diff --git a/2bc0c8-2024_spring_TA/Makefile b/2bc0c8-2024_spring_TA/Makefile
new file mode 100644
index 0000000..54ed1fb
--- /dev/null
+++ b/2bc0c8-2024_spring_TA/Makefile
@@ -0,0 +1,13 @@
+main: main_ hw16_17 function_analysis_intro lecture_9
+
+main_: main.typ
+	typst compile --font-path ../fonts/ ./main.typ ../build/2bc0c8-2024_spring_TA_main.pdf
+
+hw16_17: hw16_17.typ
+	typst compile --font-path ../fonts/ ./hw16_17.typ ../build/2bc0c8-2024_spring_TA_hw16_17.pdf
+
+function_analysis_intro: function_analysis_intro.typ
+	typst compile --font-path ../fonts/ ./function_analysis_intro.typ ../build/2bc0c8-2024_spring_TA_function_analysis_intro.pdf
+
+lecture_9: lecture_9.typ
+	typst compile --font-path ../fonts/ ./lecture_9.typ ../build/2bc0c8-2024_spring_TA_lecture_9.pdf
\ No newline at end of file
diff --git a/2bc0c8-2024_spring_TA/hw16_17.typ b/2bc0c8-2024_spring_TA/hw16_17.typ
new file mode 100644
index 0000000..d5b351a
--- /dev/null
+++ b/2bc0c8-2024_spring_TA/hw16_17.typ
@@ -0,0 +1,492 @@
+#set text(
+  font: ("linux libertine", "Source Han Serif SC", "Source Han Serif"),
+  size: 10pt,
+)
+#show math.equation: it => text(size: 10pt, math.display(it))
+#let dcases(..args) = {
+  let dargs = args.pos().map(it => math.display(it))
+  math.cases(..dargs)
+}
+
+#show image: it => align(center, it)
+#show heading.where(level: 3): it => highlight(it)
+
+#let Real = "Re"
+#let Imaginary = "Im"
+
+#set page(
+  header: context [
+    #h(1fr)
+    #if counter(page).get().first() > 1 [
+      *Homework 16 & 17 Ans*
+    ]
+  ],
+  footer: context [
+    #let headings = query(selector(heading.where(level: 2)).before(here()))
+    #if headings.len() > 0 {
+      text(size: 8pt, headings.last().body)
+    }
+    #h(1fr)
+    #counter(page).display(
+      "1/1",
+      both: true,
+    )
+  ],
+)
+
+#align(center)[
+  = Homework 16 & 17 Ans
+  2024 Spring 数学分析 B2
+
+  PB21000030 马天开
+]
+
+== 6.12 Ans
+
+=== 1(3)
+
+$
+  integral_2^(+oo) (dif x) / (x^u ln x)
+$
+
++ $u<=1$, then $1/(x^u ln x)>=1/(x ln x)$, we have:
+  $
+    integral_2^(+oo) (dif x) / (x^u ln x) >= integral_2^(+oo) (dif x) / (x ln x) = lim_(t->+oo) ln ln t = +oo
+  $
+
++ $u>1$, since $ln x$ is strictly increasing, we have:
+
+  $
+    integral_2^(+oo) (dif x) / (x^u ln x) <= integral_2^(+oo) (dif x) / (x^u ln 2) < +oo
+  $
+
+  Or more generally, since $1/(ln x) -> 0 quad (x->+oo)$, and $integral_2^(+oo) (dif x)/(x^u)<+oo$, with Dirchelet (Page 301, Theorem 13.7), we can show that:
+
+  $
+    integral_2^(+oo) (dif x) / (x^u ln x) < +oo
+  $
+
+So the integral converges for $u>1$.
+
+=== 1(5)
+
+$
+  integral_0^(+oo) (sin^2 x) / (x^alpha (1+x)) dif x
+$
+
+First split the integral into two parts:
+
+$
+  I = I_1 + I_2 =integral_0^1 (sin^2 x) / (x^alpha (1+x)) dif x + integral_1^(+oo) (sin^2 x) / (x^alpha (1+x)) dif x
+$
+
+// Note than when $I=integral_0^(+oo)dif x$ converges, a necessary & sufficient condition is that both $I_1=integral_0^1 dif x$ and $I_2=integral_1^(+oo)dif x$ converge. (This differs from splitting into, say $I_1=integral_0^(+oo) f_1 dif x$ and $I_2=integral_0^(+oo) f_2 dif x$, where the in-convergance of $I$ is not guaranteed by the convergance of $I_1$ and $I_2$, or vice versa.)
+
++ We start by discussing $I_1$:
+
+  $
+    (sin^2 x) / (x^alpha (1+x)) tilde x^2 / x^alpha = x^(2-alpha)
+  $
+
+  So integral $I_1 = integral_0^1 (sin^2 x)/(x^alpha (1+x))$ and $integral_0^1 1/(x^(alpha - 2))$ have the same convergance, and the latter converges if $alpha - 2 < 1$, i.e. $alpha < 3$.
+
++ The second part $I_2$ requires some techniques.
+
+  + Let's first show the case for $alpha>0$:
+
+    $
+      integral_1^(+oo) (sin^2 x) / (x^alpha (1+x)) dif x = 1 / 2 integral_1^(+oo) (1-cos x) / (x^alpha (1+x)) = 1 / 2(
+        integral_1^(+oo)1 / (2x^alpha (1+x)) dif x-integral_1^(+oo) (cos 2x) / (2x^alpha (1+x)) dif x
+      ) = 1 / 2 (I_3 + I_4)
+    $
+
+    // This is what I meant by in-convergance of $I_1, I_2$ can't deduce the in-convergance of $I$, luckily in this case we can prove both are under $alpha >=0$
+
+    It's trivial to prove $I_3$ converges with $alpha >0$. For $I_4$, since $integral_1^A cos 2x dif x$ is always bounded, $1/(x^alpha (1+x))$ is strictly decreasing, with Dirchelet, we can show that $I_4$ converges for $alpha >=0$.
+
+  + With $alpha=0$:
+
+    $I_3=integral_1^(+oo)1/(1+x)dif x$ no longer converges, but $I_4$ remains convergent, so $I_2$ diverges for $alpha=0$
+
+  + Then there's the case for $alpha<0$:
+
+    $
+      integral_1^(+oo) (sin^2 x) / (x^alpha (1+x)) dif x >= integral_1^(+oo) (sin^2 x) / (1+x) dif x -> +oo
+    $
+
+    Thus $I_2$ diverges for $alpha<0$.
+
+To sum up, $I$ converges for $0<alpha<3$.
+
+=== 1(6)
+
+$
+  integral_0^(+oo) ln(1+x^2) / x^alpha dif x
+$
+
+The idea is shared:
+
+$
+  I= I_1 + I_2 = integral_0^1 ln(1+x^2) / x^alpha dif x + integral_1^(+oo) ln(1+x^2) / x^alpha dif x
+$
+
++ For $I_1$, as $x->0$:
+
+  $
+    ln(1+x^2) / x^alpha tilde x^2 / x^alpha = x^(2-alpha)
+  $
+
+  So $I_1$ converges if $2-alpha > -1$, i.e. $alpha < 3$.
+
++ For $I_2$, the key is to use *logrithm is smaller than any power function when $x->+oo$*, so for $alpha> 1$
+
+  $
+    0<(ln (1+x^2)) / x^alpha <= x^((alpha-1) / 2) / x^alpha = 1 / x^((alpha+1) / 2)
+  $
+
+  We can show $I_2$ is convergent, for $alpha<=1$, use simple inequality:
+
+  $
+    ln(1+x^2) / x^alpha >= 1 / x^alpha >0
+  $
+
+  Thus $I_2$ diverges for $alpha<=1$. To sum up, $I$ converges for $1<alpha<3$.
+
+=== 2(2)
+
+$
+  integral_0^(+oo) e^(alpha x) sin(beta x) dif x
+$
+
++ When $0<alpha_0<=alpha<+oo$:
+
+  $
+    abs(e^(alpha x) sin(beta x))<=e^(alpha_0 x)
+  $
+
+  With the fact that $integral_1^(+oo)e^(alpha_0 x)dif x$ is convergent and Weierstrass M-test, we can show that $integral_0^(+oo)e^(alpha x) sin(beta x) dif x$ uniformly converges.
+
++ When $0<alpha<+oo$, except for $beta=0$ (which is a trivial case), we can show that $ integral_0^(+oo)e^(alpha x) sin(beta x) dif x $ doesn't uniformly converges (with Q3).
+
+For this question, we can actually calculate the integral:
+
+$
+  integral e^(alpha x) sin(beta x) = integral Imaginary(e^(alpha x) dot e^(i beta x)) = Imaginary(integral e^(alpha + i beta x)) = Imaginary(-1/(alpha + i beta)) = - Imaginary(alpha - i beta) / (alpha^2-beta^2) = beta / (alpha^2 - beta^2)
+$
+
+This would provide the same result as above.
+
+
+=== 2(4)
+
+$
+  integral_1^(+oo) (ln (1+x^2)) / (x^alpha) dif x quad (1< alpha<+oo)
+$
+
+This depends on Q4 as well. As shown in Q1(6), $integral_1^(+oo) (ln (1+x^2)) / (x^alpha)$ converges on $alpha>1$, but not $alpha=1$. So the integral is *not* uniformly convergent.
+
+=== 2(5)
+
+$
+  integral_1^(+oo) e^(-alpha x) (cos x) / x^p dif x quad (0<=alpha<+oo, p>0)
+$
+
+Given that $abs(e^(-alpha x)) <=1 quad forall 1<x<+oo space.quad 0<=alpha <+oo$ is decreasing and strictly bounded, we now proof $integral_1^(+oo) (cos x)/x^p dif x$ is uniformly convergent (with respect to $alpha$), which is to say we just need to prove its convergance. With these two facts, we can show that the integral is uniformly convergent using Abel's Theorem.
+
++ For $p>1$:
+
+  Given $abs(integral_1^A cos x)<2$ and $1/x^p arrow.br 0$, using Dirchelet, we can show that $integral_1^(+oo) (cos x) / x^p dif x$ converges.
+
++ For $0<p<=1$, use integrate by parts:
+
+  $
+    integral_1^(+oo) (cos x) / x^p dif x = [(sin x) / x^p]_1^(+oo) + p integral_1^(+oo) (sin x) / x^(p+1) dif x
+  $
+
+  That turns the discussion from $p in (0,1]$ to $p+1 in (1,2]$, as above.
+
++ Another much faster solution for this:
+
+  Consider splitting $x$ based on $cos x < 0$ or $cos x >0$:
+
+  $
+    integral_1^(+oo) (cos x) / (x^p) = integral_1^(pi / 2) (cos x) / (x^p) + sum_(n=1)^oo integral_(pi / 2 + (
+      n-1
+    ) pi)^(pi / 2 + n pi) (cos x) / (x^p)
+  $
+
+  This is a alternating series, after this step, don't use Mean Value Theorem for integrals, which you have no control of $xi_n$ over $[pi/2+(n-1)pi, pi/2+n pi]$. Instead, for each nearby interval, subtract them directly, then one would have:
+
+  $
+    integral_(pi / 2+(n-1)pi)^(pi / 2+n pi) (cos x) / (x^p) + (cos (x+pi / 2)) / ((x+pi / 2)^p) >0
+  $
+
+So the alternating series have a decreasing absolute value, and the integral converges. Thus for $p>0$, the integral uniformly converges.
+
+=== 2(6)
+
+$
+  integral_0^(+oo) (sin(x^2)) / (1+x^p) dif x quad (0<=p<+oo)
+$
+
+Same as above, we just prove $integral_0^(+oo) sin(x^2) dif x <+oo$, alongside the fact that $1/(1+x^p) arrow.br quad (p>0)$ (also bounded), this would enough to show uniformly convergent with Abel's Theorem.
+
+To save some time, we use the same method as above, to turn the integral to a alternating series. However directly invoking the mothod won't work:
+
+$
+  integral_0^(+oo) sin(x^2) = sum_(n=0)^(+oo) integral_(sqrt(n pi))^(sqrt((n+1) pi)) sin(x^2)
+$
+
+Consider substitute $x^2 -> u$:
+
+$
+  integral_0^(+oo) sin(x^2) dif x &= integral_0^(+oo) (sin x) / (2 sqrt(x)) dif x\
+  &= sum_(n=0)^(+oo) integral_(2 n pi)^(2 (n+1) pi) sin(x) / (2 sqrt(x)) dif x
+$
+
+With the same idea, we subtract the nearby intervals:
+
+$
+  &quad integral_(2 n pi)^(2 (n+1) pi) sin(x) / (2 sqrt(x)) dif x + integral_(2 (n+1) pi)^(2 (
+    n+1
+  ) pi) sin(x) / (2 sqrt(x)) dif x \
+  &=integral_(2 n pi)^(2 (n+1) pi) sin(x) / (2 sqrt(x)) - sin(x) / (2 sqrt(x+pi)) dif x\
+  & > 0
+$
+
+(skipping some details.)
+
+Thus the integral converges, and uniformly converges for $p>0$.
+
+=== 3
+
+If $f(x,u)$ is continous on $a<=x<+oo, alpha<=u<=beta$. For every $u in [alpha, beta)$, integral $integral_a^(+oo) f(x,u) dif x$ converges, but not when $u=beta$, which is to say $integral_a^(+oo) f(x, beta) dif x$. Prove that $integral_a^(+oo)f(x,u) dif x$ mustn't be uniformly convergent on $[alpha, beta)$.
+
+As scary it might seems, this isn't that hard to prove. Using contradiction, let's first say $integral_a^(+oo)f(x,u) dif x$ is uniformly convergent on $[alpha, beta)$, then $forall epsilon > 0, exists X > a, forall A_1,A_2 > X &$
+$
+  & abs(integral_(A_1)^(A_2) f(x,u) dif x)<epsilon / 2 quad forall u in [alpha, beta)
+$
+
+(Cauchy Criterion for uniform convergent)
+
+Given $f(x,u)$ 's continoutiy on $[a,+oo)times[alpha, beta]$, $u->beta^-$, we have:
+
+$
+  forall epsilon' > 0, exists delta_u>0, forall u in (beta - delta_u, beta) quad abs(f(x,u) - f(x,beta))<epsilon'
+$
+
+Take $epsilon' = epsilon/(2 (A_2 - A_1))$, then we have
+
+$
+  abs(integral_(A_1)^A_2 f(x,beta) dif x) &<= abs(integral_(A_1)^(A_2) f(x,u) dif x) + abs(integral_(A_1)^(A_2) f(x,beta) - f(x, u) dif x)\
+  &= epsilon / 2 + epsilon / (2 (A_2 - A_1)) dot (A_2 -A_1) < epsilon
+$
+
+This is contradictory to $integral_a^(+oo) f(x, beta)$'s divergance, thus the original assumption is wrong, and $integral_a^(+oo) f(x,u) dif x$ mustn't be uniformly convergent on $[alpha, beta)$.
+
+== 6.14 Ans
+
+=== 6
+
+Prove that $F(alpha):=integral_0^(+oo) (cos x)/(1+(x + alpha)^2)$ is continous and differentiable on $alpha in [0, +oo)$.
+
+- Continuity:
+
+  $
+    abs((cos x)/(1+(x+alpha)^2)) <= 1 / (1+x^2) quad x,alpha >=0
+  $
+
+  Since $integral_0^(+oo) 1/(1+x^2)$ is convergent, based on Weierstrass, we know that $F(alpha)$ is unifromly convergent on $[0,+oo)$, thus $F(alpha)$ is continous.
+
+- Differentiable:
+
+  $
+    diff / (diff alpha) ((cos x) / (1+(x+alpha)^2)) = cos x dot (- (2 (x+a)) / ((1+(x+alpha)^2)^2))
+  $
+
+  Control the sizes:
+
+  $
+    abs(diff/(diff alpha) ((cos x)/(1+(x+alpha)^2))) <= 2 dot (x+a) / (1+(x+alpha)^2)^2
+  $
+
+  Below is a solution I copied from someone's homework
+
+  $
+    integral_0^(+oo) 2 dot (x+a) / (1+(x+alpha)^2)^2 dif x = 2 integral_a^(+oo) x / (
+      1+x^2
+    )^2 dif x = integral_sqrt(a)^(+oo) (dif u) / (1+u)^2 <integral_0^(+oo) (dif u) / (1+u)^2 <+oo
+  $
+
+  Based on Weierstrass, we know $integral_0^(+oo) (diff)/(diff alpha)((cos x)/(1+(x+alpha)^2)) dif x$ is uniformly convergent, thus $F(alpha)$ is differentiable on $[0,+oo)$.
+
+  Note: *THIS IS WRONG*. To use Weierstrass, one would have to strictly construct a $abs(f(x,u))<g(x), forall u$, otherwise it's still pointwise than uniform.
+
+  Quick way to do fix this is by comparing $(x+a)/((1+(x+a)^2)^2)$ with $x/(1+x^2)^2$, note that $x/(1+x^2)^2$ is strictly decreasing at $x>1$, so the correct way to do so is just:
+
+  $
+    integral_1^(+oo) 2 dot (x+a) / (1+(x+alpha)^2)^2 dif x < 2 dot integral_1^(+oo) x / (1+x^2)^2 dif x
+  $
+
+  The range for $alpha$ isn't changed here, only the starting point of the integral, unifromly convergance holds. For proof, recall: uniformly convergent $<=>$
+  $
+    lim_(A->+oo) sup_(u in [alpha, beta]) abs(integral_A^(+oo) f(x,u) dif x) = 0
+  $
+
+  Or if one could figure it out, use inequality like this:
+  $
+    2 dot (x+a) / (1+(x+alpha)^2)^2 <= 2 dot (x+a) / (1+(x+alpha)^4) <= 3 / (1+(x+alpha)^3) <= 3 / (1+x^3)
+  $
+
+  The following inequality is used here:
+  $
+    3+3(x+alpha)^4>=2(x+alpha)^4+4(x+alpha)>=2(x+alpha^4)+2(x+alpha)=2(x+alpha)(1+(x+alpha)^3)
+  $
+
+=== 7(2)
+
+$
+  integral_0^(+oo) (1-e^(-a x)) / (x e^x) dif x
+$
+
+
+$
+  integral_0^(+oo) (1-e^(-a x)) / (x e^x) dif x &= integral_0^(+oo) (e^(-1 dot x) - e^((-a -1) x)) / (x)dif x\
+  &=integral_0^(+oo) dif x integral_(-a-1)^(-1) e^(k x) dif k
+$
+
+$
+  &phi(k) = integral_0^(+oo) e^(k x) dif x = -1 / k quad k <= k_0 < 0 \
+$
+
+Consider that: $e^(k x) <= e^(k_0 x)$, also $integral_0^(+oo) e^(k_0 x) <+oo$. $phi(k)$ is uniformly convergent on $[-a-1, -1]$, thus:
+
+$
+  integral_0^(+oo) (1-e^(-a x)) / (x e^x) dif x &= integral_0^(+oo) dif x integral_(-a-1)^(-1) e^(k x) dif k\
+  &= integral_(-a-1)^(-1) phi(k) dif k\
+  &= [-ln (-k)]_(k=-a-1)^(k=-1) = ln(a+1)
+$
+
+=== 7(5)
+
+$
+  integral_0^(+oo) (arctan a x) / (x (1+x^2)) dif x
+$
+
+$
+  (arctan a x) / (x (1+x^2)) &= 1 / (x (1+x^2)) arctan x u |_(u = 0)^(u=a) \
+  &=integral_0^a 1 / ((1+u^2 x^2)(1+x^2)) dif u
+$
+
+$
+  phi(u) = integral_0^(+oo) 1 / ((1+u^2 x^2)(1+x^2)) dif x
+$
+
+Consider $1 / ((1+u^2 x^2)(1+x^2)) <= 1/(1+x^2)$. With Weierstrass, $phi(u)$ is uniformly convergent on $[0,a]$, thus:
+
+$
+  integral_0^(+oo) dif x integral_0^a 1 / ((1+u^2 x^2)(1+x^2)) dif u = integral_0^(+a) dif u integral_0^(+oo) 1 / ((
+    1+u^2 x^2
+  )(1+x^2)) dif x
+$
+
+$
+  integral_0^(+oo) 1 / ((1+u^2 x^2)(
+    1+x^2
+  )) dif x &= - u^2 / (1-u^2) integral_0^(+oo) 1 / (1+u^2x^2) dif x + 1 / (1-u^2) integral_0^(+oo) 1 / (1+x^2) dif x\
+  &= - u / (1-u^2) arctan u x|_0^(+oo) + 1 / (1-u^2) arctan x|_0^(+oo) \
+  &= pi / 2(-u / (1-u^2) + 1 / (1-u^2))\
+  &= pi / 2 1 / (1+u)
+$
+
+So the original intergral becomes:
+
+$
+  integral_0^a 1 / (1+u) dif u = pi / 2 ln(a+1)
+$
+
+=== 7(6)
+
+$
+  integral_0^(+oo) [e^(-(a^2\/x^2))- e^(-(b^2\/ x^2))] dif x
+$
+
+Wihtout substituting variables, after a few tries, you can see it's impossible to make the form.
+$
+  &(diff) / (diff u) e^(-u^2\/x^2) = -(2u) / x^2 e^(-u^2\/x^2)\
+  &(diff) / (diff u) (1 / u e^(-u^2\/x^2)) = (-(2u) \/ x^2 dot e^(-u^2\/x^2)dot u-e^(-u^2\/x^2)) / (u^2) \
+  &(diff) / (diff u) (u e^(-u^2\/x^2)) = e^(-u^2\/x^2) - 2u^2 / x^2 e^(-u^2\/x^2)
+$
+
+Consider substitute $1\/x -> y$:
+
+$
+  &integral_(oo)^0 [e^(-a^2 y^2)-e^(-b^2 y^2)] (dif y) / (y^2)\
+  &=integral_0^(+oo) [e^(-a^2 y^2) - e^(-b^2 y^2)] / (-y^2) dif y\
+$
+
+$
+  (diff) / (diff v) e^(-v dot y^2) = -y^2 dot e^(-v dot y^2)
+$
+
+$
+  [e^(-a^2 y^2) - e^(-b^2 y^2)] / (-y^2) = integral_b^a e^(-v y^2) dif v
+$
+
+$
+  phi(v) = integral_0^(+oo) e^(-v y^2) dif y = sqrt(pi) / (2 sqrt(v))
+$
+
+$
+  e^(-v y^2) <= e^(-b y^2) quad forall v in [a,b]
+$
+
+$
+  integral_0^(+oo) [e^(-a^2 y^2) - e^(-b^2 y^2)] / (-y^2) dif y = integral_(a^2)^(b^2) phi(v) dif v = sqrt(pi) dot (b-a)
+$
+
+=== 8(1)
+
+$
+  &quad integral_(-oo)^(+oo) x / (sigma sqrt(2 pi)) exp(-1/2 ((x-a)/sigma)^2) dif x \
+  &=integral_(-oo)^(+oo) (x+a) / (sigma sqrt(2 pi)) exp(-1/2 ((x)/sigma)^2) dif x\
+  &=integral_(-oo)^(+oo) x / (sigma sqrt(2 pi)) exp(-1/2 ((x)/sigma)^2) dif x + integral_(-oo)^(+oo) a / (sigma sqrt(2 pi)) exp(-1/2 ((x)/sigma)^2) dif x\
+  &= 0 + a / (sigma sqrt(2pi)) integral_(-oo)^(+oo) exp(-1/2 ((x)/sigma)^2) dif x\
+  &= a
+$
+
+This is the expectation of the normal distribution, which is its mean.
+
+=== 8(3)
+
+
+$
+  integral_0^(+oo) (sin a x cos b x) / x dif x &= integral_0^(+oo) (sin (a+b)x + sin(a-b)x) / (2x) dif x\
+  &=dcases(
+  1/2(pi/2 + pi/2) &= pi/2 quad & a!=b,
+  1/2(pi/2 + 0) &= pi/4 quad &a=b,
+)
+$
+
+*Note:*
+
+$
+  integral_0^(+oo) (sin A x) / x dif x = dcases(
+  integral_0^(+oo) (sin A x)/(A x) dif (A x) = pi/2 quad& A!=0,
+  0 quad& A = 0
+)
+$
+
+
+=== 8(6)
+
+$
+  integral_0^(+oo) (sin^4 x) / (x^2) dif x &= integral_0^(+oo) 1 / x^2 (sin^2 x-1 / 4sin^2 2x) dif x\
+  &=integral_0^(+oo) ((sin^2 x) / x^2 - (sin^2 2x) / (2x)^2) dif x\
+  &=integral_0^(+oo) (sin^2 x) / x^2 dif x - 1 / 2 integral_0^(+oo) (sin^2 2x) / (2x)^2 dif (2x)\
+  &=1 / 2integral_0^(+oo) (sin^2 x) / x^2 dif x\
+  &=1 / 2integral_o^(+oo) sin^2 x dif (-1 / x)\
+  &=1 / 2[-(sin^2 x) / x |_0^(+oo) - integral_0^(+oo)2 / x sin x cos x dif x]\
+  &=1 / 2 integral_0^(+oo) (sin 2x) / (2x) dif (2x)\
+  &=pi / 4
+$
diff --git a/compile.sh b/compile.sh
index 3bfc288..10991cb 100755
--- a/compile.sh
+++ b/compile.sh
@@ -17,6 +17,12 @@ for folder in $(ls -d */); do
     if [ "$folder" == "build" ]; then
         continue
     fi
-    echo "Compiling $folder"
-    typst compile --font-path fonts/ ./$folder/main.typ build/${folder}.pdf
+    # if folder contains Makefile, run make
+    if [ -f "$folder/Makefile" ]; then
+        echo "Running make in $folder"
+        make -C $folder
+    else
+        echo "Compiling $folder"
+        typst compile --font-path fonts/ ./$folder/main.typ build/${folder}.pdf
+    fi
 done
\ No newline at end of file