Asymptotic expansion of cosine integral

Question

Can anybody help with this problem to find the full asymptotic expansion of $\int_1^\infty \frac{\cos(xt)\, d t}{t}$ from Bender & Orzsag). Does it work by Taylor expansion?

Answer 1

In this case, integration by parts is the method of choice. You can show easily that (for $n>0$)

$$\int_1^\infty\!dt\, \frac{\cos(x t)}{t^n} = -\frac{\sin(x)}{x} + \frac{n}{x} \int_1^\infty\!dt\, \frac{\sin(x t)}{t^{n+1}} .$$

Similarly $$\int_1^\infty\!dt\, \frac{\sin(x t)}{t^n} = \frac{\cos(x)}{x} - \frac{n}{x}\int_1^\infty\!dt\, \frac{\cos(x t)}{t^{n+1}}. $$

Applying these results recursively, we obtain $$\int_1^\infty\!dt\, \frac{\cos(x t)}{t^1} = -\frac{\sin(x)}{x} + \frac{1}{x} \int_1^\infty\!dt\, \frac{\sin(x t)}{t^{2}} =-\frac{\sin(x)}{x}+ \frac{\cos(x)}{x^2} - \frac{2}{x^2}\int_1^\infty\!dt\, \frac{\cos(x t)}{t^{3}} = \dots \\ = \cos(x)\left(\frac{1}{x^2} -\frac{3!}{x^4} + \frac{5!}{x^6} - \dots\right) - \sin(x) \left(\frac{1}{x} -\frac{2!}{x^3} + \frac{4!}{x^5} - \dots\right)$$

Answer 2

Considering $$\int\frac{\cos (xt)}{t}\,dt=\text{Ci}(t x)$$ $$I=\int_1^\infty \frac{\cos (xt)}{t}\,dt=-\text{Ci}( x)$$ provided $x>0$.

Now, looking here, you will find that $$I=\cos(x)\left(\frac{1}{x^2}-\frac{6}{x^4}+\frac{120}{x^6}+O\left(\frac{1}{x^7} \right) \right)-\sin(x)\left(\frac{1}{x}-\frac{2}{x^3}+\frac{24}{x^5}+O\left(\frac{1}{x^7}\right) \right)$$

For illustration purposes, let us use it for $x=100 \pi$. The above expansion would give $$\frac{3-1500 \pi ^2+2500000 \pi ^4}{25000000000 \pi ^6}\approx 0.0000101315025301179$$ while the exact value should be $$0.0000101315025300648$$