Evaluate some integrals with hypergeometric function

There are some interesting integral problems: $$\int_{0}^{\infty} \cos\left (\frac{x}{2} \right ) \left ( \cosh(x)-\frac{4x^{3/2}{}_1F_2\left ( 1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4} \right ) }{3\sqrt{\pi} } \right )\text{d}x =\frac{4}{5}.$$ $$\int_{0}^{\infty} e^{-x} \left ( \cosh(x)-\frac{4x^{3/2}{}_1F_2\left ( 1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4} \right ) }{3\sqrt{\pi} } \right )\text{d}x =\frac{3}{4}.$$ Where ${}_1F_2$ is generalized hypergeometric function.
My question is that:

• How to prove these closed expressions?
• Are there any generalizations?

Solution 1:

I present an approach using a well-known property of the Laplace Transform, which is typically a good approach for such integrals.

For the first integral: $$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$ Letting $f\left(x\right) = x \cos \left(\frac{x}{2}\right)\left(\cosh (x) -\frac{4x^{3/2} {}_{1} F_{2} \left(1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4}\right)}{3\sqrt{\pi}}\right)$ and $g(x) = \frac{1}{x}$:

$$\left( \mathcal{L} f \right) (y) = \frac{2\sqrt{2}(9-20y(y-i))}{(2y-i)^{3/2}(5-4y(y-i))^2}-\frac{2\sqrt{2}(-9+20y(y+i))}{(2y+i)^{3/2}(5-4y(y+i))^2}+\frac{1}{(-2y+(2+i))^2}+\frac{1}{(2y-(2-i))^2}+\frac{1}{(2y+(2-i))^2}+\frac{1}{(2y+(2+i))^2}$$ (I am not able to write the derivation of the Laplace transform currently, but I will update my answer with the derivation soon) $$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$

This gives then: $$I = \int_{0}^{\infty} \frac{2\sqrt{2}(9-20y(y-i))}{(2y-i)^{3/2}(5-4y(y-i))^2}-\frac{2\sqrt{2}(-9+20y(y+i))}{(2y+i)^{3/2}(5-4y(y+i))^2}+\frac{1}{(-2y+(2+i))^2}+\frac{1}{(2y-(2-i))^2}+\frac{1}{(2y+(2-i))^2}+\frac{1}{(2y+(2+i))^2}\,dy$$ $$\implies I = \frac{1}{2}\left(\frac{1}{-2y-(2-i)}+\frac{1}{-2y+(2-i)}+\frac{1}{-2y+(2+i)}-\frac{1}{2y+(2+i)}+\frac{4}{\sqrt{y+\frac{i}{2}}(-5+4y(y+i))}+\frac{4}{(-5+4y(y-i))\sqrt{y-\frac{i}{2}}}\right) \Bigg]_{0}^{\infty} =\frac{4}{5}$$

As was to be shown. The same idea works for your second integral.