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2023MIT積分競賽決賽(前4題)個(gè)人思路

2023-03-09 08:38 作者:幻想元旦  | 我要投稿

2023麻省理工積分大賽決賽!你能做出幾道題?

第一題:

%5Cint_0%5E%7B%5Cpi%2F2%7D%7B%5Cfrac%7B%5Csqrt%5B3%5D%7B%5Ctan%20x%7D%7D%7B%5Cleft(%20%5Csin%20x%2B%5Ccos%20x%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D

解答:

%5Cleft(%20%5Csin%20x%2B%5Ccos%20x%20%5Cright)%20%5E2%3D%5Cfrac%7B%5Ccos%20%5E2x%2B2%5Csin%20x%5Ccos%20x%2B%5Csin%20%5E2x%7D%7B%5Ccos%20%5E2x%2B%5Csin%20%5E2x%7D%3D%5Cfrac%7B1%2B2%5Ctan%20x%2B%5Ctan%20%5E2x%7D%7B1%2B%5Ctan%20%5E2x%7D

%5Cint_0%5E%7B%5Cpi%20%2F2%7D%7B%5Cfrac%7B%5Csqrt%5B3%5D%7B%5Ctan%20x%7D%7D%7B%5Cleft(%20%5Csin%20x%2B%5Ccos%20x%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%3D%5Cint%5E%7B%5Cpi%2F2%7D_%7B0%7D%7B%5Cfrac%7B%5Csqrt%5B3%5D%7B%5Ctan%20x%7D%5Csec%20%5E2x%7D%7B1%2B2%5Ctan%20x%2B%5Ctan%20%5E2x%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D

換元:t%3D%5Ctan%20x

%5Cint_0%5E%7B%5Cpi%20%2F2%7D%7B%5Cfrac%7B%5Csqrt%5B3%5D%7B%5Ctan%20x%7D%5Csec%20%5E2x%7D%7B1%2B2%5Ctan%20x%2B%5Ctan%20%5E2x%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Csqrt%5B3%5D%7Bt%7D%7D%7B%5Cleft(%201%2Bt%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dt%7D

再換元:u%3D%5Csqrt%5B3%5D%7Bt%7D

%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Csqrt%5B3%5D%7Bt%7D%7D%7B%5Cleft(%201%2Bt%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dt%7D%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B3u%5E3%7D%7B%5Cleft(%201%2Bu%5E3%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Du%7D

直接拆開(算了一個(gè)多小時(shí)...)

%5Cfrac%7B3u%5E3%7D%7B%5Cleft(%201%2Bu%5E3%20%5Cright)%20%5E2%7D%3D%5Cfrac%7B1%7D%7B3%5Cleft(%201%2Bu%20%5Cright)%7D-%5Cfrac%7B1%7D%7B3%5Cleft(%201%2Bu%20%5Cright)%20%5E2%7D-%5Cfrac%7Bu-3%7D%7B3%5Cleft(%20u%5E2-u%2B1%20%5Cright)%7D%2B%5Cfrac%7Bu-1%7D%7B%5Cleft(%20u%5E2-u%2B1%20%5Cright)%20%5E2%7D

然后硬算(

%5Cbegin%7Baligned%7D%0A%20%09%5Cint_0%5E%7B%5Cinfty%7D%5Cfrac%7B3u%5E3%7D%7B%5Cleft(%201%2Bu%5E3%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Du%26%3D%5Cfrac%7B1%7D%7B3%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Du%7D%7B1%2Bu%7D%7D-%5Cfrac%7B1%7D%7B3%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Du%7D%7B%5Cleft(%201%2Bu%20%5Cright)%5E2%7D%7D%5C%5C%0A%20%09%26-%5Cfrac%7B1%7D%7B3%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bu-3%7D%7Bu%5E2-u%2B1%7D%5C%2C%5Cmathrm%7Bd%7Du%7D%2B%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bu-1%7D%7B%5Cleft(%20u%5E2-u%2B1%20%5Cright)%20%5E2%7D%7D%5C%2C%5Cmathrm%7Bd%7Du%5C%5C%26%0A%20%09%3D%5Cfrac%7B1%7D%7B3%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Du%7D%7B1%2Bu%7D%7D-%5Cfrac%7B1%7D%7B3%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Du%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%7D%7D%5C%5C%26-%5Cfrac%7B1%7D%7B6%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B2u-1%7D%7Bu%5E2-u%2B1%7D%5C%2C%5Cmathrm%7Bd%7Du%7D%2B%5Cfrac%7B5%7D%7B6%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Du%7D%7Bu%5E2-u%2B1%7D%7D%5C%5C%26%2B%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B2u-1%7D%7B%5Cleft(%20u%5E2-u%2B1%20%5Cright)%20%5E2%7D%5C%2C%5Cmathrm%7Bd%7Du%7D-%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Du%7D%7B%5Cleft(%20u%5E2-u%2B1%20%5Cright)%20%5E2%7D%7D%5C%5C%26%0A%20%09%3D%5Cfrac%7B1%7D%7B3%7D%5Cleft%5B%20%5Cln%20%5Cleft%7C%20%5Cfrac%7B1%2Bu%7D%7B%5Csqrt%7Bu%5E2-u%2B1%7D%7D%20%5Cright%7C%2B%5Cfrac%7B1%7D%7B1%2Bu%7D%2B%5Csqrt%7B3%7D%5Carctan%20%5Cleft(%20%5Cfrac%7B2u-1%7D%7B%5Csqrt%7B3%7D%7D%20%5Cright)%20-%5Cfrac%7Bu%2B1%7D%7Bu%5E2-u%2B1%7D%20%5Cright%5D%20_%7B0%7D%5E%7B%5Cinfty%7D%5C%5C%26%0A%20%09%3D%5Cfrac%7B2%5Csqrt%7B3%7D%7D%7B9%7D%5Cpi%0A%20%09%5Cend%7Baligned%7D

即,

%5Cint_%7B0%7D%5E%7B%5Cpi%2F2%7D%5Cfrac%7B%5Csqrt%5B3%5D%7B%5Ctan%20x%7D%7D%7B%5Cleft(%5Csin%20x%20%2B%20%5Ccos%20x%5Cright)%5E2%7D%5C%2C%5Cmathrm%7Bd%7Dx%20%3D%20%5Cfrac%7B2%5Csqrt%7B3%7D%7D%7B9%7D%5Cpi


第二題:

%5Cint_0%5E%7B%5Cpi%7D%7B%5Cleft(%20%5Cfrac%7B%5Csin%20%5Cleft(%202x%20%5Cright)%20%5Csin%20%5Cleft(%203x%20%5Cright)%20%5Csin%20%5Cleft(%205x%20%5Cright)%20%5Csin%20%5Cleft(%2030x%20%5Cright)%7D%7B%5Csin%20%5Cleft(%20x%20%5Cright)%20%5Csin%20%5Cleft(%206x%20%5Cright)%20%5Csin%20%5Cleft(%2010x%20%5Cright)%20%5Csin%20%5Cleft(%2015x%20%5Cright)%7D%20%5Cright)%20%5E2%5Cmathrm%7Bd%7Dx%7D

解答:

%5Cbegin%7Baligned%7D%0A%26%5Cleft(%5Cfrac%7B%5Csin%20%5Cleft(%202x%20%5Cright)%20%5Csin%20%5Cleft(%203x%20%5Cright)%20%5Csin%20%5Cleft(%205x%20%5Cright)%20%5Csin%20%5Cleft(%2030x%20%5Cright)%7D%7B%5Csin%20%5Cleft(%20x%20%5Cright)%20%5Csin%20%5Cleft(%206x%20%5Cright)%20%5Csin%20%5Cleft(%2010x%20%5Cright)%20%5Csin%20%5Cleft(%2015x%20%5Cright)%7D%5Cright)%5E2%5C%5C%3D%26%5Cleft(%5Cfrac%7B%5Ccos%20%5Cleft(%20x%20%5Cright)%20%5Ccos%20%5Cleft(%2015x%20%5Cright)%7D%7B%5Ccos%20%5Cleft(%203x%20%5Cright)%20%5Ccos%20%5Cleft(%205x%20%5Cright)%7D%5Cright)%5E2%5C%5C%0A%3D%26%5Cleft(%20%5Cfrac%7B4%5Ccos%20%5E2%5Cleft(%205x%20%5Cright)%20-3%7D%7B4%5Ccos%20%5E2%5Cleft(%20x%20%5Cright)%20-3%7D%20%5Cright)%20%5E2%5C%5C%3D%26%5Cleft(%5Cfrac%7B2%5Ccos%5Cleft(10x%5Cright)-1%7D%7B2%5Ccos%5Cleft(2x%5Cright)-1%7D%5Cright)%5E2%0A%5Cend%7Baligned%7D

用復(fù)數(shù)換元:%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7D%5Ctheta%7D%3D%5Ccos%20%5Ctheta%20%2B%20%5Cmathrm%7Bi%7D%5C%2C%5Csin%20%5Ctheta

z%3D%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dx%7D

%5Ccos%20%5Cleft(kx%5Cright)%3D%20%5Cfrac%7B%5Cmathrm%7Be%7D%5E%7B%5Cmathrm%7Bi%7Dkx%7D%2B%5Cmathrm%7Be%7D%5E%7B-%5Cmathrm%7Bi%7Dkx%7D%7D%7B2%7D%3D%5Cfrac%7Bz%5Ek%2Bz%5E%7B-k%7D%7D%7B2%7D

%5Cbegin%7Baligned%7D%0A%09%5Cleft(%20%5Cfrac%7B2%5Ccos%20%5Cleft(%2010x%20%5Cright)%20-1%7D%7B2%5Ccos%20%5Cleft(%202x%20%5Cright)%20-1%7D%20%5Cright)%20%5E2%26%3D%5Cleft(%20%5Cfrac%7Bz%5E%7B10%7D%2Bz%5E%7B-10%7D%2B1%7D%7Bz%5E2%2Bz%5E%7B-2%7D%2B1%7D%20%5Cright)%20%5E2%5C%5C%26%0A%09%3D%5Cleft(%20%5Cfrac%7Bz%5E%7B20%7D%2Bz%5E%7B10%7D%2B1%7D%7Bz%5E8%5Cleft(%20z%5E4%2Bz%5E2%2B1%20%5Cright)%7D%20%5Cright)%20%5E2%5C%5C%26%0A%09%3D%5Cleft(%20z%5E8%2Bz%5E%7B-8%7D%2Bz%5E6%2Bz%5E%7B-6%7D-z%5E2-z%5E%7B-2%7D-1%20%5Cright)%20%5E2%5C%5C%26%0A%09%3D%5Cleft(%20z%5E%7B16%7D%2Bz%5E%7B-16%7D%20%5Cright)%20%2B2%5Cleft(%20z%5E%7B14%7D%2Bz%5E%7B-14%7D%20%5Cright)%20%2B%5Cleft(%20z%5E%7B12%7D%2Bz%5E%7B-12%7D%20%5Cright)%5C%5C%26%0A%09-2%5Cleft(%20z%5E%7B10%7D%2Bz%5E%7B-10%7D%20%5Cright)-4%5Cleft(%20z%5E8%2Bz%5E%7B-8%7D%20%5Cright)%20-4%5Cleft(%20z%5E6%2Bz%5E%7B-6%7D%20%5Cright)%5C%5C%26%0A%09-%5Cleft(%20z%5E4%2Bz%5E%7B-4%7D%20%5Cright)%20%2B4%5Cleft(%20z%5E2%2Bz%5E%7B-2%7D%20%5Cright)%20%2B7%5C%5C%26%0A%09%3D2%5Ccos%20%5Cleft(%2016x%20%5Cright)%20%2B4%5Ccos%20%5Cleft(%2014x%20%5Cright)%20%2B2%5Ccos%20%5Cleft(%2012x%20%5Cright)%5C%5C%26%0A%09-4%5Ccos%20%5Cleft(%2010x%20%5Cright)-8%5Ccos%20%5Cleft(%208x%20%5Cright)-8%5Ccos%20%5Cleft(%206x%20%5Cright)%5C%5C%26%0A%09-2%5Ccos%20%5Cleft(%204x%20%5Cright)%20%2B8%5Ccos%20%5Cleft(%202x%20%5Cright)%20%2B7.%0A%09%5Cend%7Baligned%7D

前面那一大堆余弦的積分都是0,所以

%5Cint_0%5E%7B%5Cpi%7D%7B%5Cleft(%20%5Cfrac%7B%5Csin%20%5Cleft(%202x%20%5Cright)%20%5Csin%20%5Cleft(%203x%20%5Cright)%20%5Csin%20%5Cleft(%205x%20%5Cright)%20%5Csin%20%5Cleft(%2030x%20%5Cright)%7D%7B%5Csin%20%5Cleft(%20x%20%5Cright)%20%5Csin%20%5Cleft(%206x%20%5Cright)%20%5Csin%20%5Cleft(%2010x%20%5Cright)%20%5Csin%20%5Cleft(%2015x%20%5Cright)%7D%20%5Cright)%20%5E2%5Cmathrm%7Bd%7Dx%7D%3D%5Cint%5E%7B%5Cpi%7D_%7B0%7D7%5C%2C%5Cmathrm%7Bd%7Dx%3D7%5Cpi


第三題:

%5Cint_%7B-1%2F2%7D%5E%7B1%2F2%7D%5Csqrt%7Bx%5E2%2B1%2B%5Csqrt%7Bx%5E4%2Bx%5E2%2B1%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx

解答:

顯然是偶函數(shù)在對稱區(qū)間積分,所以

%5Cint_%7B-1%2F2%7D%5E%7B1%2F2%7D%5Csqrt%7Bx%5E2%2B1%2B%5Csqrt%7Bx%5E4%2Bx%5E2%2B1%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx%3D2%5Cint_%7B0%7D%5E%7B1%2F2%7D%5Csqrt%7Bx%5E2%2B1%2B%5Csqrt%7Bx%5E4%2Bx%5E2%2B1%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx

處理根號里面的根號:

x%5E4%2Bx%5E2%2B1%3D%5Cleft(%20x%5E2%2B1%20%5Cright)%20%5E2-x%5E2%3D%5Cleft(%20x%5E2%2Bx%2B1%20%5Cright)%20%5Cleft(%20x%5E2-x%2B1%20%5Cright)

觀察這兩部分,可以嘗試在外層根號內(nèi)湊完全平方式%5Cbegin%7Baligned%7D%0A%09%26%5Csqrt%7Bx%5E2%2B1%2B%5Csqrt%7Bx%5E4%2Bx%5E2%2B1%7D%7D%5C%5C%0A%09%3D%26%5Csqrt%7B%5Cfrac%7Bx%5E2%2B1%2B2%5Csqrt%7B%5Cleft(%20x%5E2%2Bx%2B1%20%5Cright)%20%5Cleft(%20x%5E2-x%2B1%20%5Cright)%7D%2Bx%5E2%2B1%7D%7B2%7D%7D%5C%5C%0A%09%3D%26%5Csqrt%7B%5Cfrac%7B%5Cleft(%20x%5E2%2Bx%2B1%20%5Cright)%20%2B2%5Csqrt%7B%5Cleft(%20x%5E2%2Bx%2B1%20%5Cright)%20%5Cleft(%20x%5E2-x%2B1%20%5Cright)%7D%2B%5Cleft(%20x%5E2-x%2B1%20%5Cright)%7D%7B2%7D%7D%5C%5C%0A%09%3D%26%5Csqrt%7B%5Cfrac%7B%5Cleft(%20%5Csqrt%7Bx%5E2%2Bx%2B1%7D%2B%5Csqrt%7Bx%5E2-x%2B1%7D%20%5Cright)%20%5E2%7D%7B2%7D%7D%5C%5C%0A%09%3D%26%5Cfrac%7B%5Csqrt%7Bx%5E2%2Bx%2B1%7D%2B%5Csqrt%7Bx%5E2-x%2B1%7D%7D%7B%5Csqrt%7B2%7D%7D%2C%0A%09%5Cend%7Baligned%7D

原積分就可以化簡為這個(gè)

%5Cbegin%7Baligned%7D%0A%09%26%5Cint_%7B-1%2F2%7D%5E%7B1%2F2%7D%7B%5Csqrt%7Bx%5E2%2B1%2B%5Csqrt%7Bx%5E4%2Bx%5E2%2B1%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%3D%26%5Csqrt%7B2%7D%5Cint_0%5E%7B1%2F2%7D%7B%5Csqrt%7Bx%5E2%2Bx%2B1%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%2B%5Csqrt%7B2%7D%5Cint_0%5E%7B1%2F2%7D%7B%5Csqrt%7Bx%5E2-x%2B1%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%3D%26%5Csqrt%7B2%7D%5Cint_0%5E%7B1%2F2%7D%7B%5Csqrt%7B%5Cleft(%20x%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%2B%5Cfrac%7B3%7D%7B4%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%2B%5Csqrt%7B2%7D%5Cint_0%5E%7B1%2F2%7D%7B%5Csqrt%7B%5Cleft(%20x-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%2B%5Cfrac%7B3%7D%7B4%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%3D%26%5Cfrac%7B3%7D%7B2%5Csqrt%7B2%7D%7D%5Cint_%7B%5Cpi%2F6%7D%5E%7B%5Carctan%20%5Cleft(%202%2F%5Csqrt%7B3%7D%20%5Cright)%7D%7B%5Csec%20%5E3%5Calpha%5C%2C%20%5Cmathrm%7Bd%7D%5Calpha%7D%2B%5Cfrac%7B3%7D%7B2%5Csqrt%7B2%7D%7D%5Cint_%7B-%5Cpi%2F6%7D%5E0%7B%5Csec%20%5E3%5Cbeta%20%5C%2C%5Cmathrm%7Bd%7D%5Cbeta%7D.%09%0A%09%5Cend%7Baligned%7D

用分部積分法,可以得到

%5Cint_a%5Eb%7B%5Csec%20%5E3%5Ctheta%20%5C%2C%5Cmathrm%7Bd%7D%5Ctheta%7D%3D%5Cleft%5B%20%5Cfrac%7B1%7D%7B2%7D%5Cln%20%5Cleft%7C%20%5Csec%20%5Ctheta%20%2B%5Ctan%20%5Ctheta%20%5Cright%7C%2B%5Cfrac%7B1%7D%7B2%7D%5Ctan%20%5Ctheta%20%5Csec%20%5Ctheta%20%5Cright%5D%20_%7Ba%7D%5E%7Bb%7D

所以

%5Cbegin%7Baligned%7D%0A%09%26%5Cint_%7B-1%2F2%7D%5E%7B1%2F2%7D%7B%5Csqrt%7Bx%5E2%2B1%2B%5Csqrt%7Bx%5E4%2Bx%5E2%2B1%7D%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%3D%26%5Cfrac%7B3%7D%7B2%5Csqrt%7B2%7D%7D%5Cleft%5B%20%5Cfrac%7B1%7D%7B2%7D%5Cln%20%5Cleft%7C%20%5Csec%20%5Calpha%20%2B%5Ctan%20%5Calpha%20%5Cright%7C%2B%5Cfrac%7B1%7D%7B2%7D%5Ctan%20%5Calpha%20%5Csec%20%5Calpha%20%5Cright%5D%20_%7B%5Cpi%20%2F6%7D%5E%7B%5Carctan%20%5Cleft(%202%2F%5Csqrt%7B3%7D%20%5Cright)%7D%5C%5C%0A%09%2B%26%5Cfrac%7B3%7D%7B2%5Csqrt%7B2%7D%7D%5Cleft%5B%20%5Cfrac%7B1%7D%7B2%7D%5Cln%20%5Cleft%7C%20%5Csec%20%5Cbeta%20%2B%5Ctan%20%5Cbeta%20%5Cright%7C%2B%5Cfrac%7B1%7D%7B2%7D%5Ctan%20%5Cbeta%20%5Csec%20%5Cbeta%20%5Cright%5D%20_%7B-%5Cpi%20%2F6%7D%5E%7B0%7D%5C%5C%0A%09%3D%26%5Cfrac%7B3%7D%7B2%5Csqrt%7B2%7D%7D%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%5Cln%20%5Cleft(%20%5Cfrac%7B%5Csqrt%7B7%7D%2B2%7D%7B3%7D%20%5Cright)%20%2B%5Cfrac%7B%5Csqrt%7B7%7D-1%7D%7B3%7D%20%5Cright)%20%2B%5Cfrac%7B3%7D%7B2%5Csqrt%7B2%7D%7D%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%5Cln%20%5Csqrt%7B3%7D%2B%5Cfrac%7B1%7D%7B3%7D%20%5Cright)%5C%5C%0A%09%3D%26%5Cfrac%7B3%7D%7B4%5Csqrt%7B2%7D%7D%5Cln%20%5Cleft(%20%5Cfrac%7B%5Csqrt%7B7%7D%2B2%7D%7B%5Csqrt%7B3%7D%7D%20%5Cright)%20%2B%5Cfrac%7B%5Csqrt%7B7%7D%7D%7B2%5Csqrt%7B2%7D%7D.%0A%09%5Cend%7Baligned%7D


(最簡單的)第四題:

%5Cleft%5Clfloor%2010%5E%7B20%7D%5Cint_2%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bx%5E9%7D%7Bx%5E%7B20%7D-48x%5E%7B10%7D%2B575%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%20%5Cright%5Crfloor

解答:

顯然直接換元:t%3Dx%5E%7B10%7D,然后裂項(xiàng)求積分即可

%5Cbegin%7Baligned%7D%0A%5Cint_2%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bx%5E9%7D%7Bx%5E%7B20%7D-48x%5E%7B10%7D%2B575%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%26%3D%5Cfrac%7B1%7D%7B10%7D%5Cint_%7B2%5E%7B10%7D%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dt%7D%7Bt%5E2-48t%2B575%7D%7D%5C%5C%0A%26%3D%5Cfrac%7B1%7D%7B20%7D%5Cleft(%20%5Cint_%7B2%5E%7B10%7D%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dt%7D%7Bt-25%7D%7D-%5Cint_%7B2%5E%7B10%7D%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dt%7D%7Bt-23%7D%7D%20%5Cright)%5C%5C%26%3D%5Cfrac%7B1%7D%7B20%7D%5Cln%20%5Cleft%7C%20%5Cfrac%7Bt-25%7D%7Bt-23%7D%20%5Cright%7C%5Cbigg%7C%5E%7B%5Cinfty%7D_%7B2%5E%7B10%7D%7D%5C%5C%26%3D%0A%5Cfrac%7B1%7D%7B20%7D%5Cln%5Cleft(%5Cfrac%7B2%5E%7B10%7D-23%7D%7B2%5E%7B10%7D-25%7D%5Cright)%0A%5Cend%7Baligned%7D

所以

%5Cbegin%7Baligned%7D%0A%26%5Cleft%5Clfloor%2010%5E%7B20%7D%5Cint_2%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bx%5E9%7D%7Bx%5E%7B20%7D-48x%5E%7B10%7D%2B575%7D%5C%2C%5Cmathrm%7Bd%7Dx%7D%20%5Cright%5Crfloor%5C%5C%3D%26%5Cleft%5Clfloor%5Cfrac%7B10%5E%7B19%7D%7D%7B2%7D%5Cln%5Cleft(%5Cfrac%7B2%5E%7B10%7D-23%7D%7B2%5E%7B10%7D-25%7D%5Cright)%5Cright%5Crfloor%5C%5C%3D%2610000003333335333%0A%5Cend%7Baligned%7D


第五題:

%5Cint%5E%7B1%7D_%7B0%7D%5Cleft(%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D1%7D%5Cfrac%7B%5Cleft%5Clfloor2%5Enx%5Cright%5Crfloor%7D%7B3%5En%7D%5Cright)%5Cmathrm%7Bd%7Dx

就超出我能力范圍力,所以擺爛了




2023MIT積分競賽決賽(前4題)個(gè)人思路的評論 (共 條)

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