with it. I must admit that by now I am rather angry about this strange policy of closing and downvoting contributions without notice. The more because, as I said before, there are many closely related questions which have deserved and received great popularity, and the corresponding upvotes and answers. – Dr. Wolfgang Hintze Jun 26 '18 at 10:26