TITLE: How to evaluate $\int_0^1 {\cos(tx)\over \sqrt{1+x^2}}\,\mathrm{d}x$?
QUESTION [9 upvotes]: I would like to calculate the integral
$$\int_0^1 {\cos(tx)\over \sqrt{1+x^2}}\,\mathrm{d}x$$
I have tried to consider the integral as fourier coefficient of $f(x)=\dfrac1{\sqrt{1+x^2}}$, however no ideas. Besides, let $F(t)$ denote above integral as a function of t, after derivating $F(t)$, I could not make any breakthrough. Does anybody have ideas?

REPLY [5 votes]: Let:
$$E(t)=\int\limits_0^1\dfrac{\exp(jtx)}{\sqrt{1+x^2}}dx,\quad j=\sqrt{-1}$$
Derive it $n$ times with respect to $t$:
$$E^{(n)}(t)=j^n\int\limits_0^1\dfrac{\exp(jtx)}{\sqrt{1+x^2}}x^ndx$$
$$E^{(n)}(0)=j^n\int\limits_0^1\dfrac{x^n}{\sqrt{1+x^2}}dx$$
By parts:
$$I_n=\int\limits_0^1\dfrac{x^n}{\sqrt{1+x^2}}dx=x^{n-1}\left.\sqrt{1+x^2}\right|_0^1-(n-1)\int\limits_0^1\dfrac{1+x^2}{\sqrt{1+x^2}}x^{n-2}dx$$
$$nI_n=\sqrt2-(n-1)I_{n-2}$$
$$I_0=\int\limits_0^1\dfrac{dx}{\sqrt{1+x^2}}=\left.\ln|x+\sqrt{1+x^2}|\right|_0^1 = \ln(1+\sqrt2)$$
Maclaurin series:
$$E(t)=\sum\limits_{n=0}^{\infty}\dfrac1{n!}j^nI_nt^n,\quad J(t)=\int\limits_0^1\dfrac{\cos(tx)}{\sqrt{1+x^2}}dx = \operatorname{Re} E(t)$$
Items with $n=2k-1$ have only imaginary part, so we can take in consideration only items with $n=2k$: 
$$\boxed{J(t)=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^k}{k!}J_kt^{2k},\quad J_0=\ln(1+\sqrt2),\quad J_k=\dfrac{\sqrt2}{2k}-\dfrac{2k-1}{2k}J_{k-1}}$$