Evaluating $\int_{-5}^{\sqrt{x}}(\frac{\cos t}{t^{10}})dt$ 
Evaluate $y=\int_{-5}^{\sqrt{x}}(\frac{\cos t}{t^{10}})dt$


I've tried differentiating both sides of the fraction until the denominator was 1, and then integrating that by parts, but this was marked wrong. I know I can't just integrate by parts right off the bat because differentiating $t^{-b}$ will always end in another varied reciprocal, it'll just make things worse and worse.
How can I solve this?
 A: HINT
Expand $cos$ as a power series. Then, since the power series of the integrand is an infinite polynomial equation, you just integrate each term of the sum and you'll achieve the desired answer.
A: Hint. For some nonnegative integer $n$ consider the integral $$I_n=\int \frac{\cos t\mathrm dt}{t^n},$$ which is your integral for $n=9.$ Now proceed by parts twice to obtain $$I_n=\frac{t^{1-n}}{1-n}\cos t+\frac{t^{2-n}}{(1-n)(2-n)}\sin t+\frac{1}{(1-n)(2-n)}I_{n-2},$$ which is valid provided $n>2,$ where $I_1$ and $I_2$ may be given in terms of $\operatorname{Si},\,\operatorname{Ci}.$
A: The integral you're dealing with is
$$y=\int_{-5}^{\sqrt{x}}\left(\frac{\cos(t)}{t^{10}}\right)dt \tag{1}\label{eq1A}$$
Note $\sqrt{x} \ge 0$ and, for example, with $-5 \lt -\frac{\pi}{4} \lt t \lt 0$, you get $\cos(t) \gt \frac{1}{\sqrt{2}}$. Thus, you have in this range of $t$ that
$$\frac{\cos(t)}{t^{10}} \gt \frac{1}{\sqrt{2}t^{10}} \tag{2}\label{eq2A}$$
As such, you have for any $-\frac{\pi}{4} \lt y \lt 0$ that
$$\begin{equation}\begin{aligned}
\int_{-\frac{\pi}{4}}^{y}\left(\frac{\cos(t)}{t^{10}}\right)dt & \gt \int_{-\frac{\pi}{4}}^{y}\left(\frac{1}{\sqrt{2}t^{10}}\right)dt \\
& = \left. \frac{-1}{9\sqrt{2}t^{9}}\right\rvert_{-\frac{\pi}{4}}^{y} \\
& = \frac{-1}{9\sqrt{2}y^{9}} - \frac{1}{9\sqrt{2}\left(\frac{\pi}{4}\right)^{9}}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
However,
$$\lim_{y \to 0^{-}}\frac{-1}{9\sqrt{2}y^{9}} = \infty \tag{4}\label{eq4A}$$
This show a part of the integral in \eqref{eq1A} goes to $\infty$, so the entire integral is not Riemann integrable. If $\sqrt{x} \gt 0$, then the part of the integral from just above $0$ up to small positive values is also positive with the limit as the lower integration value goes to $0$ also being $\infty$. Thus, as far as I can tell, this means for any $\sqrt{x} \ge 0$ that \eqref{eq1A} is also not even considered to be Lebesgue integrable.
A: First note that this integral is improper because $-5 < 0<\sqrt x.$ Split off a bad piece near 0 by writing $c=\min(5,\sqrt x,\pi/3)$ so $$\int_{-5}^\sqrt x \frac{\cos t}{t^{10}}dt=\int_{-5}^{-c} \frac{\cos t}{t^{10}}dt+\int_{-c}^c \frac{\cos t}{t^{10}}dt+\int_c^\sqrt x \frac{\cos t}{t^{10}}dt$$. Of the three integrals on the right side of the equation above, the first and third can be expressed in terms of the special functions Si and Ci, called sine-integral and cosine-integral but we shall see that there is no need to do so. For now, note that in both the first and third integrals, we have a bounded function integrated over a finite interval, so both the first and third integrals are definite real numbers. Now look at the second of the three inegrals on the  rirght side. That inegral is improper with a discontinuity at 0. By definition, $$\int_{-c}^c \frac{\cos t}{t^{10}}dt=\lim_{u,v \to 0}(\int_{-c}^u \frac{\cos t}{t^{10}}dt+\int_v^c \frac{\cos t}{t^{10}}dt)$$ where $u<0,v>0.$ The function $\frac{\cos t}{t^{10}}$ is even so it is enough to look at the integral $$\int_v^c \frac{\cos t}{t^{10}}dt).$$ (Alternatively, it will turn out that all we have to do is note that $\frac{\cos t}{t^{10}}$ is positive on $[-c,u].)$ On the interval $[v,c], \cos t \ge 1/2 $ so $$\int_v^c \frac{\cos t}{t^{10}}dt \ge \frac{1}{2}\int_v^c \frac{1}{t^{10}}dt$$ $$\lim_{v \to 0}\int_v^c\frac{1}{t^{10}}dt=\lim_{v \to 0}\frac{1}{9}(\frac{1}{v^9}-\frac{1}{c^9})=\infty.$$ Thus $$\int_v^c \frac{\cos t}{t^{10}}dt=\infty.$$ We conclude that 
$$\int_{-5}^\sqrt x \frac{\cos t}{t^{10}}dt=\infty.$$
A: It is possible to find or verify the solution to the indefinite integral 
$I = {\displaystyle\int}\dfrac{\cos\left(t\right)}{t^{10}}\,dt$
by consulting or using the Integral Calculator site.
The integration solution involves several integration by parts of the known form
$\displaystyle {\int}f g' = f g - {\int}f'g$
Integrating by parts and taking
$f = \cos(t); g'=\dfrac{1}{t^{10}} \\ f' = -\sin(t); g=-\dfrac{1}{9t^9},$
we get
$I = -\dfrac{\cos\left(t\right)}{9t^9}-{\displaystyle\int}\dfrac{\sin\left(t\right)}{9t^9}\,dt \quad (1)$
Applying integration by parts to the indefinite integral in (1):
$f = \sin(t); g'= \dfrac{1}{t^9} \\ f' = \cos(t); g = -\dfrac{1}{8t^8},$
we obtain
${\displaystyle\int}\dfrac{\sin\left(t\right)}{t^9}\,dt= -\dfrac{\sin\left(t\right)}{8t^8}-{\displaystyle\int}-\dfrac{\cos\left(t\right)}{8t^8}\,dt$
Applying integration by parts in a similar way seven more times, we arrive to the integral
${\displaystyle\int}\dfrac{\sin\left(t\right)}{t}\,dt,$
which represents a definition of the sine integral $\operatorname{Si}\left(t\right).$
Adding the integration results, calculating and developing, we obtain the following solution to the indefinite integral:
$\displaystyle I = -\frac{t^9 \text{Si}(t)+\left(t^6-6 t^4+120 t^2-5040\right) t \sin (t)+\left(t^8-2 t^6+24 t^4-720 t^2+40320\right) \cos (t)}{362880 t^9} +C$
Applying the given limits or bounds of integration, we obtain the solution to the definite integral (verified with Mathematica):
$\displaystyle \int_{-5}^{\sqrt{x}} \frac{\cos (t)}{t^{10}} \, dt =\\ \displaystyle \left(\frac{1}{362880 x^{9/2}}\right)\left(x^{9/2} \left(-\text{Si}\left(\sqrt{x}\right)\right)-\sqrt{x} \left(x^3-6 x^2+120 x-5040\right) \sin \left(\sqrt{x}\right)\\ \displaystyle -\left(x^4-2 x^3+24 x^2-720 x+40320\right) \cos \left(\sqrt{x}\right)\right) \\ \displaystyle+\frac{-390625 \text{Si}(5)-9835 \sin (5)-79339 \cos (5)}{141750000000}$
It can be verified with Mathematica that the indefinite integral has the following more "compact" solution:
$\displaystyle I = -\frac{E_{10}(-i t)+E_{10}(i t)}{2 t^9}+C,$
where $E_n(z)$ is the exponential integral function.
And the given definite integral has the following solution:
$\displaystyle \int_{-5}^{\sqrt{x}} \frac{\cos (t)}{t^{10}} \, dt =\frac{-E_{10}(-5 i)-E_{10}(5 i)}{3906250}-\frac{E_{10}\left(-i \sqrt{x}\right)+E_{10}\left(i \sqrt{x}\right)}{2 x^{9/2}}$
The graphical representation below shows the real part (in blue) and the imaginary part (in red) of the last solution above to the definite integral (made with Mathematica):

